The first reason to invent $L^p$ spaces is that pointwise control is too rigid for many analytic problems: an integral can stay small while a function develops narrow spikes, and a sequence of functions can converge in average size while refusing to converge at any fixed point. The $L^p$ viewpoint measures the size of a function by integrating a power of its magnitude, then treats functions that differ only on a null set as the same analytic object. This change of viewpoint is not cosmetic. It is the move that lets limits, approximation, Fourier analysis, probability, and partial differential equations live in common function spaces. Instead of asking whether $f_n(x)$ approaches $f(x)$ for every $x$, we ask whether the total $p$-power error tends to zero. That question is stable under the operations that analysis uses most often. [example: Narrow Spikes With Small $L^p$ Mass] Let $X=[0,1]$ with [Lebesgue measure](/page/Lebesgue%20Measure), fix $1\leq p<\infty$, and define \begin{align*} f_n(x)=n^{1/p}\mathbb{1}_{(0,1/n)}(x). \end{align*} For $x\in(0,1/n)$, one has $|f_n(x)|^p=(n^{1/p})^p=n$, while for $x\notin(0,1/n)$ one has $|f_n(x)|^p=0$. Hence \begin{align*} \int_0^1 |f_n(x)|^p\,dx=\int_0^{1/n} n\,dx. \end{align*} Since the integrand is constant on $(0,1/n)$, \begin{align*} \int_0^{1/n} n\,dx=n\left(\frac{1}{n}-0\right)=1. \end{align*} Therefore $\|f_n\|_p^p=1$, and so $\|f_n\|_p=1$ for every $n$. Now define \begin{align*} g_n(x)=n^{1/p}\mathbb{1}_{(0,1/n^2)}(x). \end{align*} For $x\in(0,1/n^2)$, one has $|g_n(x)|^p=n$, and outside this interval the value is $0$. Thus \begin{align*} \|g_n\|_p^p=\int_0^1 |g_n(x)|^p\,dx=\int_0^{1/n^2} n\,dx. \end{align*} Again using constancy of the integrand, \begin{align*} \int_0^{1/n^2} n\,dx=n\left(\frac{1}{n^2}-0\right)=\frac{1}{n}. \end{align*} Thus $\|g_n\|_p^p=1/n$ and $\|g_n\|_p=n^{-1/p}\to 0$. The two families both form tall spikes near $0$, but $L^p$ size records the product of height-to-the-$p$ and supporting measure, not the visual height alone. [/example] The example shows the central tradeoff. $L^p$ norms do not remember where every value occurs, but they remember how much $p$-power is distributed across the underlying [measure space](/page/Measure%20Space). That is exactly the information needed for integral estimates. ## Definition To make the preceding measurement into a space, we must decide two things: which functions have finite size, and when two functions should be identified. The null-set identification is forced by integration, since the integral cannot distinguish functions that disagree only on a set of measure zero. [definition: $L^p$ Space] Let $(X,\mathcal{A},\mu)$ be a measure space, let $\mathbb{F}$ be either $\mathbb{R}$ or $\mathbb{C}$, and let $1 \leq p \leq \infty$. For $1 \leq p < \infty$, $L^p(X,\mathcal{A},\mu;\mathbb{F})$ is the set of equivalence classes of [measurable functions](/page/Measurable%20Functions) $f: X \to \mathbb{F}$ such that \begin{align*} \int_X |f|^p \, d\mu < \infty. \end{align*} For $p=\infty$, $L^\infty(X,\mathcal{A},\mu;\mathbb{F})$ is the set of equivalence classes of measurable functions $f: X \to \mathbb{F}$ for which there exists $M \in [0,\infty)$ such that $|f(x)| \leq M$ for $\mu$-almost every $x \in X$. Two measurable functions represent the same element when they agree $\mu$-almost everywhere. For $1 \leq p < \infty$, the norm is the map $\|\cdot\|_p: L^p(X,\mathcal{A},\mu;\mathbb{F}) \to [0,\infty)$ given by \begin{align*} \|f\|_p = \left(\int_X |f|^p \, d\mu\right)^{1/p}. \end{align*} For $p=\infty$, the norm is the map $\|\cdot\|_\infty: L^\infty(X,\mathcal{A},\mu;\mathbb{F}) \to [0,\infty)$ given by \begin{align*} \|f\|_\infty = \inf\{M \in [0,\infty): |f(x)| \leq M \text{ for } \mu\text{-almost every } x \in X\}. \end{align*} [/definition] After passing to equivalence classes, the notation $f(x)$ is a convenient representative-level phrase rather than a value attached to the equivalence class. This convention is harmless as long as statements are invariant under changes on null sets. It is also the source of many beginner mistakes: an element of $L^p$ need not have a preferred value at a point. [remark: Naming the Underlying Measure] When the measure space is understood, analysts write $L^p(X)$ or $L^p(\mu)$. On a domain $\Omega \subset \mathbb{R}^n$, the default usually means $L^p(\Omega,\mathcal{B}(\Omega),\mathcal{L}^n;\mathbb{F})$, where $\mathcal{L}^n$ is Lebesgue measure. [/remark] The definition includes several familiar spaces at once. For counting measure on $\mathbb{N}$ it produces sequence spaces; for [probability measure](/page/Probability%20Measure) it describes random variables with finite moments; for Lebesgue measure on a domain it gives the standard function spaces of analysis. ## Integral Size and First Examples The same formula behaves differently on finite, infinite, atomic, and non-atomic measure spaces. This section records the main tests that tell the reader what the norm is actually seeing. ### Sequence Spaces Counting measure turns integration into summation, so $L^p$ becomes the space where the $p$-power sum converges. This example is the cleanest way to see that $p$ controls a tail condition, not merely a pointwise bound. [example: The Sequence Space $\ell^p$] Let $X=\mathbb{N}$, let $\mathcal{A}=2^{\mathbb{N}}$, and let $\mu$ be counting measure. A [measurable function](/page/Measurable%20Function) $a:\mathbb{N}\to\mathbb{F}$ is the same thing as a sequence $a=(a_n)_{n=1}^{\infty}$, because every subset of $\mathbb{N}$ is measurable. For $1\leq p<\infty$, integration with respect to counting measure gives \begin{align*} \int_{\mathbb{N}} |a|^p\,d\mu=\sum_{n=1}^{\infty} |a(n)|^p=\sum_{n=1}^{\infty}|a_n|^p. \end{align*} Therefore $a\in L^p(\mathbb{N},2^{\mathbb{N}},\mu;\mathbb{F})$ exactly when \begin{align*} \sum_{n=1}^{\infty}|a_n|^p<\infty. \end{align*} In that case, \begin{align*} \|a\|_p=\left(\sum_{n=1}^{\infty}|a_n|^p\right)^{1/p}. \end{align*} For $p=\infty$, the only counting-measure null set is $\varnothing$, since every nonempty subset of $\mathbb{N}$ has positive counting measure. Thus $|a_n|\leq M$ for almost every $n$ is the same as $|a_n|\leq M$ for every $n$, and hence \begin{align*} \|a\|_\infty=\inf\{M\in[0,\infty): |a_n|\leq M\text{ for every }n\in\mathbb{N}\}=\sup_{n\in\mathbb{N}}|a_n|. \end{align*} So membership in $L^\infty$ is exactly boundedness of the sequence. For the concrete sequence $a_n=n^{-\alpha}$ with $\alpha>0$ and $1\leq p<\infty$, we have \begin{align*} |a_n|^p=|n^{-\alpha}|^p=n^{-\alpha p}. \end{align*} Thus \begin{align*} \sum_{n=1}^{\infty}|a_n|^p=\sum_{n=1}^{\infty}n^{-\alpha p}. \end{align*} By the *[p-series test](/theorems/3834)*, the series $\sum_{n=1}^{\infty}n^{-s}$ converges exactly when $s>1$. Applying this with $s=\alpha p$, the sequence $(n^{-\alpha})$ belongs to $\ell^p$ exactly when $\alpha p>1$. For example, if $\alpha=3/4$, then $\alpha\cdot 2=3/2>1$ but $\alpha\cdot 1=3/4\leq 1$, so $(n^{-3/4})\in\ell^2$ and $(n^{-3/4})\notin\ell^1$. The exponent $p$ therefore controls how much tail decay is required, not merely whether the terms are individually small. [/example] Sequence spaces show why no single exponent is universally best. Larger $p$ penalizes large terms more strongly, while smaller $p$ is more sensitive to long tails. On an infinite measure space these pressures are not ordered in a simple inclusion chain. ### Finite Measure Spaces On a finite measure space the total amount of room is bounded, so high integrability forces low integrability. This is a first example of a recurring theme: the measure of the ambient space matters as much as the formula. [quotetheorem:9904] The theorem explains why probability theory often treats higher moments as stronger information. When $\mu(X)=1$, the estimate becomes $\|f\|_p \leq \|f\|_q$. The statement fails on many infinite measure spaces, so finite measure is not a decorative hypothesis. [example: Infinite Measure Breaks the Inclusion] Fix $1\leq p<q\leq\infty$. On $X=(1,\infty)$ with Lebesgue measure, take $f(x)=x^{-\alpha}$ with $\alpha>0$. For any finite exponent $r\geq 1$, \begin{align*} \int_1^\infty |f(x)|^r\,dx=\int_1^\infty x^{-\alpha r}\,dx. \end{align*} Writing $s=\alpha r$, if $s\neq 1$, then for $R>1$, \begin{align*} \int_1^R x^{-s}\,dx=\frac{R^{1-s}-1}{1-s}. \end{align*} If $s>1$, then $1-s<0$ and $R^{1-s}\to 0$, so \begin{align*} \lim_{R\to\infty}\frac{R^{1-s}-1}{1-s}=\frac{1}{s-1}. \end{align*} If $s<1$, then $1-s>0$ and $R^{1-s}\to\infty$, so the integral diverges. If $s=1$, then \begin{align*} \int_1^R x^{-1}\,dx=\log R, \end{align*} and $\log R\to\infty$. Hence \begin{align*} \int_1^\infty x^{-\alpha r}\,dx<\infty \end{align*} exactly when $\alpha r>1$. Now choose $\alpha$ with $1/q<\alpha\leq 1/p$, interpreting $1/\infty=0$. If $q<\infty$, then $\alpha q>1$, so $f\in L^q(1,\infty)$; if $q=\infty$, then $0<x^{-\alpha}<1$ on $(1,\infty)$, so $f\in L^\infty(1,\infty)$. But $\alpha p\leq 1$, so the calculation above gives $f\notin L^p(1,\infty)$. Thus on an infinite measure space, higher integrability need not force lower integrability. The reverse failure already occurs near a point singularity. On $X=(0,1)$, take $h(x)=x^{-\beta}$ with $\beta>0$. For any finite exponent $r\geq 1$, \begin{align*} \int_0^1 |h(x)|^r\,dx=\int_0^1 x^{-\beta r}\,dx. \end{align*} Writing $t=\beta r$, if $t\neq 1$, then for $0<\varepsilon<1$, \begin{align*} \int_\varepsilon^1 x^{-t}\,dx=\frac{1-\varepsilon^{1-t}}{1-t}. \end{align*} If $t<1$, then $1-t>0$ and $\varepsilon^{1-t}\to 0$, so \begin{align*} \lim_{\varepsilon\downarrow 0}\frac{1-\varepsilon^{1-t}}{1-t}=\frac{1}{1-t}. \end{align*} If $t>1$, then $1-t<0$ and $\varepsilon^{1-t}\to\infty$, so the integral diverges. If $t=1$, then \begin{align*} \int_\varepsilon^1 x^{-1}\,dx=-\log \varepsilon, \end{align*} and $-\log\varepsilon\to\infty$ as $\varepsilon\downarrow 0$. Hence \begin{align*} \int_0^1 x^{-\beta r}\,dx<\infty \end{align*} exactly when $\beta r<1$. Choose $\beta$ with $1/q\leq\beta<1/p$, again interpreting $1/\infty=0$. Then $\beta p<1$, so $h\in L^p(0,1)$. If $q<\infty$, then $\beta q\geq 1$, so $h\notin L^q(0,1)$; if $q=\infty$, then $x^{-\beta}\to\infty$ as $x\downarrow 0$, so no finite essential bound exists and $h\notin L^\infty(0,1)$. These two constructions show that, without a finite-measure hypothesis, the relation between $L^p$ and $L^q$ depends on where the mass sits: spread out at infinity or concentrated near a singularity. [/example] The finite-measure inclusion is therefore not a statement about exponents alone. It is a statement about exponents together with the size of the underlying measure space. ## Inequalities That Make $L^p$ Work The definition gives a candidate norm, but the theory depends on inequalities that let integrals of products and sums be controlled. Without these estimates, $L^p$ would be a collection of finite integrals rather than a usable analytic setting. ### Dual Exponents Products appear everywhere: testing [weak convergence](/page/Weak%20Convergence), pairing a function with a source term, writing an energy integral, or estimating an expectation. The right way to split a product is to choose exponents whose reciprocals add to one. [definition: Conjugate Exponents] Let $1 \leq p \leq \infty$. A number $q \in [1,\infty]$ is conjugate to $p$ when \begin{align*} \frac{1}{p}+\frac{1}{q}=1, \end{align*} using the convention $1/\infty=0$. [/definition] With this convention, $1$ and $\infty$ are conjugate. This definition packages the arithmetic that makes products integrable. The point is not the equation itself; the point is that it exactly balances the powers in [Young's inequality](/theorems/244). The estimate needed next turns two separate size bounds into one product bound, which is what makes dual testing and weak formulations possible. [quotetheorem:516] Holder's inequality turns $L^q$ functions into bounded linear tests on $L^p$. It is the bridge from integral size to functional analysis, and it is the estimate behind the weak formulations used throughout [Cambridge III Functional Analysis](/page/Cambridge%20III%20Functional%20Analysis). ### The Norm Inequality Once products are under control, sums can be handled. This is the moment when the formula for $\|f\|_p$ becomes a norm rather than just a size functional. The needed estimate is Minkowski's inequality: for $1\leq p\leq \infty$, measurable functions $f$ and $g$ satisfy $\|f+g\|_p\leq \|f\|_p+\|g\|_p$. This is exactly the triangle inequality for $L^p$. Together with scalar homogeneity and definiteness modulo null sets, it makes $L^p$ a [normed vector space](/page/Normed%20Vector%20Space). [remark: Why Null Sets Matter for Definiteness] If $\|f\|_p=0$, then $f=0$ almost everywhere, not necessarily at every point of $X$. The quotient by almost-everywhere equality is what changes this from a seminorm on measurable functions into a norm on equivalence classes. [/remark] The next analytic question is whether Cauchy sequences converge inside the same space. For applications, this is decisive: approximation methods produce limits, and a useful function space must contain those limits. [quotetheorem:892] Completeness is the reason $L^p$ spaces are Banach spaces. In the special case $p=2$, the norm also comes from an [inner product](/page/Inner%20Product), and this extra structure changes the geometry of the space. ## Geometry at $p=2$ and Duality The exponent $2$ is special because products and squares interact through an inner product. This section separates the Hilbert-space geometry of $L^2$ from the Banach-space geometry of the other exponents. ### The Inner Product Case In $L^2$, orthogonality replaces many pointwise arguments. The integral of a product becomes an inner product, so projection, Fourier expansion, and least-squares approximation become