The [Lebesgue integral](/page/Lebesgue%20Integral) solves a basic problem of approximation theory: if a sequence of functions converges in a reasonable sense, does the integral of the limit equal the limit of the integrals? The Riemann integral fails this test badly --- there exist sequences of Riemann-integrable functions, converging pointwise to a limit that is not Riemann integrable. The Lebesgue integral, built on [measure theory](/page/Hausdorff%20Measure), handles such limits gracefully through the [Monotone Convergence Theorem](/theorems/509) and the [Dominated Convergence Theorem](/theorems/4). But a deeper problem remains. In applications to partial differential equations, to the [calculus of variations](/page/Calculus%20of%20Variations), and to probability theory, we do not merely need to integrate individual functions --- we need a **space** of functions that behaves like a well-structured infinite-dimensional vector space. Specifically, we need: 1. **Completeness.** [Cauchy sequences](/page/Cauchy%20Sequence) of functions must converge to a function *in the same space*. Without this, iterative methods (Picard iteration, Galerkin approximation, variational minimization) cannot be guaranteed to converge. 2. **A norm.** We need a quantitative measure of the "size" of a function that is compatible with the vector space structure. 3. **Duality.** We need a concrete description of the bounded linear functionals on the space, so that weak solutions of differential equations and distributional formulations have a precise meaning. The classical space $C([0,1])$ of [continuous functions](/page/Continuity%20(Metric%20Spaces)), equipped with the $L^2$-type norm, fails the first requirement catastrophically. [example: Failure of Completeness in $C([0,1])$] Let $U = (0,1) \subset \mathbb{R}$ and consider the space $C([0,1])$ equipped with the norm \begin{align*} \|f\|_2 := \left( \int_0^1 |f(t)|^2 \, d\mathcal{L}^1(t) \right)^{1/2}. \end{align*} Define the sequence of [continuous](/page/Continuity) functions $f_m: [0,1] \to \mathbb{R}$ by \begin{align*} f_m(t) := \begin{cases} 0 & \text{if } 0 \le t \le \tfrac{1}{2}, \\ m(t - \tfrac{1}{2}) & \text{if } \tfrac{1}{2} < t < \tfrac{1}{2} + \tfrac{1}{m}, \\ 1 & \text{if } \tfrac{1}{2} + \tfrac{1}{m} \le t \le 1. \end{cases} \end{align*} Each $f_m$ is continuous, so $f_m \in C([0,1])$. We verify that $\{f_m\}$ is Cauchy. For $m < k$, the functions $f_m$ and $f_k$ differ only on the interval $(\tfrac{1}{2}, \tfrac{1}{2} + \tfrac{1}{m})$, where both take values in $[0,1]$. Thus \begin{align*} \|f_m - f_k\|_2^2 = \int_{1/2}^{1/2 + 1/m} |f_m(t) - f_k(t)|^2 \, d\mathcal{L}^1(t) \le \int_{1/2}^{1/2 + 1/m} 1 \, d\mathcal{L}^1(t) = \frac{1}{m} \to 0. \end{align*} The pointwise limit is $f(t) = \mathbb{1}_{(1/2, 1]}(t)$, the indicator function of $(1/2, 1]$. This function is *not* continuous, so $f \notin C([0,1])$. The sequence is Cauchy but does not converge within the space. The norm $\|\cdot\|_2$ is perfectly well-defined on $C([0,1])$, but the resulting [normed space](/page/Normed%20Vector%20Space) is incomplete. [/example] The resolution is to enlarge the space by passing from continuous functions to [measurable functions](/page/Lebesgue%20Integral) and using the Lebesgue integral. The $L^p$ spaces are the result of this construction: they are the completions of spaces of integrable functions under the $p$-th power norm. They form the backbone of modern analysis --- the natural domain and codomain for differential operators in [Sobolev space](/page/Sobolev%20Space) theory, the setting for the [Fourier transform](/page/Fourier%20Transform), and the foundation of probability theory through $L^2(\Omega, \mathcal{F}, \mathbb{P})$. ## Definition The construction of $L^p$ requires two steps: first defining the *seminormed* space $\mathcal{L}^p$ of $p$-integrable functions, then passing to the quotient by the kernel of the seminorm (functions that are zero almost everywhere). This quotient is essential --- without it, the $L^p$ "norm" would fail to be a true norm, since $\|f\|_p = 0$ would not imply $f = 0$ but only that $f = 0$ $\mu$-almost everywhere. [definition: $L^p$ Space] Let $(E, \mathcal{E}, \mu)$ be a measure space and let $p \in [1, \infty)$. The **pre-$L^p$ space** $\mathcal{L}^p(E, \mathcal{E}, \mu)$ consists of all $\mathcal{E}$-measurable functions $f: E \to \mathbb{R}$ (or $\mathbb{C}$) satisfying \begin{align*} \|f\|_p := \left( \int_E |f|^p \, d\mu \right)^{1/p} < \infty. \end{align*} For $p = \infty$, the space $\mathcal{L}^\infty(E, \mathcal{E}, \mu)$ consists of all $\mathcal{E}$-measurable functions $f: E \to \mathbb{R}$ satisfying \begin{align*} \|f\|_\infty := \operatorname{ess\,sup}_E |f| := \inf \bigl\{ M \ge 0 : \mu(\{x \in E : |f(x)| > M\}) = 0 \bigr\} < \infty. \end{align*} The functional $\|\cdot\|_p$ is a seminorm on $\mathcal{L}^p$: it satisfies $\|f\|_p \ge 0$, $\|\lambda f\|_p = |\lambda| \|f\|_p$, and the triangle inequality (Minkowski's inequality, stated below), but $\|f\|_p = 0$ forces only $f = 0$ $\mu$-almost everywhere. The **$L^p$ space** is the quotient \begin{align*} L^p(E, \mathcal{E}, \mu) := \mathcal{L}^p(E, \mathcal{E}, \mu) \,/\, \sim, \qquad \text{where } f \sim g \iff f = g \;\; \mu\text{-a.e.} \end{align*} An element of $L^p$ is an [equivalence class](/page/Equivalence%20Relation) $[f]$ of $\mu$-a.e. equal functions. The $L^p$ norm $\|[f]\|_{L^p} := \|f\|_p$ is well-defined on the quotient (independent of representative) and is a true norm. [/definition] The passage from $\mathcal{L}^p$ to $L^p$ is more than a formality. Working with [equivalence classes](/page/Equivalence%20Relation) means that statements about $L^p$ functions hold only $\mu$-almost everywhere, not pointwise. In particular, "let $f \in L^p(E)$ and suppose $f(x_0) = 3$" is meaningless unless we have a specific continuous representative or a [trace operator](/page/Sobolev%20Space). [remark: Notation Convention] Throughout this page and on Androma generally, we write $f \in L^p(E)$ rather than $[f] \in L^p(E)$, identifying a function with its equivalence class. When the measure space is $(\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n), \mathcal{L}^n)$ and the context is clear, we abbreviate $L^p(\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n), \mathcal{L}^n)$ as $L^p(\mathbb{R}^n)$. [/remark] The following example illustrates how the integrability threshold for $L^p$ depends on the interplay between the singularity strength and the exponent $p$, and reveals the nesting of $L^p$ spaces on bounded domains. [example: Power Singularities in $L^p(\mathbb{R}^n)$] A basic but instructive class of $L^p$ functions on $\mathbb{R}^n$ consists of radial power functions. Let $U = B(0,1) \subset \mathbb{R}^n$ with $n \ge 1$, and consider the function $f: U \to \mathbb{R}$ defined by $f(x) = |x|^{-\alpha}$ for a