Analysis requires a way to measure the "size" of the objects it studies — [functions](/page/Function), [sequences](/page/Sequence), operators, distributions. A **norm** is the abstraction of size: it assigns to each element of a vector space a non-negative real number satisfying the triangle inequality, positive-definiteness, and homogeneity. Equipped with a norm, a vector space becomes a **normed vector space** — a setting where notions of convergence, [continuity](/page/Continuity), and completeness are governed by a single, explicit quantity $\|x\|$. Normed spaces sit at a critical junction in the hierarchy of mathematical structures. They are special cases of [metric spaces](/page/Metric%20Space) (the metric is $d(x,y) = \|x - y\|$), of [topological vector spaces](/page/Topological%20Vector%20Space) (the topology is the metric topology), and of locally convex spaces (a single norm is a single seminorm). But they carry more structure than any of these: the norm is compatible with the linear structure in a way that a general metric or topology need not be, and this compatibility is what makes the theories of [bounded linear operators](/page/Linear%20Operators%20on%20Banach%20Spaces), [dual spaces](/page/Topological%20Dual), and spectral theory possible. The central dichotomy in the theory is between **complete** normed spaces ([Banach spaces](/page/Banach%20Space)) and incomplete ones. Completeness — the convergence of every Cauchy sequence — is the property that makes [limits](/page/Limit) exist, and its presence or absence controls whether fundamental theorems like the [Open Mapping Theorem](/theorems/631), the [Closed Graph Theorem](/theorems/217), and the [Uniform Boundedness Principle](/theorems/549) hold. [motivation] ## Motivation ### From Absolute Value to Norm On $\mathbb{R}$, the absolute value $|x|$ measures the size of a number and gives a distance $d(x, y) = |x - y|$. On $\mathbb{R}^n$, the Euclidean length $|x| = (\sum_{i=1}^n x_i^2)^{1/2}$ does the same. Both satisfy the triangle inequality, scale linearly ($|\alpha x| = |\alpha| |x|$), and vanish only at the origin. The question is: can we extend this structure to infinite-dimensional vector spaces — spaces of functions, sequences, or distributions — in a way that supports the convergence arguments of analysis? A **metric** alone is not enough. Metrics on vector spaces need not be compatible with the linear structure: there exist translation-invariant metrics on $\mathbb{R}^\infty$ for which scalar multiplication is discontinuous. A **topology** alone is too general: the topology on $\mathcal{D}(\Omega)$ ([test functions](/page/Test%20Function)) cannot be generated by any single norm, and the full machinery of [topological vector spaces](/page/Topological%20Vector%20Space) is needed to handle such spaces. What we want is the simplest structure that simultaneously provides a notion of size, induces a topology compatible with the algebra, and supports quantitative estimates — bounds of the form $\|Tx\| \le C\|x\|$ that are the language of analysis. ### What a Norm Must Do A norm on a vector space $X$ is a function $\|\cdot\|: X \to [0, \infty)$ that captures the essential properties of length: it is zero only at the origin (so distinct points are distinguishable), it scales linearly with the scalar field (so $\|2x\| = 2\|x\|$), and it satisfies the triangle inequality (so the "direct path" is always shortest). These three axioms are precisely what is needed to define a metric $d(x,y) = \|x - y\|$ that is translation-invariant ($d(x+z, y+z) = d(x,y)$) and homogeneous ($d(\alpha x, \alpha y) = |\alpha| d(x,y)$) — and these two properties ensure that the induced topology makes addition and scalar multiplication continuous, so the resulting topological vector space is well-behaved. ### The Completeness Question Having a norm gives us [Cauchy sequences](/page/Cauchy%20Sequence). But Cauchy sequences need not converge: the rational numbers $\mathbb{Q}$ with the absolute value form a normed space in which $\sum_{k=0}^n 1/k!