available. [definition: $L^2$ Inner Product] Let $(X,\mathcal{A},\mu)$ be a measure space. For $f,g \in L^2(X,\mathcal{A},\mu;\mathbb{C})$, the $L^2$ inner product is the map \begin{align*} \langle \cdot,\cdot\rangle: L^2(X,\mathcal{A},\mu;\mathbb{C}) \times L^2(X,\mathcal{A},\mu;\mathbb{C}) \to \mathbb{C}, \quad (f,g) \mapsto \int_X f\overline{g} \, d\mu. \end{align*} For real-valued functions, the codomain is $\mathbb{R}$ and the integrand is $fg$. [/definition] The inner product is more than notation. It creates angles and orthogonal projections in a space of equivalence classes of functions. The bound needed next is the estimate that makes this pairing continuous, which is why $L^2$ is the natural home for Fourier methods and many weak formulations. [quotetheorem:432] This estimate is Holder's inequality at $p=q=2$, but it deserves separate attention because it drives geometric reasoning. Orthogonality, projections, and energy methods all start from this bound. [example: Fourier Coefficients as $L^2$ Pairings] On $[0,2\pi]$ with Lebesgue measure, fix $n\in\mathbb{Z}$ and write \begin{align*} \varphi_n(x)=e^{inx}. \end{align*} Since $|e^{inx}|=1$ for every $x$, we have \begin{align*} \|\varphi_n\|_2^2=\int_0^{2\pi}|e^{inx}|^2\,dx=\int_0^{2\pi}1\,dx=2\pi. \end{align*} Thus $\|\varphi_n\|_2=(2\pi)^{1/2}$, so $\varphi_n\in L^2([0,2\pi])$. For $f\in L^2([0,2\pi])$, the $n$th Fourier coefficient is the normalized $L^2$ pairing \begin{align*} \widehat{f}(n)=\frac{1}{2\pi}\int_0^{2\pi}f(x)e^{-inx}\,dx=\frac{1}{2\pi}\int_0^{2\pi}f(x)\overline{\varphi_n(x)}\,dx. \end{align*} By *[Cauchy-Schwarz Inequality](/theorems/432) in $L^2$*, \begin{align*} \left|\int_0^{2\pi}f(x)\overline{\varphi_n(x)}\,dx\right|\leq \|f\|_2\|\varphi_n\|_2. \end{align*} Substituting $\|\varphi_n\|_2=(2\pi)^{1/2}$ gives \begin{align*} |\widehat{f}(n)|\leq \frac{1}{2\pi}\|f\|_2(2\pi)^{1/2}=\frac{1}{(2\pi)^{1/2}}\|f\|_2. \end{align*} The map $f\mapsto \widehat{f}(n)$ is linear because integration is linear, and the displayed bound shows it is continuous with [operator norm](/page/Operator%20Norm) at most $(2\pi)^{-1/2}$. Therefore a Fourier coefficient is not a pointwise sample of $f$; it is a bounded linear measurement of the $L^2$ equivalence class of $f$. [/example] ### Dual Pairings For exponents other than $2$, there may be no inner product, but there is still a robust pairing between conjugate spaces. This pairing is the basic language of weak convergence and distributional formulations. [quotetheorem:9905] The theorem gives a canonical isometric embedding from $L^q$ into the continuous dual of $L^p$; surjectivity is a separate representation theorem, not a consequence of Holder's inequality alone. On the usual $\sigma$-finite measure spaces, the embedding is onto for $1\leq p<\infty$, so $(L^p)^*\cong L^q$ with $q$ conjugate to $p$. For arbitrary measure spaces, this slogan should be read only after the relevant measure-theoretic hypotheses have been stated, and the general form is usually expressed using localizable measure spaces. The endpoint $p=\infty$ is different even in familiar settings: the dual of $L^\infty$ contains bounded linear functionals not given by integration against an $L^1$ function. [remark: Endpoint Caution] The statement that every bounded linear functional on $L^p$ is represented by an $L^q$ function depends on both the exponent and the measure space. The reflexive duality picture belongs to $1<p<\infty$ under standard