parameter $\alpha > 0$. We determine for which values of $p$ and $\alpha$ the function $f$ belongs to $L^p(U)$. Using polar coordinates with $r = |x|$ and the surface area element $d\mathcal{H}^{n-1}$ on the unit sphere $\mathbb{S}^{n-1}$ (where $\mathcal{H}^{n-1}(\mathbb{S}^{n-1}) = n\omega_n$ with $\omega_n$ the volume of the unit ball): \begin{align*} \int_{B(0,1)} |x|^{-\alpha p} \, d\mathcal{L}^n(x) &= n\omega_n \int_0^1 r^{-\alpha p} \cdot r^{n-1} \, dr = n\omega_n \int_0^1 r^{n - 1 - \alpha p} \, dr. \end{align*} The integrand is $r^{n - 1 - \alpha p}$, which is integrable near $r = 0$ if and only if the exponent exceeds $-1$: \begin{align*} n - 1 - \alpha p > -1 \iff \alpha p < n \iff p < \frac{n}{\alpha}. \end{align*} Therefore $|x|^{-\alpha} \in L^p(B(0,1))$ if and only if $p < n/\alpha$. For instance, in dimension $n = 3$: the Coulomb potential singularity $|x|^{-1}$ lies in $L^p(B(0,1))$ for $p < 3$; the stronger singularity $|x|^{-2}$ lies in $L^p(B(0,1))$ only for $p < 3/2$. As $\alpha \to n$, the critical exponent $n/\alpha$ approaches $1$, and the function barely fails to be integrable. This example also reveals the **nesting failure** on bounded domains: higher integrability is a *stronger* condition. If $f \in L^q(U)$ for some $q > p$ and $\mu(E) < \infty$, then $f \in L^p(U)$ (by Holder's inequality, as we show below), but not conversely. [/example] ## The Fundamental Inequalities Three inequalities form the analytic backbone of $L^p$ theory. Each serves a distinct structural role: Holder's inequality establishes duality, Minkowski's inequality proves the triangle inequality for the $L^p$ norm (making $L^p$ a normed space), and Young's inequality for products is the pointwise engine that drives both. ### Young's Inequality for Products The proof of Holder's inequality reduces, after normalizing both functions to have unit $L^p$ and $L^q$ norms, to bounding a product $ab$ by a sum of powers $a^p/p + b^q/q$. This reduction is the heart of the argument: if we can control a product by a convex combination of pure powers, then integrating both sides yields the desired integral bound. The need for such a "product-to-sum" conversion arises repeatedly --- in proving Minkowski's inequality, in establishing convolution bounds, and in interpolation estimates --- making the following numerical inequality the load-bearing ingredient of the entire $L^p$ framework. [quotetheorem:244] The conjugate exponent relation $1/p + 1/q = 1$ appears throughout $L^p$ theory. Given $p \in (1, \infty)$, we write $q = p/(p-1)$ for its conjugate. The endpoint cases are $p = 1, q = \infty$ and $p = \infty, q = 1$, where the duality pairing takes a different form (see the section on Duality below). The map $p \mapsto q = p/(p-1)$ is a decreasing involution on $(1, \infty)$ that fixes $p = 2$: the space $L^2$ is self-dual, a fact with far-reaching consequences. ### Holder's Inequality The question that Holder's inequality answers is: **When is the pointwise product $fg$ integrable?** If $f$ and $g$ are both in $L^2$, the Cauchy-Schwarz inequality gives $\int |fg| \, d\mu \le \|f\|_2 \|g\|_2$. Holder's inequality generalizes this to arbitrary conjugate pairs. [quotetheorem:516] Holder's inequality has a natural interpretation in terms of duality: the map $g \mapsto \int_E fg \, d\mu$ defines a bounded linear functional on $L^p(E)$, with operator norm at most $\|g\|_{L^q}$. The [Riesz Representation Theorem for $L^p$](/theorems/901) shows that this bound is sharp and that *every* bounded linear functional on $L^p$ (for $1 \le p < \infty$) arises in this way. The case $p = q = 2$ is the Cauchy-Schwarz inequality. The endpoint case $p = 1, q = \infty$ reads $\int |fg| \, d\mu \le \|f\|_{L^1} \|g\|_{L^\infty}$, which follows directly from $|f(x)g(x)| \le |f(x)| \cdot \|g\|_{L^\infty}$ for $\mu$-a.e. $x$. The two-function version extends naturally to products of multiple functions, and this generalization is frequently needed in nonlinear PDE. [explanation: The Generalized Holder Inequality] If $p_1, \ldots, p_k \in [1, \infty]$ satisfy $1/p_1 + \cdots + 1/p_k = 1/r$ for some $r \ge 1$, and $f_i \in L^{p_i}(E)$ for each $i$, then the product $f_1 \cdots f_k \in L^r(E)$ with \begin{align*} \|f_1 \cdots f_k\|_{L^r} \le \|f_1\|_{L^{p_1}} \cdots \|f_k\|_{L^{p_k}}. \end{align*} The most common instance is the three-function version with $1/p + 1/q + 1/r = 1$, which appears in the analysis of nonlinear PDE (e.g., bounding trilinear forms in Navier-Stokes theory). The two-function Holder inequality is the case $k = 2$, $r = 1$. [/explanation] The generalized inequality has an immediate application to the nesting of $L^p$ spaces on finite measure spaces, where the constant function $1$ serves as one of the factors. [example: Inclusion $L^q(E) \subset L^p(E)$ on Finite Measure Spaces] On a measure space with $\mu(E) < \infty$, higher integrability implies lower integrability: if $1 \le p < q \le \infty$, then $L^q(E) \subset L^p(E)$ with a quantitative bound. This is a direct application of Holder's inequality. Let $f \in L^q(E)$ with $q < \infty$. Write $|f|^p = |f|^p \cdot 1$ and apply Holder's inequality with exponents $q/p > 1$ and $(q/p)' = q/(q-p)$: \begin{align*} \int_E |f|^p \, d\mu = \int_E |f|^p \cdot 1 \, d\mu \le \left( \int_E |f|^{p \cdot q/p} \, d\mu \right)^{p/q} \left( \int_E 1^{q/(q-p)} \, d\mu \right)^{(q-p)/q} = \|f\|_{L^q}^p \cdot \mu(E)^{(q-p)/q}. \end{align*} Taking $p$-th roots: \begin{align*} \|f\|_{L^p(E)} \le \mu(E)^{1/p - 1/q} \|f\|_{L^q(E)}. \end{align*} The constant $\mu(E)^{1/p - 1/q}$ is finite precisely because $\mu(E) < \infty$ and $1/p - 1/q > 0$. When $q = \infty$, the argument simplifies: $\int_E |f|^p \, d\mu \le \|f\|_{L^\infty}^p \mu(E)$, giving $\|f\|_{L^p} \le \mu(E)^{1/p} \|f\|_{L^\infty}$. This inclusion **fails** on infinite measure spaces, and it fails in both directions. On $(\mathbb{R}, \mathcal{B}(\mathbb{R}), \mathcal{L}^1)$, the function $f(x) = (1 + |x|)^{-1}$ belongs to $L^q(\mathbb{R})$ for $q > 1$ but not to $L^1(\mathbb{R})$, because $\int_{\mathbb{R}} (1+|x|)^{-q} \, d\mathcal{L}^1$ converges for $q > 1$ by comparison with $|x|^{-q}$, while the case $q = 1$ yields $\int_0^\infty (1+x)^{-1} \, dx = \infty$. Conversely, the function $g(x) = |x|^{-1/2} \mathbb{1}_{|x| \ge 1}$ belongs to $L^q(\mathbb{R})$ for $q > 2$ but not for $q \le 2$: the integral $\int_1^\infty x^{-q/2} \, d\mathcal{L}^1$ converges if and only if $q/2 > 1$, i.e., $q > 2$. In particular, $g \in L^3(\mathbb{R}) \setminus L^2(\mathbb{R})$, violating $L^3 \subset L^2$. On $\mathbb{R}^n$ with Lebesgue measure, neither $L^p \subset L^q$ nor $L^q \subset L^p$ holds for any $p \neq q$. [/example] ### Minkowski's Inequality The triangle inequality for the $L^p$ norm is not at all immediate from the definition. For $p = 1$ it follows from the ordinary triangle inequality $|f + g| \le |f| + |g|$ after integration. For $p = 2$ it is a consequence of the Cauchy-Schwarz inequality. For general $p$, the proof requires Holder's inequality applied to $|f+g|^{p-1}$, which is the step that forces the conjugate exponent relation to appear. Without a triangle inequality, $\|\cdot\|_p$ would be merely a homogeneous, positive functional, insufficient to define a metric space structure on $L^p$. [quotetheorem:517] Minkowski's inequality, combined with the absolute homogeneity and positivity of $\|\cdot\|_p$, establishes that $L^p(E)$ is a **normed vector space** for every $p \in [1, \infty]$. The completeness of this normed space (the Riesz-Fischer theorem, below) upgrades it to a [Banach space](/page/Banach%20Space). There is also an integral version of Minkowski's inequality, which extends the triangle inequality from finite sums to continuous "sums" (integrals). This continuous analogue is indispensable whenever one needs to bound the $L^p$ norm of a function defined by integrating a family of functions against a parameter --- a situation that arises naturally in convolution estimates and the study of integral operators. [quotetheorem:464] The $\sigma$-finiteness of both measure spaces is needed to apply Fubini's theorem in the proof, which requires interchanging the order of integration. On non-$\sigma$-finite spaces, the integral on the left may not even be measurable as a function of $x$. The inequality can be read as a continuous analogue of the ordinary triangle inequality: just as $\|f_1 + \cdots + f_m\|_{L^p} \le \|f_1\|_{L^p} + \cdots + \|f_m\|_{L^p}$, the integral version replaces the finite sum with an integral against $\nu$. The most important application is the proof of [Young's convolution inequality](/page/Convolution): bounding $\|f * g\|_{L^r}$ requires estimating the $L^r_x$ norm of $\int f(x-y)g(y) \, d\mathcal{L}^n(y)$, and Minkowski's integral inequality provides exactly the right tool. ## Completeness and the Riesz-Fischer Theorem A normed space is useful for functional analysis only if it is complete --- otherwise, limits of Cauchy sequences "fall outside" the space, and fixed-point theorems, spectral theory, and variational methods all break down. The incompleteness of $C([0,1])$ under the $L^p$ norm (demonstrated in the opening example) was precisely the motivation for introducing $L^p$ spaces. The Riesz-Fischer theorem confirms that the enlargement from continuous functions to equivalence classes of measurable functions resolves this deficiency. The key structural feature of the proof is a criterion specific to series in [normed spaces](/page/Normed%20Vector%20Space): a normed space is complete if and only if every **absolutely convergent series** converges. In $L^p$, absolute convergence means $\sum_k \|f_k\|_{L^p} < \infty$, and the proof constructs the limit function pointwise using the [Monotone Convergence Theorem](/theorems/509), then verifies norm convergence using the [Dominated Convergence Theorem](/theorems/4). [quotetheorem:892] The Riesz-Fischer theorem also yields a useful byproduct: every $L^p$-convergent sequence has a **subsequence that converges pointwise $\mu$-a.e.** This is not true of the full sequence --- $L^p$ convergence does not imply pointwise convergence, and pointwise a.e. convergence does not imply $L^p$ convergence (without a dominating function). The following two examples make these distinctions precise. [example: $L^p$ Convergence Without Pointwise Convergence] Let $E = [0,1]$ with Lebesgue measure. Enumerate the dyadic intervals: for each $m \in \mathbb{N}$, let $k$ range over $0, 1, \ldots, 2^m - 1$, and define \begin{align*} f_{m,k} := \mathbb{1}_{[k/2^m, (k+1)/2^m]}. \end{align*} Order these functions into a single sequence $(g_j)_{j=1}^\infty$ by $g_1 = f_{0,0} = \mathbb{1}_{[0,1]}$, $g_2 = f_{1,0} = \mathbb{1}_{[0,1/2]}$, $g_3 = f_{1,1} = \mathbb{1}_{[1/2,1]}$, $g_4 = f_{2,0} = \mathbb{1}_{[0,1/4]}$, and so on. For any $p \in [1, \infty)$: \begin{align*} \|g_j\|_{L^p}^p = \mathcal{L}^1(\operatorname{supp}(g_j)) = 2^{-m} \to 0 \quad \text{as } j \to \infty, \end{align*} so $g_j \to 0$ in $L^p([0,1])$. However, for every $x \in [0,1]$, the value $g_j(x) = 1$ for infinitely many $j$ (since the supports "sweep" repeatedly across $[0,1]$), and $g_j(x) = 0$ for infinitely many $j$. Therefore, $\limsup_{j \to \infty} g_j(x) = 1$ and $\liminf_{j \to \infty} g_j(x) = 0$ for every $x$. The sequence $g_j$ does not converge pointwise at any point. This demonstrates that $L^p$ convergence is strictly weaker than pointwise convergence: the "mass" of each $g_j$ dissipates (its $L^p$ norm shrinks to zero), but the function keeps "visiting" every point. [/example] The converse failure is equally important: a sequence can converge pointwise everywhere yet fail to converge in $L^p$, provided no single integrable function dominates the entire sequence. [example: Pointwise Convergence Without $L^p$ Convergence] Let $E = (0, \infty)$ with Lebesgue measure and fix $p \in [1, \infty)$. Define $f_m: (0, \infty) \to \mathbb{R}$ by \begin{align*} f_m(x) := m^{1/p} \mathbb{1}_{(0, 1/m]}(x). \end{align*} For every fixed $x > 0$, we have $f_m(x) = 0$ for all $m > 1/x$, so $f_m(x) \to 0$ pointwise. However: \begin{align*} \|f_m\|_{L^p}^p = \int_0^{1/m} m \, d\mathcal{L}^1 = 1 \quad \text{for every } m. \end{align*} So $\|f_m\|_{L^p} = 1$ for all $m$, and in particular $\|f_m - 0\|_{L^p} = 1 \not\to 0$. The sequence converges pointwise to $0$ but does not converge to $0$ in $L^p$. The culprit is the lack of a dominating function: $|f_m(x)| \le m^{1/p}$, but no single $L^p$ function dominates the entire sequence. This is precisely the hypothesis that the [Dominated Convergence Theorem](/theorems/4) requires for interchanging the limit and integral. [/example] ## Duality and the Pairing with $L^q$ The dual space of a Banach space $X$ --- the space $X^*$ of bounded linear functionals $\varphi: X \to \mathbb{R}$ --- is a fundamental object in functional analysis. Weak formulations of PDE require knowing the dual explicitly: when we write "find $u \in H^1_0(U)$ such that $B[u,v] = \langle f, v \rangle$ for all $v \in H^1_0(U)$," the datum $f$ lives in the dual space $H^{-1}(U) = (H^1_0(U))^*$, and we need to know what elements of this dual look like. For $L^p$ spaces, the dual is identified concretely by Holder's inequality. The pairing \begin{align*} \Phi: L^q(E) &\to (L^p(E))^* \\ g &\mapsto \Phi_g, \quad \text{where } \Phi_g(f) := \int_E fg \, d\mu, \end{align*} is well-defined and bounded (with $\|\Phi_g\| \le \|g\|_{L^q}$) by Holder's inequality. The deep content of the Riesz Representation Theorem for $L^p$ is that this map is an **isometric isomorphism** --- every bounded linear functional on $L^p$ arises from integration against an $L^q$ function, and the operator norm equals the $L^q$ norm. [quotetheorem:901] Several aspects