$ is Cauchy but does not converge (its limit is $e \notin \mathbb{Q}$). In infinite-dimensional function spaces, the same phenomenon is more severe: a sequence of smooth functions can be Cauchy in one norm but converge to a non-smooth limit (or to no element of the space at all). The question of whether every Cauchy sequence converges — **completeness** — is the dividing line between spaces where analysis works well (Banach spaces) and spaces where it does not. [/motivation] ## Definition A norm must capture three properties of the absolute value: positive-definiteness (only zero has zero size), absolute homogeneity (scaling the vector scales the size), and the triangle inequality (the size of a sum is at most the sum of the sizes). These are the minimal axioms that produce a useful metric from a linear structure. [definition:Norm] Let $X$ be a vector space over $\mathbb{F}$, where $\mathbb{F} = \mathbb{R}$ or $\mathbb{F} = \mathbb{C}$. A **norm** on $X$ is a function $\|\cdot\|: X \to [0, \infty)$ satisfying: 1. **Positive-definiteness:** $\|x\| = 0$ if and only if $x = 0$. 2. **Absolute homogeneity:** $\|\alpha x\| = |\alpha| \|x\|$ for all $\alpha \in \mathbb{F}$ and $x \in X$. 3. **Triangle inequality:** $\|x + y\| \le \|x\| + \|y\|$ for all $x, y \in X$. [/definition] [definition:Normed Vector Space] A **normed vector space** (or **normed space**) is a pair $(X, \|\cdot\|)$ consisting of a vector space $X$ over $\mathbb{F}$ and a norm $\|\cdot\|$ on $X$. [/definition] Every norm induces a metric $d: X \times X \to [0, \infty)$ by $d(x, y) := \|x - y\|$. This metric is translation-invariant ($d(x+z, y+z) = d(x,y)$) and absolutely homogeneous ($d(\alpha x, \alpha y) = |\alpha| d(x,y)$). Conversely, a translation-invariant, absolutely homogeneous metric on a vector space arises from a norm via $\|x\| = d(x, 0)$. The metric induces a topology (the metric topology), making $(X, \|\cdot\|)$ a [topological vector space](/page/Topological%20Vector%20Space). This topology is Hausdorff (distinct points can be separated by open sets, since $\|x - y\| > 0$ for $x \neq y$), first countable (open balls $B(x, 1/n)$ form a countable neighbourhood basis), and locally convex (the open balls are convex [sets](/page/Set)). The open ball of radius $r$ centred at $x$ is $B(x, r) := \{y \in X : \|y - x\| < r\}$. ## Examples [example:The Euclidean Norm On $\mathbb{R}^n$] The **Euclidean norm** on $\mathbb{R}^n$ is \begin{align*} \|x\|_2 := \left(\sum_{i=1}^n x_i^2\right)^{1/2}. \end{align*} Positive-definiteness and homogeneity are immediate. The triangle inequality $\|x + y\|_2 \le \|x\|_2 + \|y\|_2$ is the **Cauchy-Schwarz inequality** applied to the inner product $x \cdot y = \sum_{i=1}^n x_i y_i$: since $(x \cdot y)^2 \le \|x\|_2^2 \|y\|_2^2$, one obtains $\|x + y\|_2^2 = \|x\|_2^2 + 2(x \cdot y) + \|y\|_2^2 \le (\|x\|_2 + \|y\|_2)^2$. More generally, $\mathbb{R}^n$ carries the family of **$p$-norms** for $1 \le p \le \infty$: \begin{align*} \|x\|_p &:= \left(\sum_{i=1}^n |x_i|^p\right)^{1/p} \quad (1 \le p < \infty), \\ \|x\|_\infty &:= \max_{1 \le i \le n} |x_i|. \end{align*} The triangle inequality for $\|\cdot\|_p$ is the **Minkowski inequality**, which follows from the **Hölder inequality** $\sum |x_i y_i| \le \|x\|_p \|y\|_q$ with $1/p + 1/q = 1$. [/example] [example:Sequence Spaces $\ell^p$] For $1 \le p \le \infty$, the **sequence space** $\ell^p$ consists of sequences $x = (x_k)_{k=1}^\infty$ in $\mathbb{C}$ with finite $p$-norm: \begin{align*} \ell^p &:= \left\{(x_k) \in \mathbb{C}^{\mathbb{N}} : \|x\|_{\ell^p} := \left(\sum_{k=1}^\infty |x_k|^p\right)^{1/p} < \infty\right\} \quad (1 \le p < \infty), \\ \ell^\infty &:= \left\{(x_k) \in \mathbb{C}^{\mathbb{N}} : \|x\|_{\ell^\infty} := \sup_{k \ge 1} |x_k| < \infty\right\}. \end{align*} Each $\ell^p$ is a Banach space (completeness is proved below). The inclusions $\ell^p \subset \ell^q$ for $p \le q$ hold with $\|x\|_{\ell^q} \le \|x\|_{\ell^p}$. The space $\ell^2$ is a [Hilbert space](/page/Hilbert%20Space) with inner product $(x, y)_{\ell^2} = \sum_{k=1}^\infty x_k \overline{y_k}$. [/example] [example:Function Spaces $L^p$] For a measure space $(E, \mathcal{E}, \mu)$ and $1 \le p \le \infty$, the **Lebesgue space** $L^p(E, \mu)$ consists of (equivalence classes of) measurable functions $f: E \to \mathbb{C}$ with finite $p$-norm: \begin{align*} \|f\|_{L^p} &:= \left(\int_E |f|^p \, d\mu\right)^{1/p} \quad (1 \le p < \infty), \\ \|f\|_{L^\infty} &:= \operatorname{ess\,sup}_{x \in E} |f(x)| = \inf\{C \ge 0 : |f(x)| \le C \text{ $\mu$-a.e.}\}. \end{align*} Two functions $f, g$ with $f = g$ $\mu$-a.e. are identified, so elements of $L^p$ are equivalence classes — this is necessary for positive-definiteness ($\|f\|_{L^p} = 0$ implies $f = 0$ a.e., hence $f = 0$ in $L^p$). The triangle inequality for $L^p$ is the **Minkowski inequality** for integrals. Each $L^p$ space is complete — the Riesz-Fischer theorem — making them Banach spaces. The space $L^2(E, \mu)$ is a Hilbert space with inner product $(f, g)_{L^2} = \int_E f \bar{g} \, d\mu$. For $1 \le p < \infty$ and $E \subseteq \mathbb{R}^n$ open, $L^p(E, \mathcal{L}^n)$ is [separable](/theorems/548). The space $L^\infty$ is not [separable](/page/Separable) when $\mu$ is non-atomic. [/example] [example:The Supremum Norm On $C(K)$] For a compact Hausdorff space $K$, the space $C(K)$ of continuous functions $f: K \to \mathbb{C}$ carries the **supremum norm** \begin{align*} \|f\|_\infty := \max_{x \in K} |f(x)|. \end{align*} The maximum exists by compactness and continuity. This makes $C(K)$ a Banach space: if $(f_n)$ is Cauchy in $C(K)$, the pointwise limit $f(x) = \lim_n f_n(x)$ exists (since $\mathbb{C}$ is complete) and is continuous (since the convergence is uniform), so $f \in C(K)$ and $\|f_n - f\|_\infty \to 0$. For $K = [0,1]$, the [topological dual](/page/Topological%20Dual) $C([0,1])^*$ is the space of complex Radon measures on $[0,1]$. [/example] [example:Sobolev Norms] The [Sobolev space](/page/Sobolev%20Space) $W^{k,p}(U)$ for an open set $U \subseteq \mathbb{R}^n$ carries the norm \begin{align*} \|u\|_{W^{k,p}(U)} := \left(\sum_{|\alpha| \le k} \|D^\alpha u\|_{L^p(U)}^p\right)^{1/p} \quad (1 \le p < \infty), \end{align*} which measures both the function and its weak [derivatives](/page/Derivative) up to order $k$ in the $L^p$ sense. The Sobolev norm is strictly stronger than the $L^p$ norm: $\|u\|_{L^p} \le \|u\|_{W^{k,p}}$. The space $W^{k,p}(U)$ is a Banach space, and $H^k(U) = W^{k,2}(U)$ is a Hilbert space with inner product $(u, v)_{H^k} = \sum_{|\alpha| \le k} (D^\alpha u, D^\alpha v)_{L^2}$. [/example] ## Completeness and Banach Spaces In a normed space, every convergent sequence is Cauchy (since $\|x_m - x_n\| \le \|x_m - x\| + \|x - x_n\| \to 0$). The converse — every Cauchy sequence converges — need not hold, and its validity is the most important structural property a normed space can have. [definition:Banach Space] A normed vector space $(X, \|\cdot\|)$ is a **Banach space** if it is **complete**: every Cauchy sequence in $X$ converges to an element of $X$. Explicitly, if $(x_n)_{n=1}^\infty \subset X$ satisfies $\|x_m - x_n\| \to 0$ as $m, n \to \infty$, then there exists $x \in X$ with $\|x_n - x\| \to 0$. [/definition] Completeness is not an intrinsic property of the vector space — it depends on the norm. The same vector space can be complete in one norm and incomplete in another. [example:An Incomplete Normed Space] Let $X = C([0,1])$ equipped with the $L^1$ norm $\|f\|_{L^1} = \int_0^1 |f(x)| \, d\mathcal{L}^1(x)$ instead of the supremum norm. The space $(C([0,1]), \|\cdot\|_{L^1})$ is a normed space but is **not** complete. Define the sequence of continuous functions \begin{align*} f_n(x) := \begin{cases} 0 & \text{if } 0 \le x \le 1/2 - 1/n, \\ \frac{n}{2}(x - 1/2) + 1/2 & \text{if } 1/2 - 1/n < x < 1/2 + 1/n, \\ 1 & \text{if } 1/2 + 1/n \le x \le 1. \end{cases} \end{align*} Each $f_n$ is continuous, so $f_n \in C([0,1])$. For $m > n$, the functions $f_m$ and $f_n$ differ only on the interval $(1/2 - 1/n, 1/2 + 1/n)$ where $|f_m - f_n| \le 1$, so \begin{align*} \|f_m - f_n\|_{L^1} \le \mathcal{L}^1\big((1/2 - 1/n, 1/2 + 1/n)\big) = \frac{2}{n} \to 0. \end{align*} The sequence is Cauchy in $L^1$. But the pointwise limit is $f(x) = \mathbb{1}_{[1/2, 1]}(x)$ — the Heaviside step function — which is discontinuous and does not belong to $C([0,1])$. No continuous function is the $L^1$-limit of $(f_n)$, so $(C([0,1]), \|\cdot\|_{L^1})$ is incomplete. The completion of $(C([0,1]), \|\cdot\|_{L^1})$ is $L^1([0,1], \mathcal{L}^1)$: the space of Lebesgue-[integrable](/page/Integral) functions (identified up to a.e. equality). This is the prototypical example of how the passage to Lebesgue theory *completes* classical function spaces. [/example] ### The Completeness Criterion via Absolute Convergence A useful equivalent formulation of completeness avoids Cauchy sequences entirely. [theorem:Absolute Convergence Characterisation Of Banach Spaces] A normed vector space $(X, \|\cdot\|)$ is a Banach space if and only if every absolutely convergent series converges: whenever $(x_k)_{k=1}^\infty \subset X$ satisfies $\sum_{k=1}^\infty \|x_k\| < \infty$, the partial sums $S_N = \sum_{k=1}^N x_k$ converge in $X$. [/theorem] **Proof.** ($\Rightarrow$) If $X$ is complete and $\sum \|x_k\| < \infty$, then the partial sums form a Cauchy sequence: for $M > N$, $\|S_M - S_N\| = \|\sum_{k=N+1}^M x_k\| \le \sum_{k=N+1}^M \|x_k\| \to 0$ as $N \to \infty$ (tail of a convergent series). By completeness, $(S_N)$ converges. ($\Leftarrow$) Let $(y_n)$ be Cauchy in $X$. Choose a subsequence $(y_{n_k})$ with $\|y_{n_{k+1}} - y_{n_k}\| < 2^{-k}$. Define $x_1 = y_{n_1}$ and $x_k = y_{n_k} - y_{n_{k-1}}$ for $k \ge 2$. Then $\sum \|x_k\| \le \|y_{n_1}\| + \sum_{k=1}^\infty 2^{-k} < \infty$, so $S_K = y_{n_K}$ converges to some $y \in X$. Since $(y_n)$ is Cauchy and has a convergent subsequence, $y_n \to y$. This criterion is often the most convenient way to prove completeness of a specific function space: to show $L^p$ is complete, one takes an absolutely convergent series in $L^p$, constructs the pointwise limit via monotone convergence, and verifies it belongs to $L^p$. ### The Completion Theorem Every incomplete normed space can be embedded into a Banach space in an essentially unique way. This is the analogue of constructing $\mathbb{R}$ from $\mathbb{Q}$ by completing with respect to the absolute value. [theorem:Completion Of A Normed Space] Let $(X, \|\cdot\|)$ be a normed vector space. There exists a Banach space $(\bar{X}, \|\cdot\|_{\bar{X}})$ and an isometric linear embedding $\iota: X \hookrightarrow \bar{X}$ such that $\iota(X)$ is dense in $\bar{X}$. The pair $(\bar{X}, \iota)$ is unique up to isometric isomorphism. [/theorem] The construction proceeds by taking equivalence classes of Cauchy sequences in $X$ (two Cauchy sequences $(x_n)$ and $(y_n)$ are equivalent if $\|x_n - y_n\| \to 0$), defining the norm of a class as $\|[(x_n)]\| := \lim_n \|x_n\|$, and embedding $X$ via constant sequences $x \mapsto [(x, x, x, \ldots)]$. [example:Completions Of Classical Spaces] The following completions are foundational: | Incomplete space | Norm | Completion | |---|---|---| | $\mathbb{Q}$ | $\|x\| = \|x\|$ | $\mathbb{R}$ | | $(C([0,1]), \|\cdot\|_{L^p})$ | $\|f\|_{L^p} = (\int |f|^p)^{1/p}$ | $L^p([0,1])$ | | $(C_c^\infty(U), \|\cdot\|_{W^{k,p}})$ | Sobolev norm | $W^{k,p}_0(U)$ | | $c_{00}$ (finite sequences) | $\|\cdot\|_{\ell^p}$ | $\ell^p$ | The third line is the definition of the Sobolev space $W^{k,p}_0(U)$: it is the closure (hence completion) of $C_c^\infty(U)$ in the $W^{k,p}$ norm. This is the space of Sobolev functions that "vanish on the [boundary](/page/Boundary)" in the trace sense. [/example] ## Equivalence of Norms A vector space can carry many different norms, each giving a different metric, different open sets, and potentially different notions of convergence. The natural question is: when do two norms give the "same" analysis? [definition:Equivalent Norms] Two norms $\|\cdot\|_a$ and $\|\cdot\|_b$ on a vector space $X$ are **equivalent** if there exist constants $c, C > 0$ such that \begin{align*} c \|x\|_a \le \|x\|_b \le C \|x\|_a \quad \text{for all } x \in X. \end{align*} [/definition] Equivalent norms generate the same topology: the identity map $(X, \|\cdot\|_a) \to (X, \|\cdot\|_b)$ is a homeomorphism. In particular, they have the same [open sets](/page/Open%20Set), the same convergent sequences, the same Cauchy sequences, and one is complete if and only if the other is. Equivalence of norms is an [equivalence relation](/page/Equivalence%20Relation). ### Finite-Dimensional Norm Equivalence In finite dimensions, the choice of norm is irrelevant for [topological](/page/Topology) purposes — all norms are equivalent. This is the fundamental reason why finite-dimensional analysis is "easy": convergence, continuity, and compactness do not depend on which norm one uses. [theorem:All Norms On A Finite-Dimensional Space Are Equivalent] Let $X$ be a finite-dimensional vector space over $\mathbb{R}$ or $\mathbb{C}$ with $\dim X = n < \infty$. Then any two norms on $X$ are equivalent. [/theorem] **Proof sketch.** Fix a basis $\{e_1, \ldots, e_n\}$ and the reference norm $\|x\|_1 = \sum_{i=1}^n |\alpha_i|$ where $x = \sum \alpha_i e_i$. For any norm $\|\cdot\|$, the triangle inequality gives $\|x\| \le (\max_i \|e_i\|) \sum |\alpha_i| = C\|x\|_1$. The reverse bound $c\|x\|_1 \le \|x\|$ uses compactness: the unit sphere $S = \{x : \|x\|_1 = 1\}$ is compact in the $\|\cdot\|_1$ topology (since $X$ is finite-dimensional), and $x \mapsto \|x\|$ is continuous and positive on $S$, so it attains a positive minimum $c > 0$. A crucial consequence: **every finite-dimensional normed space is complete** (hence a Banach space). This follows because $\mathbb{R}^n$ with the Euclidean norm is complete, and completeness is preserved under norm equivalence. In particular, every finite-dimensional subspace of a normed space is closed. ### Failure in Infinite Dimensions In infinite dimensions, norms can be genuinely inequivalent — they give different topologies, different convergent sequences, and different completions. [example:Inequivalent Norms On $C([0,1])$] On $X = C([0,1])$, the supremum norm $\|f\|_\infty = \max |f|$ and the $L^1$ norm $\|f\|_{L^1} = \int_0^1 |f| \, d\mathcal{L}^1$ are not equivalent. The $L^1$ norm is dominated by the supremum norm: $\|f\|_{L^1} \le \|f\|_\infty$. But no reverse bound exists. To see this, define $f_n(x) = x^n$ for $x \in [0,1]$. Then $\|f_n\|_\infty = 1$ for all $n$, but \begin{align*} \|f_n\|_{L^1} = \int_0^1 x^n \, d\mathcal{L}^1(x) = \frac{1}{n+1} \to 0. \end{align*} If the norms were equivalent, $\|f_n\|_\infty \le C \|f_n\|_{L^1}$ would give $1 \le C/(n+1) \to 0$, a contradiction. The two norms give different completions: $C([0,1])$ is complete in $\|\cdot\|_\infty$ but incomplete in $\|\cdot\|_{L^1}$ (as shown in the earlier example). The topology of the $L^1$ norm is strictly coarser than that of $\|\cdot\|_\infty$. [/example] ## Bounded [Linear Maps](/page/Linear%20Map) and the Operator Norm The morphisms between normed spaces are the **bounded linear maps**: linear maps that do not blow up the norm by more than a fixed multiplicative constant. The collection of all bounded linear maps between two normed spaces is itself a normed space — this is the operator norm construction, and it is the mechanism by which the [topological dual](/page/Topological%20Dual) $X^*$ inherits a Banach space structure. A linear map $T: X \to Y$ between normed spaces is continuous if and only if it is bounded — that is, the quantity $\sup_{\|x\| \le 1} \|Tx\|_Y$ is finite. This equivalence is special to normed spaces (for general topological vector spaces, continuity does not reduce to a single bound). [definition:Bounded Linear Map And Operator Norm] Let $(X, \|\cdot\|_X)$ and $(Y, \|\cdot\|_Y)$ be normed vector spaces. A linear map $T: X \to Y$ is **bounded** if \begin{align*} \|T\|_{\mathcal{L}(X,Y)} := \sup_{\substack{x \in X \\ \|x\|_X \le 1}} \|Tx\|_Y < \infty. \end{align*} The quantity $\|T\|_{\mathcal{L}(X,Y)}$ is the **operator norm** of $T$. The space of all bounded linear maps from $X$ to $Y$ is denoted $\mathcal{L}(X, Y)$. [/definition] The operator norm satisfies $\|Tx\|_Y \le \|T\|_{\mathcal{L}(X,Y)} \|x\|_X$ for all $x \in X$ — this is the fundamental estimate that makes bounded operators useful. The space $\mathcal{L}(X, Y)$ is a normed vector space under the operator norm, and if $Y$ is a Banach space, then $\mathcal{L}(X, Y)$ is also a Banach space (regardless of whether $X$ is complete). In particular, the [topological dual](/page/Topological%20Dual) $X^* = \mathcal{L}(X, \mathbb{C})$ is always a Banach space. [example:Computing Operator Norms] **The identity map.** Let $\mathrm{Id}: (X, \|\cdot\|) \to (X, \|\cdot\|)$ be the identity. Then $\|\mathrm{Id}\| = \sup_{\|x\| \le 1} \|x\| = 1$ (for $X \neq \{0\}$). **Change of norm.** If $X$ carries two equivalent norms $\|\cdot\|_a$ and $\|\cdot\|_b$ with $c\|x\|_a \le \|x\|_b \le C\|x\|_a$, then the identity map $\mathrm{Id}: (X, \|\cdot\|_a) \to (X, \|\cdot\|_b)$ has operator norm $\|\mathrm{Id}\| \le C$, and $\|\mathrm{Id}^{-1}\| \le 1/c$. **Integration as a bounded functional.** The map $I: (C([0,1]), \|\cdot\|_\infty) \to \mathbb{C}$ defined by $I(f) = \int_0^1 f(x) \, d\mathcal{L}^1(x)$ is bounded: $|I(f)| \le \int_0^1 |f| \le \|f\|_\infty$, so $\|I\| \le 1$. Testing on $f \equiv 1$: $I(1) = 1 = \|1\|_\infty$, so $\|I\| = 1$. The functional $I$ belongs to $C([0,1])^*$. **The left-shift on $\ell^p$.** Define $L: \ell^p \to \ell^p$ by $L(x_1, x_2, x_3, \ldots) = (x_2, x_3, x_4, \ldots)$. Then $\|Lx\|_{\ell^p}^p = \sum_{k=2}^\infty |x_k|^p \le \sum_{k=1}^\infty |x_k|^p = \|x\|_{\ell^p}^p$, so $\|L\| \le 1$. Testing on $e_2 = (0, 1, 0, \ldots)$: $Le_2 = e_1 = (1, 0, \ldots)$, $\|Le_2\| = 1 = \|e_2\|$, so $\|L\| = 1$. [/example] ## The Parallelogram Law and Inner Products Not every norm comes from an inner product. A norm arises from an inner product $(x, y) \mapsto (x, y)_H$ via $\|x\| = (x, x)_H^{1/2}$ if and only if it satisfies a special identity. [theorem:Parallelogram Law] A norm $\|\cdot\|$ on a vector space $X$ arises from an inner product if and only if it satisfies the **parallelogram law**: for all $x, y \in X$, \begin{align*} \|x + y\|^2 + \|x - y\|^2 = 2\|x\|^2 + 2\|y\|^2. \end{align*} When this holds, the inner product is recovered by the **polarisation identity**: \begin{align*} (x, y) = \frac{1}{4}\left(\|x+y\|^2 - \|x-y\|^2\right) \quad (\text{real case}). \end{align*} In the complex case: $(x, y) = \frac{1}{4}\sum_{k=0}^3 i^k \|x + i^k y\|^2$. [/theorem] Geometrically, the parallelogram law says that the sum of the squares of the diagonals of a parallelogram equals the sum of the squares of the sides — a property