hypotheses; the $L^1$ endpoint requires its own measure-theoretic representation theorem, and $L^\infty$ is the main warning sign because its dual contains finitely additive phenomena beyond integration against $L^1$ functions. [/remark] ## Approximation and Convergence The value of $L^p$ spaces comes from the way they interact with limiting processes. A function may be rough, discontinuous, or only defined up to a null set, yet still be approximated by simpler functions in the norm that matters. ### Simple Functions The first approximation principle says that measurable functions can be built from finite-valued pieces. This is the measure-theoretic analogue of approximating a complicated object by step functions. [definition: Simple Function] Let $(X,\mathcal{A})$ be a measurable space and let $\mathbb{F}$ be either $\mathbb{R}$ or $\mathbb{C}$. A [simple function](/page/Simple%20Function) is a measurable map $s:X\to\mathbb{F}$ whose image $s(X)$ is a finite subset of $\mathbb{F}$. [/definition] Simple functions are the algebraic atoms of the [Lebesgue integral](/page/Lebesgue%20Integral). They are concrete enough that many estimates can first be checked on finite sums. The approximation theorem needed next says that, for finite $p$, these finite-valued atoms are still rich enough to recover every $L^p$ element in norm. [quotetheorem:1079] Density of simple functions is the entry point to more refined approximation. On Euclidean domains, additional regularity of the measure and domain lets one approximate many $L^p$ functions by continuous or smooth functions. The restriction $p<\infty$ is real: the essential supremum norm is too rigid to allow this kind of approximation for arbitrary measurable functions. [example: A Step Function Is Not Uniformly Approximable by Continuous Functions] Let $X=[0,1]$ with Lebesgue measure and let $f=\mathbb{1}_{(0,1/2)}$. We show that no [continuous function](/page/Continuous%20Function) $g\in C([0,1])$ can satisfy $\|f-g\|_\infty<1/2$. Suppose, for contradiction, that $g\in C([0,1])$ and $\|f-g\|_\infty<1/2$. Choose a number $\alpha$ such that \begin{align*} \|f-g\|_\infty<\alpha<\frac{1}{2}. \end{align*} By the definition of the $L^\infty$ norm, there is a null set $N\subset[0,1]$ such that $|f(x)-g(x)|\leq\alpha$ for every $x\in[0,1]\setminus N$. If $x\in(0,1/2)\setminus N$, then $f(x)=1$, so \begin{align*} |1-g(x)|\leq\alpha. \end{align*} This implies \begin{align*} g(x)\geq 1-\alpha. \end{align*} If $x\in(1/2,1)\setminus N$, then $f(x)=0$, so \begin{align*} |g(x)|\leq\alpha, \end{align*} and therefore \begin{align*} g(x)\leq\alpha. \end{align*} The set $(0,1/2)\setminus N$ is dense in $(0,1/2)$: if some open interval $I\subset(0,1/2)$ were disjoint from it, then $I\subset N$, which would force $\lambda(N)\geq\lambda(I)>0$, contradicting that $N$ is null. Hence for each $k$ we can choose \begin{align*} x_k\in\left(\frac{1}{2}-\frac{1}{k},\frac{1}{2}\right)\setminus N \end{align*} for all sufficiently large $k$, and then $x_k\to 1/2$ with $g(x_k)\geq 1-\alpha$. Similarly, $(1/2,1)\setminus N$ is dense in $(1/2,1)$, so for all sufficiently large $k$ we can choose \begin{align*} y_k\in\left(\frac{1}{2},\frac{1}{2}+\frac{1}{k}\right)\setminus N, \end{align*} and then $y_k\to 1/2$ with $g(y_k)\leq\alpha$. Since $g$ is continuous at $1/2$, both sequences must satisfy \begin{align*} g(x_k)\to g(1/2) \end{align*} and \begin{align*} g(y_k)\to g(1/2). \end{align*} But $g(x_k)\geq 1-\alpha$ for every chosen $k$ implies $g(1/2)\geq 1-\alpha$, while $g(y_k)\leq\alpha$ for every chosen $k$ implies $g(1/2)\leq\alpha$. This is impossible, because $\alpha<1/2$ gives $1-\alpha>\alpha$. Therefore $\|f-g\|_\infty\geq 1/2$ for every $g\in C([0,1])$, so the step function cannot be approximated in $L^\infty$ by continuous functions. [/example] This example explains why approximation theorems in $L^\infty$ usually need extra structure or weaker topologies. Finite $p$ norms can tolerate small transition layers; the essential supremum norm cannot hide a jump by making the bad set narrow. ### Modes of Convergence Norm convergence is stronger than [convergence in measure](/page/Convergence%20in%20Measure), and both are distinct from pointwise convergence. The $L^p$ norm chooses a quantitative mode: the total $p$-power error must vanish. [definition: Convergence in $L^p$] Let $(X,\mathcal{A},\mu)$ be a measure space and let $1 \leq p \leq \infty$. Let $(f_n)_{n=1}^{\infty}$ be a sequence of equivalence classes in $L^p(X,\mathcal{A},\mu;\mathbb{F})$, represented by measurable functions $f_n:X\to\mathbb{F}$, and let $f \in L^p(X,\mathcal{A},\mu;\mathbb{F})$ be represented by a measurable function $f:X\to\mathbb{F}$. The sequence $(f_n)_{n=1}^{\infty}$ converges to $f$ in $L^p$ if \begin{align*} \lim_{n\to\infty}\|f_n-f\|_p=0. \end{align*} [/definition] This definition is strong enough to pass to many integral estimates, but weak enough to ignore behavior on small sets. The spike examples at the start are warnings that pointwise intuition and norm intuition do not always move together. [quotetheorem:4] The theorem gives a common route from pointwise convergence to norm convergence. Without domination, pointwise convergence can miss escaping mass, moving spikes, or growth on small sets. ## Beyond and Connected Topics $L^p$ spaces are a common base layer for several parts of analysis. In [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology), they sit near the transition from metric completeness to functional spaces. In [Cambridge II Analysis of Functions](/page/Cambridge%20II%20Analysis%20of%20Functions), they connect measure-theoretic convergence to approximation and Fourier analysis. In [Cambridge III Functional Analysis](/page/Cambridge%20III%20Functional%20Analysis), they become examples of Banach spaces, Hilbert spaces, dual spaces, and weak topologies. The next natural direction is [Sobolev Space](/page/Sobolev%20Space), where $L^p$ control is imposed not only on a function but also on its weak derivatives. Another direction is probability, where elements of $L^p$ are random variables with finite $p$th moment and the inequalities above become moment estimates. A third direction is harmonic analysis, where boundedness of operators between $L^p$ spaces becomes the central question. ## References Androma, [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology). Androma, [Cambridge II Analysis of Functions](/page/Cambridge%20II%20Analysis%20of%20Functions). Androma, [Cambridge III Functional Analysis](/page/Cambridge%20III%20Functional%20Analysis). Androma, [Sobolev Space](/page/Sobolev%20Space). Androma, [Cambridge IA Analysis Notes](/page/Cambridge%20IA%20Analysis%20Notes). Gerald B. Folland, *Real Analysis: Modern Techniques and Their Applications* (1999). Walter Rudin, *Real and Complex Analysis* (1987). Elias M. Stein and Rami Shakarchi, *Real Analysis: Measure Theory, Integration, and Hilbert Spaces* (2005).