of this theorem deserve attention. The $\sigma$-finiteness hypothesis is essential for the case $p = 1$. On a non-$\sigma$-finite measure space, the dual of $L^1$ is strictly larger than $L^\infty$: it includes "singular" functionals that cannot be represented by integration. For $1 < p < \infty$, the theorem holds without $\sigma$-finiteness, but the standard proof via the [Radon-Nikodym theorem](/page/Absolutely%20Continuous%20Measures) requires $\sigma$-finiteness. The case $p = \infty$ is excluded for good reason: $(L^\infty)^*$ is **not** isomorphic to $L^1$. The dual of $L^\infty$ is the space of finitely additive signed measures that are absolutely continuous with respect to $\mu$, which is much larger than $L^1$. This failure is intimately connected to the non-[separability](/page/Separable) of $L^\infty$ (see below). The identification $(L^p)^* \cong L^q$ for $1 < p < \infty$ means that $L^p$ is **[reflexive](/page/Reflexive%20Space)**: the natural embedding $L^p \hookrightarrow (L^p)^{**} = (L^q)^* \cong L^p$ is surjective. Reflexivity has powerful consequences --- it implies that every bounded sequence in $L^p$ has a [weakly convergent](/page/Weak%20Convergence) subsequence (by the [Banach-Alaoglu theorem](/page/Weak%20Convergence), since weak and weak* topologies coincide on reflexive spaces). This is the starting point for the direct method in the calculus of variations. The space $L^1$ is not reflexive: its bidual $(L^1)^{**} = (L^\infty)^*$ is strictly larger than $L^1$. As a consequence, bounded sequences in $L^1$ need not have weakly convergent subsequences. This failure is a fundamental source of difficulty in nonlinear PDE, where the natural energy space is often $L^1$ (e.g., for conservation laws or the theory of [bounded variation](/page/Bounded%20Variation)). The following example shows concretely how the Riesz representation works and illustrates what happens at the boundary of integrability. [example: A Bounded Linear Functional on $L^p$ and Its Representing Function] Let $E = (0,1)$ with Lebesgue measure and $p = 3$, so $q = 3/2$. Define the functional $F: L^3(0,1) \to \mathbb{R}$ by \begin{align*} F(f) := \int_0^1 f(t) \sqrt{t} \, d\mathcal{L}^1(t). \end{align*} We verify that $F \in (L^3(0,1))^*$ and compute its norm. The representing function is $g(t) = \sqrt{t} = t^{1/2}$. We check $g \in L^{3/2}(0,1)$: \begin{align*} \|g\|_{L^{3/2}}^{3/2} = \int_0^1 |t^{1/2}|^{3/2} \, d\mathcal{L}^1(t) = \int_0^1 t^{3/4} \, dt = \left[ \frac{t^{7/4}}{7/4} \right]_0^1 = \frac{4}{7}. \end{align*} Therefore $\|g\|_{L^{3/2}} = (4/7)^{2/3}$. By the Riesz Representation theorem, $\|F\|_{(L^3)^*} = \|g\|_{L^{3/2}} = (4/7)^{2/3}$. The functional $F$ is bounded because the integration kernel $g = t^{1/2}$ has finite $L^{3/2}$ norm. Now consider what happens at the boundary. If we replace $g$ by $h(t) = t^{-2/3}$, we need $h \in L^{3/2}(0,1)$ for the corresponding functional to be bounded on $L^3$. But \begin{align*} \|h\|_{L^{3/2}}^{3/2} = \int_0^1 |t^{-2/3}|^{3/2} \, d\mathcal{L}^1(t) = \int_0^1 t^{-1} \, dt = \infty. \end{align*} Since $h \notin L^{3/2}(0,1)$, the functional $f \mapsto \int_0^1 f(t) t^{-2/3} \, d\mathcal{L}^1(t)$ is *not* a bounded linear functional on $L^3(0,1)$. More precisely, $t^{-2/3} \in L^r(0,1)$ if and only if $r \cdot 2/3 < 1$, i.e., $r < 3/2$. Since $r = 3/2$ is the critical exponent for duality with $L^3$, the function $t^{-2/3}$ is exactly at the boundary: it fails to define a bounded functional on $L^3$. [/example] ### Uniform Convexity and Reflexivity For $1 < p < \infty$, the [reflexivity](/page/Reflexive%20Space) of $L^p$ can also be established through a geometric route that reveals additional structure: $L^p$ spaces are **uniformly convex**, and every uniformly convex Banach space is reflexive by the [Milman-Pettis theorem](/theorems/899). Uniform convexity means that if two unit vectors $f, g \in L^p$ are "far apart" ($\|f - g\| \ge \varepsilon$), then their midpoint is bounded away from the unit sphere ($\|(f+g)/2\| \le 1 - \delta(\varepsilon)$ for some $\delta > 0$ depending only on $\varepsilon$ and $p$). This geometric property is quantified by Clarkson's inequalities. [quotetheorem:900] Clarkson's inequalities provide the quantitative version of a fundamental geometric fact: the unit ball in $L^p$ (for $1 < p < \infty$) is "round" --- it has no flat edges. This fails at both endpoints: the unit balls in $L^1$ and $L^\infty$ have flat faces (in $L^1(\{1,2\})$, the segment from $\mathbb{1}_{\{1\}}$ to $\mathbb{1}_{\{2\}}$ lies entirely on the unit sphere), and neither space is reflexive. ## Dense Subspaces and Approximation In working with $L^p$ spaces, one frequently needs to approximate an arbitrary $L^p$ function by "nicer" functions --- simple functions, continuous functions, or smooth compactly supported functions. These approximation results are the $L^p$ analogues of the density of smooth functions in [Sobolev spaces](/page/Sobolev%20Space), and they serve the same purpose: proving a property for all $L^p$ functions by verifying it for smooth functions and passing to the limit. The question of which classes of functions are dense in $L^p$ is closely tied to the **separability** of the space --- whether it contains a [countable](/page/Countable%20Set) dense subset. This topological property has significant functional-analytic consequences: separable Banach spaces have [metrizable](/page/Metrizable%20Space) [weak topologies](/page/Weak%20Topology) on bounded sets, and their duals satisfy stronger sequential [compactness](/page/Compact%20Space) properties. [quotetheorem:893] The density of simple functions follows from the definition of the Lebesgue integral: a non-negative measurable function is, by construction, the pointwise limit of an increasing sequence of simple functions, and the [Monotone Convergence Theorem](/theorems/509) converts this pointwise convergence into $L^p$ convergence. The density of $C_c^\infty(U)$ is deeper --- it requires the full machinery of [mollification](/page/Standard%20Mollifier). Given $f \in L^p(U)$, one first truncates (multiplying by a cut-off function to obtain compact support), then mollifies (convolving with the standard [mollifier](/page/Standard%20Mollifier) $\eta_\varepsilon$ to obtain smoothness). The $L^p$ convergence of mollifications is guaranteed by the [$L^p_{\mathrm{loc}}$ convergence of mollification](/theorems/48) theorem. The restriction $p < \infty$ is essential: $C_c^\infty(U)$ is **not** dense in $L^\infty(U)$. A [continuous function](/page/Continuity%20(Metric%20Spaces)) that is $L^\infty$-close to $\mathbb{1}_{(0,1/2)}$ would need to approximate a jump discontinuity uniformly, which is impossible. The $L^\infty$-closure of $C_c^\infty(\mathbb{R}^n)$ is the space $C_0(\mathbb{R}^n)$ of continuous functions vanishing at infinity, which is a proper closed subspace of $L^\infty$. The separability of $L^p$ for finite $p$, and its failure for $p = \infty$, is another fundamental distinction. The proof of non-separability constructs an uncountable family of functions that are pairwise far apart in norm. [quotetheorem:548] The difference between separability and non-separability has concrete analytic consequences. In a separable Banach space, the weak topology on bounded sets is metrizable, which means that weak convergence can be studied using sequences rather than nets. In $L^\infty$, one must work with the [weak-* topology](/page/Weak*%20Topology) (viewing $L^\infty$ as the dual of $L^1$), which is sequentially compact on bounded sets (by the separability of $L^1$) but whose full topology is not metrizable. [example: Non-Separability of $L^\infty$] We exhibit an uncountable family in $L^\infty(0,1)$ with mutual distance bounded below, which prevents the existence of a countable dense subset. For each $t \in (0,1)$, define $f_t := \mathbb{1}_{(0,t)} \in L^\infty(0,1)$. If $s \neq t$, say $s < t$, then \begin{align*} \|f_t - f_s\|_{L^\infty} = \|\mathbb{1}_{(s,t)}\|_{L^\infty} = 1. \end{align*} So the uncountable family $\{f_t\}_{t \in (0,1)}$ satisfies $\|f_t - f_s\|_{L^\infty} = 1$ for all $s \neq t$. The open balls $B(f_t, 1/2)$ are pairwise disjoint, so any dense subset must contain at least one element in each ball. Since there are uncountably many such balls, no countable set can be dense. This argument does not work in $L^p$ for $p < \infty$: the same family satisfies $\|f_t - f_s\|_{L^p} = |t - s|^{1/p}$, which can be made arbitrarily small. The $L^p$ topology is "coarser" than the $L^\infty$ topology --- it can distinguish functions by their "total mass" but not by pointwise behavior. [/example] ## The Special Landscape: $L^1$, $L^2$, and $L^\infty$ The three spaces $L^1$, $L^2$, and $L^\infty$ each have distinctive properties that set them apart from the general $L^p$ family. Understanding their individual behavior is essential for applications, since each arises naturally in a different context. ### The Space $L^1$: Integrability and Its Limitations The space $L^1(E, \mu)$ consists of integrable functions --- those for which $\int_E |f| \, d\mu < \infty$. It is the natural setting for the Lebesgue integral itself and for the [Fourier transform](/page/Fourier%20Transform) (which maps $L^1(\mathbb{R}^n)$ into $C_0(\mathbb{R}^n)$ by the Riemann-Lebesgue lemma). However, $L^1$ has significant structural deficiencies relative to $L^p$ with $1 < p < \infty$: 1. **Non-reflexivity.** As noted above, $(L^1)^{**} \supsetneq L^1$. This means bounded sequences in $L^1$ need not have weakly convergent subsequences. 2. **The Dunford-Pettis phenomenon.** Weak convergence in $L^1$ is characterized by the Dunford-Pettis theorem: $f_m \rightharpoonup f$ in $L^1$ if and only if $\int_A f_m \, d\mu \to \int_A f \, d\mu$ for every measurable $A$, and the sequence is **uniformly integrable** (the integrals $\int_{\{|f_m| > K\}} |f_m| \, d\mu$ are uniformly small for large $K$). 3. **Concentration and oscillation.** Bounded sequences in $L^1$ can fail to be weakly compact through two mechanisms: **concentration** (mass accumulates at a point, producing a Dirac delta in the limit) and **oscillation** (rapid sign changes prevent the integrals from converging). Neither mechanism is possible in $L^p$ for $p > 1$, thanks to uniform convexity and reflexivity. [example: Concentration in $L^1$ --- Approximation of the Dirac Delta] Let $E = \mathbb{R}$ and define $f_m: \mathbb{R} \to \mathbb{R}$ by $f_m(x) = m \cdot \mathbb{1}_{[0, 1/m]}(x)$. Then: \begin{align*} \|f_m\|_{L^1} = \int_{\mathbb{R}} m \cdot \mathbb{1}_{[0,1/m]} \, d\mathcal{L}^1 = m \cdot \frac{1}{m} = 1 \quad \text{for all } m. \end{align*} The sequence is bounded in $L^1(\mathbb{R})$ with constant norm $1$. For any continuous and bounded test function $\phi: \mathbb{R} \to \mathbb{R}$: \begin{align*} \int_{\mathbb{R}} f_m(x) \phi(x) \, d\mathcal{L}^1(x) = m \int_0^{1/m} \phi(x) \, d\mathcal{L}^1(x) \to \phi(0) \quad \text{as } m \to \infty, \end{align*} by the continuity of $\phi$. The sequence $f_m$ converges to the Dirac delta $\delta_0$ in the sense of [distributions](/page/Schwartz%20Space), but $\delta_0$ is not an $L^1$ function --- it is a measure. The mass "concentrates" at the origin. This phenomenon cannot occur for bounded sequences in $L^p$ with $p > 1$: if $\|f_m\|_{L^p} \le C$, then \begin{align*} \int_{B(x_0, r)} |f_m| \, d\mathcal{L}^n \le \|f_m\|_{L^p} \cdot \mathcal{L}^n(B(x_0,r))^{1/q} \le C \cdot (c_n r^n)^{1/q} \to 0 \quad \text{as } r \to 0, \end{align*} uniformly in $m$. So bounded $L^p$ sequences (with $p > 1$) cannot concentrate mass at a point. [/example] ### The Space $L^2$: Hilbert Space Structure Among all $L^p$ spaces, $L^2$ is distinguished by its inner product structure. The map \begin{align*} (\cdot, \cdot)_{L^2}: L^2(E) \times L^2(E) &\to \mathbb{R} \\ (f, g) &\mapsto \int_E fg \, d\mu \end{align*} is a well-defined inner product: it is bilinear, symmetric, and positive definite (since $(f, f)_{L^2} = \|f\|_{L^2}^2 = 0$ implies $f = 0$ in $L^2$). The $L^2$ norm satisfies $\|f\|_{L^2} = \sqrt{(f,f)_{L^2}}$, making $L^2$ a [Hilbert space](/page/Hilbert%20Space). No other $L^p$ space with $p \neq 2$ admits an inner product compatible with its norm. The reason is the **parallelogram identity**: a norm $\|\cdot\|$ on a vector space arises from an inner product if and only if \begin{align*} \|f + g\|^2 + \|f - g\|^2 = 2\|f\|^2 + 2\|g\|^2 \quad \text{for all } f, g. \end{align*} This identity holds in $L^2$ (expand the squares using the inner product) but fails in $L^p$ for $p \neq 2$. [example: Failure of the Parallelogram Identity in $L^4$] Let $E = (0,1)$ with Lebesgue measure, and take $f = \mathbb{1}_{(0,1)}$ (the constant function $1$) and $g = 2 \cdot \mathbb{1}_{(0,1/2)}$ in $L^4(0,1)$. We compute each norm: \begin{align*} \|f\|_{L^4}^4 &= \int_0^1 1 \, d\mathcal{L}^1 = 1, \quad \text{so } \|f\|_{L^4} = 1, \\ \|g\|_{L^4}^4 &= \int_0^{1/2} 2^4 \, d\mathcal{L}^1 = 16 \cdot \frac{1}{2} = 8, \quad \text{so } \|g\|_{L^4} = 8^{1/4}. \end{align*} For the sum $f + g$: on $(0, 1/2)$, $f + g = 3$; on $(1/2, 1)$, $f + g = 1$. So \begin{align*} \|f + g\|_{L^4}^4 = \int_0^{1/2} 81 \, d\mathcal{L}^1 + \int_{1/2}^1 1 \, d\mathcal{L}^1 = \frac{81}{2} + \frac{1}{2} = 41. \end{align*} For the difference $f - g$: on $(0, 1/2)$, $f - g = -1$; on $(1/2, 1)$, $f - g = 1$. So \begin{align*} \|f - g\|_{L^4}^4 = \int_0^{1/2} 1 \, d\mathcal{L}^1 + \int_{1/2}^1 1 \, d\mathcal{L}^1 = 1. \end{align*} Now we check the parallelogram identity (with squares of norms): \begin{align*} \|f + g\|_{L^4}^2 + \|f - g\|_{L^4}^2 &= 41^{1/2} + 1 \approx 7.40, \\ 2\|f\|_{L^4}^2 + 2\|g\|_{L^4}^2 &= 2 \cdot 1 + 2 \cdot 8^{1/2} = 2 + 2\sqrt{8} \approx 7.66. \end{align*} These are not equal, so the $L^4$ norm does not satisfy the parallelogram identity and cannot arise from an inner product. [/example] The Hilbert space structure of $L^2$ enables the full machinery of orthogonal projections, [orthonormal bases](/page/Hilbert%20Space), and the [Riesz Representation Theorem](/theorems/221). When $E = [0, 2\pi]$ with normalized Lebesgue measure, the trigonometric system $\{e^{inx}/\sqrt{2\pi}\}_{n \in \mathbb{Z}}$ forms a complete orthonormal system in $L^2$ --- the [completeness of this system](/theorems/585) is the foundation of [Fourier analysis](/page/Fourier%20Transform). The self-duality $(L^2)^* \cong L^2$ means that weak and strong formulations of PDE coincide in the $L^2$ setting, which is why the [Sobolev space](/page/Sobolev%20Space) $H^k(U) = W^{k,2}(U)$ and the $H^1_0(U)$ framework for [elliptic equations](/page/Second%20Order%20Elliptic%20Equations) are so natural. [remark: Probability and $L^2$] The space $L^2(\Omega, \mathcal{F}, \mathbb{P})$ over a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ is the space of random variables with finite variance: $X \in L^2(\Omega, \mathcal{F}, \mathbb{P})$ if and only if $\mathbb{E}[X^2] < \infty$, which is equivalent to $\operatorname{Var}(X) < \infty$ (since $\operatorname{Var}(X) = \mathbb{E}[X^2] - (\mathbb{E}[X])^2$). The inner product $(X, Y)_{L^2} = \mathbb{E}[XY]$ encodes the covariance structure: $\operatorname{Cov}(X, Y) = (X - \mathbb{E}[X], Y - \mathbb{E}[Y])_{L^2}$, and orthogonality in $L^2$ corresponds to uncorrelatedness. Conditional expectation $\mathbb{E}[X \mid \mathcal{G}]$ is the orthogonal projection of $X$ onto the closed subspace $L^2(\Omega, \mathcal{G}, \mathbb{P})$. This Hilbert-space perspective is the foundation of the $L^2$ theory of [martingales](/page/Martingale) and [stochastic integration](/page/Cambridge%20III%20Stochastic%20Calculus%20and%20Applications). [/remark] ### The Space $L^\infty$: Essential Supremum and Pathologies The space $L^\infty(E, \mu)$ consists of essentially bounded measurable functions. Its norm, the essential [supremum](/page/Supremum%20and%20Infimum), measures the "pointwise size" of a function up to null sets. This space serves as the natural home for coefficients of differential operators (e.g., $a_{ij} \in L^\infty(U)$ in [elliptic PDE theory](/page/Second%20Order%20Elliptic%20Equations)) and for multiplier operators. However, $L^\infty$ is the "outlier" in the $L^p$ family. It lacks almost every desirable property that $L^p$ enjoys for finite $p$: 1. **Non-separability.** As demonstrated above, $L^\infty(U)$ is not separable for any $U$ with $\mathcal{L}^n(U) > 0$. 2. **$C_c^\infty$ is not dense.** The closure of $C_c^\infty(U)$ in $L^\infty(U)$ is $C_0(U)$, the space of continuous functions vanishing at infinity, which is a proper subspace. 3. **Non-reflexivity.** The dual of $L^\infty$ is not $L^1$ (as it would be if the duality extended from $1 < p < \infty$). Instead, $(L^\infty)^*$ is the space $\mathrm{ba}(E, \mathcal{E}, \mu)$ of bounded finitely additive signed measures absolutely continuous with respect to $\mu$, which is vastly larger than $L^1$. 4. **No uniform convexity.** The unit ball of $L^\infty$ has "flat edges." These pathologies explain why $L^\infty$ enters PDE theory primarily through coefficients and test functions, rather than as a solution space. The appropriate [weak-*](/page/Weak*%20Topology) topology on $L^\infty$ (viewing it as the dual of $L^1$) is much better behaved than the norm topology: the Banach-Alaoglu theorem guarantees that bounded sets in $L^\infty$ are weak-* compact and weak-* sequentially compact (the latter by separability of $L^1$). ## Convolution and Regularization One of the most powerful operations in $L^p$ theory is [convolution](/page/Convolution) --- the "averaging" of one function against translates of another. [Convolution](/page/Convolution) is the primary tool for approximating $L^p$ functions by smooth functions, and it plays a central role in the theory of [Sobolev spaces](/page/Sobolev%20Space) (through mollification), in harmonic analysis (through singular integrals), and in probability theory (where the convolution of densities corresponds to the distribution of a sum of independent random variables). The fundamental question is: **If $f \in L^p$ and $g \in L^q$, to which $L^r$ space does $f * g$ belong?** The answer is given by Young's convolution inequality, which provides the exact relationship between the exponents. Before stating it, we recall the convolution operation itself. [definition: Convolution] Let $f, g: \mathbb{R}^n \to \mathbb{R}$ be measurable functions. The **convolution** of $f$ and $g$ is the function $f * g: \mathbb{R}^n \to \mathbb{R}$ defined by \begin{align*} (f * g)(x) := \int_{\mathbb{R}^n} f(x - y) g(y) \, d\mathcal{L}^n(y), \end{align*} whenever the integral is well-defined (i.e., $y \mapsto f(x-y)g(y)$ is in $L^1(\mathbb{R}^n)$ for a.e. $x$). [/definition] Young's convolution inequality quantifies the "smoothing" effect of convolution: convolving functions from two $L^p$ spaces produces a function in a *better* (higher-exponent) $L^r$ space. [quotetheorem:463] The exponent relation $1/p + 1/q = 1 + 1/r$ reflects a scaling constraint: convolution is a "smoothing" operation that improves integrability. The most important special cases are: - **$p = 1$, $q = r$:** Convolution with an $L^1$ function preserves $L^q$ integrability, with $\|f * g\|_{L^q} \le \|f\|_{L^1} \|g\|_{L^q}$. This is the case relevant to mollification: the standard mollifier $\eta_\varepsilon$ satisfies $\|\eta_\varepsilon\|_{L^1} = 1$, so $\|f * \eta_\varepsilon\|_{L^q} \le \|f\|_{L^q}$. - **$1/p + 1/q = 1$ (conjugate exponents), $r = \infty$:** The convolution $f * g$ is bounded and (in fact) uniformly continuous. This follows from Holder's inequality applied pointwise. - **$p = q = 1$, $r = 1$:** $L^1(\mathbb{R}^n)$ is closed under convolution, and $\|f * g\|_{L^1} \le \|f\|_{L^1} \|g\|_{L^1}$. This makes $L^1(\mathbb{R}^n)$ a **Banach algebra** under convolution. The mollification operation, which is the primary approximation tool in $L^p$ and Sobolev space theory, is a special case of convolution where one factor is a smooth, compactly supported approximation to the identity. [example: Mollification as an $L^p$-Bounded Operator] The **standard mollifier** is constructed from the function $\eta \in C_c^\infty(\mathbb{R}^n)$ defined by \begin{align*} \eta(x) := \begin{cases} C \exp\!\left(\frac{1}{|x|^2 - 1}\right) & \text{if } |x| < 1, \\ 0 & \text{if } |x| \ge 1, \end{cases} \end{align*} where $C > 0$ is the normalizing constant chosen so that $\int_{\mathbb{R}^n} \eta \, d\mathcal{L}^n = 1$. The rescaled mollifier is $\eta_\varepsilon(x) := \varepsilon^{-n} \eta(x/\varepsilon)$, which satisfies $\operatorname{supp}(\eta_\varepsilon) = \overline{B}(0, \varepsilon)$, $\eta_\varepsilon \ge 0$, and $\|\eta_\varepsilon\|_{L^1} = 1$. For any $f \in L^p(\mathbb{R}^n)$ with $1 \le p \le \infty$, define the mollification $f_\varepsilon := \eta_\varepsilon * f$. Since $\eta_\varepsilon \ge 0$ and $\|\eta_\varepsilon\|_{L^1} = 1$, Young's inequality with $p_1 = 1$, $p_2 = p$, $r = p$ (so $1/1 + 1/p = 1 + 1/p$) gives \begin{align*} \|f_\varepsilon\|_{L^p(\mathbb{R}^n)} \le \|\eta_\varepsilon\|_{L^1(\mathbb{R}^n)} \|f\|_{L^p(\mathbb{R}^n)} = \|f\|_{L^p(\mathbb{R}^n)}. \end{align*} The mollification operator $f \mapsto f_\varepsilon$ is a contraction on $L^p(\mathbb{R}^n)$. Moreover, $f_\varepsilon \in C^\infty(\mathbb{R}^n)$ because $\eta_\varepsilon$ is smooth and the smoothness passes through the integral (differentiation under the integral sign, justified by the compact support and smoothness of $\eta_\varepsilon$). The convergence $f_\varepsilon \to f$ in $L^p(\mathbb{R}^n)$ as $\varepsilon \to 0$ (for $1 \le p < \infty$) follows from the density of $C_c^\infty$ in $L^p$ and the contractivity of mollification: for any $\varphi \in C_c^\infty$, \begin{align*} \|f_\varepsilon - f\|_{L^p} &\le \|f_\varepsilon - \varphi_\varepsilon\|_{L^p} + \|\varphi_\varepsilon - \varphi\|_{L^p} + \|\varphi - f\|_{L^p} \\ &\le \|f - \varphi\|_{L^p} + \|\varphi_\varepsilon - \varphi\|_{L^p} + \|\varphi - f\|_{L^p} \\ &= 2\|f - \varphi\|_{L^p} + \|\varphi_\varepsilon - \varphi\|_{L^p}. \end{align*} The first term is made small by choosing $\varphi$ close to $f$ in $L^p$ (density of $C_c^\infty$). The second term converges to $0$ because $\varphi$ is uniformly continuous with compact support, so $\varphi_\varepsilon \to \varphi$ uniformly. [/example] ## Interpolation and the Riesz-Thorin Theorem A recurring question in analysis is: **If a linear operator is bounded on two different $L^p$ spaces, is it bounded on intermediate spaces?** For instance, if $T: L^1 \to L^1$ and $T: L^\infty \to L^\infty$ are both bounded, can we conclude $T: L^p \to L^p$ is bounded for $1 < p < \infty$? The answer is affirmative, and the precise quantitative result is the Riesz-Thorin interpolation theorem. The key idea is that the $L^p$ norm itself can be "interpolated" between two endpoint norms: if $f \in L^{p_0} \cap L^{p_1}$ and $1/p = (1-\theta)/p_0 + \theta/p_1$ for some $\theta \in [0,1]$, then \begin{align*} \|f\|_{L^p} \le \|f\|_{L^{p_0}}^{1-\theta} \|f\|_{L^{p_1}}^{\theta}. \end{align*} This "log-convexity" of the $L^p$ norm is a consequence of Holder's inequality applied with a carefully chosen splitting of $|f|^p$. [example: Log-Convexity of the $L^p$ Norm] Let $f \in L^{p_0}(E) \cap L^{p_1}(E)$ with $1 \le p_0 < p_1 \le \infty$ and $p_0 < \infty$. For $\theta \in (0,1)$, define $p$ by $1/p = (1-\theta)/p_0 + \theta/p_1$. We prove $\|f\|_{L^p} \le \|f\|_{L^{p_0}}^{1-\theta} \|f\|_{L^{p_1}}^{\theta}$. Write $|f|^p = |f|^{p(1-\theta)} \cdot |f|^{p\theta}$. The exponents satisfy: $p(1-\theta) \cdot (p_0 / (p(1-\theta))) = p_0$ and $p\theta \cdot (p_1/(p\theta)) = p_1$. Setting $a = p_0/(p(1-\theta))$ and $b = p_1/(p\theta)$, we verify $1/a + 1/b = p(1-\theta)/p_0 + p\theta/p_1 = 1$ by the definition of $p$. Therefore $a$ and $b$ are conjugate exponents. Applying Holder's inequality: \begin{align*} \int_E |f|^p \, d\mu &= \int_E |f|^{p(1-\theta)} \cdot |f|^{p\theta} \, d\mu \\ &\le \left( \int_E |f|^{p(1-\theta) \cdot a} \, d\mu \right)^{1/a} \left( \int_E |f|^{p\theta \cdot b} \, d\mu \right)^{1/b} \\ &= \left( \int_E |f|^{p_0} \, d\mu \right)^{p(1-\theta)/p_0} \left( \int_E |f|^{p_1} \, d\mu \right)^{p\theta/p_1} \\ &= \|f\|_{L^{p_0}}^{p(1-\theta)} \cdot \|f\|_{L^{p_1}}^{p\theta}. \end{align*} Taking $p$-th roots: $\|f\|_{L^p} \le \|f\|_{L^{p_0}}^{1-\theta} \cdot \|f\|_{L^{p_1}}^{\theta}$. [/example] The log-convexity estimate above is a pointwise interpolation inequality — it bounds an intermediate norm using two endpoint norms. The Riesz-Thorin theorem elevates this idea to the level of linear operators: if an operator is bounded at two endpoints, it is automatically bounded at all intermediate exponents, with a logarithmically interpolated bound on the operator norm. [quotetheorem:949] The Riesz-Thorin theorem is a key tool in harmonic analysis. Its most celebrated application is the proof that the [Fourier transform](/page/Fourier%20Transform) extends from $L^1 \cap L^2$ to a bounded operator $L^p(\mathbb{R}^n) \to L^{p'}(\mathbb{R}^n)$ for $1 \le p \le 2$ (the Hausdorff-Young inequality). The endpoint bounds are: on $L^1$, the Fourier transform maps into $L^\infty$ with norm $1$; on $L^2$, it is an isometry (Plancherel's theorem). Riesz-Thorin interpolation fills in the intermediate exponents. The restriction to **linear** operators is fundamental --- the theorem fails for sublinear or multilinear maps, which require the Marcinkiewicz interpolation theorem (a "weak-type" analogue) instead. ## Modes of Convergence In $L^p$ spaces, several notions of convergence coexist, and understanding their relationships is critical for applications. The interplay between these modes --- and the conditions under which one implies another --- is a recurring theme in measure theory and PDE. Let $(E, \mathcal{E}, \mu)$ be a measure space and let $f_m, f: E \to \mathbb{R}$ be measurable functions with $f_m \in L^p(E)$. [definition: Modes of Convergence in $L^p$] 1. **$L^p$ convergence** (convergence in norm): $f_m \to f$ in $L^p(E)$ means $\|f_m - f\|_{L^p} \to 0$. 2. **Pointwise a.e. convergence**: $f_m \to f$ a.e. means $\mu(\{x : f_m(x) \not\to f(x)\}) = 0$. 3. **Convergence in measure**: $f_m \to f$ in measure means $\mu(\{|f_m - f| > \varepsilon\}) \to 0$ for every $\varepsilon > 0$. 4. **Weak convergence in $L^p$** (for $1 < p < \infty$): $f_m \rightharpoonup f$ means $\int_E f_m g \, d\mu \to \int_E f g \, d\mu$ for every $g \in L^q(E)$. [/definition] The relationships between these modes form the following hierarchy (none of the reverse implications holds in general): - $L^p$ convergence implies convergence in measure (by [Markov's inequality](/theorems/514): $\mu(\{|f_m - f| > \varepsilon\}) \le \varepsilon^{-p} \|f_m - f\|_{L^p}^p$). - $L^p$ convergence implies weak convergence in $L^p$ (by Holder: $|\int (f_m - f)g \, d\mu| \le \|f_m - f\|_{L^p} \|g\|_{L^q}$). - Pointwise a.e. convergence implies convergence in measure on sets of finite measure ([Egorov's theorem](/theorems/896)). - Convergence in measure implies a subsequence converges pointwise a.e. (not the full sequence). - Pointwise a.e. convergence **plus** domination ($|f_m| \le h \in L^p$) implies $L^p$ convergence (by the [Dominated Convergence Theorem](/theorems/4)). [example: Weak Convergence Without Strong Convergence in $L^2$] Let $E = [0, 2\pi]$ with normalized Lebesgue measure $d\mu = (2\pi)^{-1} d\mathcal{L}^1$, and define $f_m(x) = \sqrt{2}\sin(mx)$. Then $\|f_m\|_{L^2} = 1$ for every $m$ (by orthonormality of the trigonometric