---- > [!definition]+ Definition. ([[Banach space]]) > A **Banach Space** is a [[norm|normed]] [[vector space]] that is [[complete]] with respect to the [[metric space|metric]] [[metric induced by norm|induced]] by the [[norm]]. > > There are many (nonequivalent) ways to define a [[category]] $\mathsf{Ban}$ whose objects are Banach spaces. Two standard ones are the [[subcategory|full subcategory]] $\mathsf{Ban_{T}}$ of $\mathsf{Norm_{T}}$ and the [[subcategory|full subcategory]] $\mathsf{Ban_{M}}$ [[norm|of]] $\mathsf{Norm_{M}}$.[^8] Note that $\mathsf{Ban_{M}}$ is a [[subcategory]] of $\mathsf{Ban_{T}}$. [^8]: Laid out explicitly, here are the two ways to define $\mathsf{Ban}$: $(1)$ as a [[subcategory]] $\mathsf{Ban_{T}}$ of $\mathsf{Top}$ and $\mathsf{Vect}$, where morphisms are [[continuous]] [[linear map|linear maps]] (whence an [[isomorphism]] is a [[linear map|linear]] [[homeomorphism]]). $(2)$ As a [[subcategory]] $\mathsf{Ban_{M}}$ [[metric space|of]] $\mathsf{Met}$ and $\mathsf{Vect}$, where morphisms are [[Lipschitz continuous|nonexpansive]] [[linear map|linear mappings]] (whence an [[isomorphism]] is a [[surjection|surjective]] ($\iff$ [[bijection|bijective]]) [[bijection|]] [[linear map|linear]] [[Lipschitz continuous|isometry]]). > [!note] Note. > By the equivalence in [[complete]], a **sub-Banach space** would just be a [[closed set|closed]] [[linear subspace]] of Banach space. The term 'sub-Banach space' does not really get used, though, since 'closed linear subspace' explicitly conveys the necessary meaning. ^note > [!equivalence] > The property of [[series|absolute convergence]] implying [[sequence|convergence]] for [[series]] characterizes [[Banach space|Banach spaces]] among [[normed vector space|normed vector spaces]]. > That is: suppose $V$ is a [[norm|normed]] [[vector space]]. Then $V$ is a [[Banach space]] if and only if $\sum_{k=1}^{\infty} g_{k}$ converges for every sequence $g_{1},g_{2},\dots$ in $V$ satisfying $\sum_{k=1}^{\infty} \|g_{k}\|<\infty$. ^equivalence > [!basicexample] > > > - [[Finite-dimensional normed vector spaces are complete|Finite-dimensional normed vector spaces are Banach]] > - $L^{p}(X, \Sigma, \mu)$ for $1 \leq p \leq \infty$ and $(X, \Sigma, \mu)$ a [[measure|measure space]], per [[Lp spaces are Banach]]. > - This includes the standard spaces of infinite [[sequence|sequences]] $\ell^{2}$, $\ell^{1}$, $\ell^{\infty}$ etc. > - $C(K)$ for a [[compact]] [[topological space|space]] $K$ endowed with the [[uniform metric|uniform/sup norm]] $\|\cdot\|_{\infty}$.[^1] > - This includes the collection $C([a,b])$ of [[continuous]] maps $f:[a,b] \to \mathbb{F}$ > > - [[closed set|Closed linear subspaces]] of [[Banach space|Banach spaces]] [[complete|are Banach]]. > - For example, $c_{0} \leq \ell^{\infty}$ [[space of null sequences|is Banach]]. > > [!proof]- Proof. > > > > We show the collection $C([a,b])$ of [[continuous]] maps $f:[a,b] \to \mathbb{F}$, $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$ is a [[Banach space]] under the [[Lp-norm|uniform norm]] $\|f\|_{\infty}:=\sup_{[a,b]} |f|.