that holds in Euclidean geometry but not in general normed geometry. [example:Failure Of The Parallelogram Law] The $\ell^1$ norm on $\mathbb{R}^2$ does not satisfy the parallelogram law. Take $x = (1, 0)$ and $y = (0, 1)$. Then: \begin{align*} \|x + y\|_1^2 + \|x - y\|_1^2 &= \|(1,1)\|_1^2 + \|(1,-1)\|_1^2 = 2^2 + 2^2 = 8, \\ 2\|x\|_1^2 + 2\|y\|_1^2 &= 2 \cdot 1^2 + 2 \cdot 1^2 = 4. \end{align*} Since $8 \neq 4$, the parallelogram law fails. Therefore $\|\cdot\|_1$ does not arise from any inner product on $\mathbb{R}^2$. The same computation shows that $\|\cdot\|_p$ on $\mathbb{R}^n$ (for $n \ge 2$) arises from an inner product only when $p = 2$. [/example] The failure of the parallelogram law has concrete consequences. In a [Hilbert space](/page/Hilbert%20Space) (complete inner product space), every closed convex set has a unique nearest point (the projection theorem), every bounded linear functional is represented by an inner product ([Riesz Representation Theorem](/theorems/221)), and the spectral theorem decomposes [self-adjoint operators](/page/Self-Adjoint%20Operators). None of these hold in general Banach spaces: the nearest-point projection may fail to exist or be non-unique, the dual space $X^*$ may have no canonical identification with $X$, and the spectral theory is substantially more involved. ## The Place of Normed Spaces in the Hierarchy Normed spaces occupy a specific position in the hierarchy of spaces used in analysis: \begin{align*} \text{Sets} \supset \text{Topological spaces} \supset \text{TVS} \supset \text{LCS} \supset \text{Normed spaces} \supset \text{Banach spaces} \supset \text{Hilbert spaces}, \end{align*} where TVS = [topological vector spaces](/page/Topological%20Vector%20Space) and LCS = locally convex spaces. Moving leftward (toward more general spaces) gains flexibility: $\mathcal{D}(\Omega)$ and $\mathcal{S}(\mathbb{R}^n)$ are LCS but not normed, and their duals ([distributions](/page/Distribution) and [tempered distributions](/page/Tempered%20Distributions)) require the TVS/LCS framework. Moving rightward (toward more special spaces) gains theorems: the Open Mapping Theorem, Closed Graph Theorem, and Uniform Boundedness Principle require completeness (Banach); the projection theorem, Riesz representation, and spectral theorem require inner products (Hilbert). The choice of where to work in this hierarchy is dictated by the problem at hand. PDE existence theory typically works in Hilbert or reflexive Banach spaces (where the Banach-Alaoglu and Eberlein-Šmulian theorems give weak compactness). Harmonic analysis works in $L^p$ spaces for varying $p$ (Banach but not Hilbert for $p \neq 2$). Distribution theory works in locally convex spaces that are not normed. The theory of normed spaces — and especially Banach spaces — is the common foundation underlying all of these. ## Problems [problem] Prove that $\ell^p$ is complete for $1 \le p < \infty$ using the absolute convergence criterion. [/problem] [solution] **Step 1: Setup.** Let $(x^{(k)})_{k=1}^\infty$ be a sequence in $\ell^p$ with $\sum_{k=1}^\infty \|x^{(k)}\|_{\ell^p} < \infty$. Write $x^{(k)} = (x^{(k)}_j)_{j=1}^\infty$ for the components. We must show $S_N = \sum_{k=1}^N x^{(k)}$ converges in $\ell^p$. **Step 2: Pointwise convergence.** For each fixed $j$, $\sum_{k=1}^\infty |x^{(k)}_j| \le \sum_{k=1}^\infty \|x^{(k)}\|_{\ell^p} < \infty$ (since $|x^{(k)}_j| \le \|x^{(k)}\|_{\ell^p}$). So the [series](/page/Series) $s_j := \sum_{k=1}^\infty x^{(k)}_j$ converges absolutely in $\mathbb{C}$ for every $j$. Define $s = (s_j)_{j=1}^\infty$. **Step 3: $s \in \ell^p$.