system). For any $g \in L^2(0, 2\pi)$, the Fourier coefficient $\hat{g}(m) = \frac{1}{2\pi}\int_0^{2\pi} g(x) \sqrt{2}\sin(mx) \, d\mathcal{L}^1(x)$ satisfies $\hat{g}(m) \to 0$ as $m \to \infty$ (by the Riemann-Lebesgue lemma, or equivalently, because $\sum_m |\hat{g}(m)|^2 \le \|g\|_{L^2}^2 < \infty$ by Bessel's inequality, so the terms must vanish). Therefore: \begin{align*} \int_0^{2\pi} f_m(x) g(x) \, d\mu(x) = \sqrt{2}\,\hat{g}(m) \to 0 \quad \text{for every } g \in L^2. \end{align*} The sequence $f_m$ converges weakly to $0$ in $L^2$. However, $\|f_m - 0\|_{L^2} = 1 \not\to 0$, so the convergence is not strong. This is the prototype of the "oscillation" mechanism: the functions $f_m$ oscillate ever more rapidly, so their averages against any fixed $g$ vanish, but their $L^2$ mass does not dissipate. [/example] ## Common Techniques in $L^p$ Theory Working effectively with $L^p$ spaces requires a toolkit of standard proof strategies. The following techniques appear repeatedly across analysis, PDE theory, and probability, and mastering them is essential for applying $L^p$ theory in practice. ### Truncation and Cut-off Many $L^p$ arguments require reducing to functions with bounded support or bounded range. The standard tool is **truncation**: given $f \in L^p(E)$ and $K > 0$, define \begin{align*} f_K(x) := \max(-K, \min(f(x), K)) = \begin{cases} -K & \text{if } f(x) < -K, \\ f(x) & \text{if } |f(x)| \le K, \\ K & \text{if } f(x) > K. \end{cases} \end{align*} Since $|f_K| \le |f|$ pointwise, we have $f_K \in L^p(E)$, and the Dominated Convergence Theorem gives $\|f_K - f\|_{L^p} \to 0$ as $K \to \infty$ (the dominating function is $2|f| \in L^p$). The truncated function $f_K$ is bounded (hence in $L^\infty$), so it lies in $L^p \cap L^\infty$. This is useful when an argument requires bounded functions (e.g., to apply Fubini's theorem or to justify algebraic manipulations like squaring). For spatial localization, multiply by a **cut-off function**: a function $\zeta \in C_c^\infty(\mathbb{R}^n)$ with $0 \le \zeta \le 1$, $\zeta = 1$ on a compact set $K$, and $\operatorname{supp}(\zeta) \subset U$. The product $\zeta f$ has compact support in $U$, and $\|\zeta f - f\|_{L^p(K)} = 0$. By choosing larger and larger compact sets $K_j \nearrow U$, the products $\zeta_j f \to f$ in $L^p(U)$. ### Density Arguments The structure of a density argument in $L^p$ is: (1) prove the desired property for a dense subclass (e.g., $C_c^\infty$ or simple functions), (2) show the property is stable under $L^p$-limits, (3) conclude for all $f \in L^p$ by approximation. The critical step is (2): the property must pass to limits, which typically requires a continuity or closedness estimate. [example: Density Argument for Translation Continuity] The map $\tau_h: L^p(\mathbb{R}^n) \to L^p(\mathbb{R}^n)$ defined by $\tau_h f(x) = f(x - h)$ is an isometry: $\|\tau_h f\|_{L^p} = \|f\|_{L^p}$. We prove the **translation continuity** property: for $1 \le p < \infty$, $\|\tau_h f - f\|_{L^p} \to 0$ as $h \to 0$. **Step 1.** For $\varphi \in C_c^\infty(\mathbb{R}^n)$, the function $\varphi$ is uniformly continuous, so $|\varphi(x-h) - \varphi(x)| \to 0$ uniformly as $h \to 0$. Since $\varphi(x-h) - \varphi(x)$ is supported in the $|h|$-neighborhood of $\operatorname{supp}(\varphi)$ (a bounded set), the Dominated Convergence Theorem gives $\|\tau_h \varphi - \varphi\|_{L^p} \to 0$. **Step 2.** For general $f \in L^p(\mathbb{R}^n)$, pick $\varphi \in C_c^\infty$ with $\|f - \varphi\|_{L^p} < \varepsilon/3$. Then \begin{align*} \|\tau_h f - f\|_{L^p} &\le \|\tau_h f - \tau_h \varphi\|_{L^p} + \|\tau_h \varphi - \varphi\|_{L^p} + \|\varphi - f\|_{L^p} \\ &= \|f - \varphi\|_{L^p} + \|\tau_h \varphi - \varphi\|_{L^p} + \|\varphi - f\|_{L^p} \\ &< \frac{2\varepsilon}{3} + \|\tau_h \varphi - \varphi\|_{L^p}. \end{align*} Choosing $|h|$ small enough that $\|\tau_h \varphi - \varphi\|_{L^p} < \varepsilon/3$ completes the proof. This fails for $p = \infty$: $\tau_h \mathbb{1}_{(0,1)} - \mathbb{1}_{(0,1)}$ has $L^\infty$ norm $1$ for every $h \neq 0$. [/example] ### Duality Testing To prove a norm bound on an $L^p$ function, it is often easier to test against all functions in $L^q$ rather than compute the $L^p$ norm directly. The **duality characterization** of the $L^p$ norm states that for $1 \le p \le \infty$, \begin{align*} \|f\|_{L^p} = \sup \left\{ \left| \int_E fg \, d\mu \right| : g \in L^q(E), \, \|g\|_{L^q} \le 1 \right\}, \end{align*} where $q$ is the conjugate exponent. This is a consequence of the Riesz Representation theorem (the supremum is attained by $g = |f|^{p-2}f / \|f\|_{L^p}^{p/q}$ when $1 < p < \infty$). The duality characterization is particularly useful when $f$ is defined implicitly (e.g., as the solution of an equation) and its $L^p$ norm is hard to compute directly, but testing against arbitrary $g$ is tractable. ### Interpolation Strategy When proving that a function belongs to some $L^p$ space, it is sometimes easier to establish membership in two endpoint spaces and interpolate. By the log-convexity inequality \begin{align*} \|f\|_{L^p} \le \|f\|_{L^{p_0}}^{1-\theta} \|f\|_{L^{p_1}}^{\theta} \quad \text{where } \frac{1}{p} = \frac{1-\theta}{p_0} + \frac{\theta}{p_1}, \end{align*} it suffices to bound $\|f\|_{L^{p_0}}$ and $\|f\|_{L^{p_1}}$ separately. This is particularly effective when the two endpoint estimates use different techniques (e.g., an $L^1$ bound from monotonicity and an $L^\infty$ bound from a maximum principle). ### Uniform Integrability Criterion In $L^1$, the absence of reflexivity means that bounded sequences need not have weakly convergent subsequences. The correct substitute is **uniform integrability**: a family $\mathcal{F} \subset L^1(E, \mu)$ with $\mu(E) < \infty$ is uniformly integrable if \begin{align*} \lim_{K \to \infty} \sup_{f \in \mathcal{F}} \int_{\{|f| > K\}} |f| \, d\mu = 0. \end{align*} The following theorem provides a route to $L^1$ convergence that avoids finding a pointwise dominating function — which is often impossible in nonlinear problems. [quotetheorem:950] This is the natural replacement for the Dominated Convergence Theorem in settings where no single dominating function exists. The hypothesis of uniform integrability can be verified in practice through the following criterion. [quotetheorem:951] ## References 1. Folland, G. B., *Real Analysis: Modern Techniques and Their Applications*, 2nd ed. (1999). 2. Brezis, H., *Functional Analysis, [Sobolev Spaces](/page/Sobolev%20Space) and Partial Differential Equations* (2011). 3. Lieb, E. H. and Loss, M., *Analysis*, 2nd ed. (2001). 4. Stein, E. M. and Shakarchi, R., *Real Analysis: Measure Theory, Integration, and [Hilbert Spaces](/page/Hilbert%20Space)* (2005). 5. Evans, L. C., *Partial Differential Equations*, 2nd ed. (2010). 6. Rudin, W., *Real and Complex Analysis*, 3rd ed. (1987).