$ > > Let $(f_{k})$ be a [[Cauchy sequence]] in $C([a,b])$. > > For a fixed $t \in [a,b]$, $|f_{k}(t)-f_{j}(t)|\leq \sup_{t \in [a,b]} |f_{k}(t)-f_{j}(t)|$ for all $k,j \in \mathbb{N}$, from which it follows that the [[sequence]] $(f_{k}(t))$ is [[Cauchy sequence|Cauchy]]. As a [[sequence]] of real numbers, then, $(f_{k}(t))$ [[converge|converges]]. Thus $(f_{k}) \to f$ [[pointwise converge|pointwise]]. We are done once we can show $(f_{k}) \to f$ [[uniform convergence|uniformly]]: recall that $\|\cdot\|_{\infty}$ induces the [[topological space|topology]] [[uniform metric|of uniform convergence]], and the [[uniform limit theorem]] will guarantee the limit $f$ is in $C([a,b])$. So fix $\varepsilon>0$. Choose $N$ with $\|f_{m}-f_{n}\|_{\infty}<\varepsilon$ for all $m,n \geq N$. Fix $n \geq N$. Then for every $t$, $\begin{align} > > |f_{n}(t)-f(t)| &\leq |f_{n}(t)-f_{m}(t)| + |f_{m}(t)-f(t)| \\ > > & \leq \underbrace{ \|f_{n}-f_{m}\|_{\infty} }_{ < \varepsilon } + \underbrace{ |f_{m}(t)-f(t)| }_{ \to 0 \text{ as }m \to \infty \, d\mu }. > > \end{align}$ > > As $m \to \infty$ we obtain $|f_{n}(t)-f(t)| <\varepsilon$ > > as desired. > > > > > [!basicnonexample] > > > > - The [[vector space]] $X=\{ p : [a,b] \to \mathbb{F} : p \in \mathbb{F}[T] \}$, $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$ and $a<b$, of [[polynomial 4|polynomials]] $[a,b] \to \mathbb{F}$ endowed with the [[uniform metric|uniform (sup) norm]] $\|p\|_{\infty}=\sup_{t \in [a,b]} |p(t)|$ is *not* a [[Banach space]]. Here are two arguments why: > > 1. $X$ is [[dimension|infinite-dimensional]], with one [[basis|Hamel basis]] being the [[countably infinite|countable]] set $\{ T^{n}: n \in \mathbb{N} \cup \{ 0 \} \}$. If it were [[complete]], this would contradict [[Hamel basis of infinite-dimensional Banach space is uncountable]]. > 2. $X$ is not [[closed set|closed]] as a [[subspace topology|subspace]] of the [[Banach space]] $C([a,b])$, [[complete|hence]] not [[complete]]. Indeed, $X \neq C([a,b])$ and the [[Weierstrauss Approximation Theorem]] [[dense|says]] $\overline{X}=C([a,b])$. So $X \neq \overline{X}$. > > >- $C([0,1])$ is not complete when endowed with $\|\cdot\|_{2}$ (see [[Banach space]]) > [!note] Remark. > A single underlying [[dimension|infinite-dimensional]] [[vector space]] $X$ can be turned into a [[Banach space]] in multiple, non-equivalent ways by choosing different [[complete]] [[norm|norms]]. In finite dimensions all norms are equivalent, but in infinite dimensions one can have many [[complete]] norms [[metric topology|inducing]] distinct [[topological space|topologies]]. > > Specifically, let $(X,\| \cdot\|)$ be an [[dimension|infinite-dimensional]] [[Banach space]], and let $f:X \to \mathbb{F}$ be an [[operator norm|unbounded]] [[linear functional]]. Given $x_{0} \in X$ with $f(x_{0})=1$, consider the [[linear operator]] $T \in \operatorname{End}X$ defined by $Tx:=x-2f(x)x_{0} \ (x \in X).$Then the map $\begin{align} > \left|\!\left|\!\left| \, \cdot \,\right|\!\right|\!\right|:X &\to [0, \infty) \\ > x & \mapsto \|Tx\| > \end{align}$ > defines a [[complete]] [[norm]] on $X$ which is not [[norm|equivalent]] to $\|\cdot\|$. > > > > [!proof]+ Proof. (FFA Q1.B.2.) > > > > > > > > *$\left|\!\left|\!\left|\, \cdot \,\right|\!\right|\!\right|$ is a norm.* This follows from the fact that $\|\cdot\|$ is a [[norm]] and $T$ is [[linear map|linear]] and [[injection|injective]]. > > > > *Completeness.