** For any finite $J$, by Minkowski's inequality applied to the partial sums: \begin{align*} \left(\sum_{j=1}^J |s_j|^p\right)^{1/p} = \left(\sum_{j=1}^J \left|\sum_{k=1}^\infty x^{(k)}_j\right|^p\right)^{1/p} \le \sum_{k=1}^\infty \left(\sum_{j=1}^J |x^{(k)}_j|^p\right)^{1/p} \le \sum_{k=1}^\infty \|x^{(k)}\|_{\ell^p} < \infty. \end{align*} Taking $J \to \infty$: $\|s\|_{\ell^p} \le \sum_{k=1}^\infty \|x^{(k)}\|_{\ell^p} < \infty$, so $s \in \ell^p$. **Step 4: $S_N \to s$ in $\ell^p$.** The same Minkowski argument gives \begin{align*} \|s - S_N\|_{\ell^p} = \left\|\sum_{k=N+1}^\infty x^{(k)}\right\|_{\ell^p} \le \sum_{k=N+1}^\infty \|x^{(k)}\|_{\ell^p} \to 0 \end{align*} as $N \to \infty$ (tail of a convergent series). By the absolute convergence criterion, $\ell^p$ is complete. [/solution] [problem] Let $(X, \|\cdot\|)$ be a normed space and $M \subset X$ a finite-dimensional subspace. Prove that $M$ is closed. [/problem] [solution] **Step 1: $M$ is complete.** Since $M$ is finite-dimensional, all norms on $M$ are equivalent. In particular, the restriction of $\|\cdot\|$ to $M$ is equivalent to the Euclidean norm (after choosing a basis and identifying $M \cong \mathbb{R}^n$). Since $\mathbb{R}^n$ is complete in the Euclidean norm, $M$ is complete in $\|\cdot\|$. **Step 2: Complete subspaces of metric spaces are closed.** Let $(x_n) \subset M$ with $x_n \to x$ in $X$. Then $(x_n)$ is Cauchy in $X$, hence Cauchy in $M$ (since $M \subset X$ inherits the metric). By completeness of $M$, there exists $y \in M$ with $x_n \to y$. Since limits in a metric space are unique, $x = y \in M$. [/solution] [problem] Let $T: (\ell^1, \|\cdot\|_{\ell^1}) \to (\ell^\infty, \|\cdot\|_{\ell^\infty})$ be the inclusion map $T(x) = x$. Compute $\|T\|_{\mathcal{L}(\ell^1, \ell^\infty)}$. [/problem] [solution] **Step 1: Upper bound.** For $x = (x_k) \in \ell^1$, $\|Tx\|_{\ell^\infty} = \sup_k |x_k| \le \sum_k |x_k| = \|x\|_{\ell^1}$. Therefore $\|T\| \le 1$. **Step 2: Lower bound.** Test on $e_1 = (1, 0, 0, \ldots)$: $\|Te_1\|_{\ell^\infty} = 1$ and $\|e_1\|_{\ell^1} = 1$, so $\|T\| \ge \|Te_1\|/\|e_1\| = 1$. **Step 3: Conclusion.** $\|T\| = 1$. The inclusion $\ell^1 \hookrightarrow \ell^\infty$ is a contraction (it does not increase norms) and the bound is sharp. [/solution] [problem] Show that on $C([0,1])$, the norms $\|\cdot\|_\infty$ and $\|\cdot\|_{L^2}$ are not equivalent, but the identity map $(C([0,1]), \|\cdot\|_\infty) \to (C([0,1]), \|\cdot\|_{L^2})$ is bounded. Compute its operator norm. [/problem] [solution] **Step 1: One-sided bound.** For $f \in C([0,1])$: \begin{align*} \|f\|_{L^2}^2 = \int_0^1 |f(x)|^2 \, d\mathcal{L}^1(x) \le \|f\|_\infty^2 \int_0^1 d\mathcal{L}^1 = \|f\|_\infty^2, \end{align*} so $\|f\|_{L^2} \le \|f\|_\infty$. The identity map $\mathrm{Id}: (C([0,1]), \|\cdot\|_\infty) \to (C([0,1]), \|\cdot\|_{L^2})$ is bounded with $\|\mathrm{Id}\| \le 1$. Testing on $f \equiv 1$: $\|1\|_{L^2} = 1 = \|1\|_\infty$, so $\|\mathrm{Id}\| = 1$. **Step 2: No reverse bound.** Define $f_n(x) = x^n$. Then $\|f_n\|_\infty = 1$ but $\|f_n\|_{L^2} = (2n+1)^{-1/2} \to 0$. If there were a constant $c > 0$ with $c\|f\|_\infty \le \|f\|_{L^2}$, then $c \le (2n+1)^{-1/2}$ for all $n$, which is impossible. The norms are not equivalent. **Step 3: Interpretation.** The identity map in the *reverse* direction — $(C([0,1]), \|\cdot\|_{L^2}) \to (C([0,1]), \|\cdot\|_\infty)$ — is *not bounded*: the sequence $f_n = \sqrt{2n+1} \cdot x^n$ has $\|f_n\|_{L^2} = 1$ but $\|f_n\|_\infty = \sqrt{2n+1} \to \infty$. The $\|\cdot\|_\infty$ topology is strictly finer than the $\|\cdot\|_{L^2}$ topology on $C([0,1])$. [/solution] ## References - W. Rudin, *Functional Analysis*, 2nd ed. (1991). - J. B. Conway, *A Course in Functional Analysis*, 2nd ed. (1990). - H. Brezis, *Functional Analysis, Sobolev Spaces and Partial Differential Equations* (2011). - E. Kreyszig, *Introductory Functional Analysis with Applications* (1978). - Y. Benyamini and J. Lindenstrauss, *Geometric Nonlinear Functional Analysis, Volume 1* (2000).