* [[involution|First note]] $T^2=I$, for given $x \in X$ one has $\begin{align} > > T(Tx) &= T\big( x - 2f(x) x_{0} \big) \\ > > &=\big( x - 2f(x) x_{0} \big) - 2f\big(x - 2f(x) x_{0} \big)x_{0} \\ > > &= \big( x - 2f(x) x_{0} \big) - 2f(x)x_{0} + 4f(f(x)x_{0})x_{0} \\ > > &= x - 2f(x) x_{0} - 2f(x)x_{0} + 4f(x) \cancel{ f(x_{0}) }^{=1}x_{0} \\ > > &= x - 4f(x) x_{0} + 4f(x)x_{0} \\ > > &= x. > > \end{align}$ > > Let $(x_{k})$ be a [[Cauchy sequence]] in $(X, \left|\!\left|\!\left|\, \cdot \,\right|\!\right|\!\right|)$. The identity $\left|\!\left|\!\left|x_{k}-x_{j}\right|\!\right|\!\right|=\|Tx_{k}-Tx_{j}\|$ implies that $(Tx_{k})$ is then a [[Cauchy sequence]] in $(X, \|\cdot\|)$. Since $(X, \|\cdot\|)$ is [[complete]], $(Tx_{k})$ [[converge|converges]] in $(X, \|\cdot\|)$, say, to $L \in X$. Note that, since $T^{2}=I$, we have $\left|\!\left|\!\left|Tx\right|\!\right|\!\right|=\|x\|$ for all $x \in X$. From this it follows that $\|Tx_{k}-L\|= \left|\!\left|\!\left|T(Tx_{k}-L)\right|\!\right|\!\right|=\left|\!\left|\!\left|x_{k}-T(L)\right|\!\right|\!\right|.$ > > Thus $(x_{k}) \to T(L)$ in $(X, \left|\!\left|\!\left|\, \cdot \,\right|\!\right|\!\right|)$. > > > > > > *Nonequivalence.* Note that $T$ is unbounded, since $f$ is unbounded. Let $C>0$. Then we cannot have $\left|\!\left|\!\left|\,\cdot\,\right|\!\right|\!\right| \leq C \|\cdot\|,$ > > since the [[operator norm|unboundedness]] of $T$ wrt $\|\cdot\|$ means there exists $x \in X$ such that $\left|\!\left|\!\left|x\right|\!\right|\!\right|= \|Tx\|> C \|x\|$. > > ^remark [^1]: The [[extreme value theorem]] guarantees that $\|\cdot\|_{\infty}$ is finite-valued (hence indeed a [[norm]]) on $C(K)$. > [!proof] Proof of Equivalence. > Suppose $V$ is a [[Banach space]], and $g_{1},g_{2},\dots$ is a [[sequence]] in $V$ satisfying $\sum_{k=1}^{\infty}\|g_{k}\|<\infty$. Since $V$ is Banach, it is enough to show that the sequence $s_{1},s_{2},\dots$ of partial sums, i.e. $s_{j}=g_{1}+\dots+g_j$, is [[Cauchy sequence|Cauchy]]. Fix $\varepsilon>0$. Let $N \in \mathbb{N}$ be such that $\sum_{m=N}^{\infty} \|g_{m}\|<\varepsilon$. Then for all $n_{1},n_{2} \geq N$ we have > $\|s_{n_{1}}-s_{n_{2}}\|= \| g_{n}+\dots+g_{m} \| \leq \sum_{m=N}^{\infty} \|g_{m}\|<\varepsilon,$ > as desired. > > Conversely, suppose $\sum_{k=1}^{\infty} g_{k}$ converges for every sequence $g_{1},g_{2},\dots$ in $V$ satisfying $\sum_{k=1}^{\infty}\|g_{k}\|<\infty$. We are then done if we can show that every [[Cauchy sequence]] in $V$ converges absolutely. So suppose $f_{1},f_{2},\dots$ is a [[Cauchy sequence]] in $V$. [[Cauchy sequence is convergent iff has convergent subsequence|It suffices to show that some]] [[subsequence]] of $f_{1},f_{2},\dots$ [[sequence|converges]]. Dropping to a subsequence (but not relabeling), and setting $f_{0}=0$, we can assume (see the properties in [[Cauchy sequence]]) $\sum_{k=1}^{\infty}\|f_{k}-f_{k-1}\|<\infty ;$ > then the assumption says that the [[telescoping series]] $\sum_{k=1}^{\infty}(f_{k}-f_{k-1})$ > converges. Of course, the partial sums in this telescoping series are precisely $f_{1},f_{2},\dots$, and this completes the proof. > > > ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` #reformatreviseBbatch [^2]: