---- > [!theorem] Theorem. ([[closed graph theorem]]) > Suppose $V,W$ are [[Banach space|Banach spaces]] and $T:V \to W$ is a function. Then $T$ is a [[operator norm|bounded]] [[linear map|linear map]] if and only if it is [[closed linear operator|closed]] (that is, its [[graph]] is a [[closed set|closed]] [[linear subspace]] of the [[direct sum of Banach spaces|Banach direct sum]] $V \oplus W$ with the [[product topology]]). ^theorem > [!note] Remark. > This result is one of many in mathematics characterizing function [[continuous|continuity]] ([[characterizing continuity of linear maps|see]]) or some other notion of regularity in terms of [[graph|graphs]]. Another example is [[continuous function graphs are closed for Hausdorff codomains]]. Terence Tao has a bunch more: https://terrytao.wordpress.com/2012/11/20/the-closed-graph-theorem-in-various-categories/. ^note > [!basicnonexample] Remark. > Neither [[complete|completeness]] of $X$ nor $Y$ can be dropped as an assumption of the [[closed graph theorem]]. > > **The completeness assumption on $X$ can't be dropped.** > - Let $X$ be the [[polynomial 4|polynomial space]] $X=\{ p : [0,1] \to \mathbb{R} : p \in \mathbb{R}[T] \} \subset C([0,1])$ with the [[uniform metric|uniform (sup) norm]]. [[Banach space|We know]] $X$ is not [[complete]]. > - Let $Y$ be $C([0,1])$. [[Banach space|We know]] $Y$ *is* [[complete]]. > > Then the [[derivative|differentiation operator]] $(p \in X) \mapsto (p' \in Y)$ is [[closed linear operator|closed]] but not [[operator norm|bounded]] as a [[linear map|linear map]]. > > > > [!proof]- Proof. > > It is unbounded as witnessed by the collection $(p_{n}(x)=x^{n})_{n\in \mathbb{N}}$ which satisfy $\|p_{n}\|_{\infty}=1$ with $|p_{n}'(x) |= n|p^{n-1}(x) |$ so that $\|p_{n}'\|_{\infty}=n$. However, the differentiation operator *is* [[closed linear operator|closed]]. To see this, we'll show $\text{Gr}:= \{ (p,p'): p:[0,1] \to \mathbb{R} \text{ a polynomial} \}$is [[closure is set together with limit points|sequentially closed]]. So let $(p_{n}, p_{n}')_{n \in \mathbb{N}}$ be a [[sequence]] in $\text{Gr}$ [[converge|converging]] to some $(f,g) \in X \times Y$.[^1] We have to show and $g=f'$. [[direct sum of Banach spaces|We know]] $(p_{n}) \to f$ [[uniform convergence|uniformly]] and $(p_{n}') \to g$ [[uniform convergence|uniformly]] ([[uniform topology|recall]]). > > > > By [[The Fundamental Theorem of Calculus]], > > > > $p_{n}-p_{n}(0)=\int _{0}^{\cdot} p_{n}' \, .$ > > Taking $n \to \infty$ on both sides gives $f - f(0)= \lim_{n \to \infty} \int _{0}^{\cdot} p_{n}'= \int_{0}^{\cdot} g, $ > > where we have used uniform convergence to interchange the limit and integral (via [[Dominated Convergence Theorem|DCT]]). Then [[derivative|differentiating]] yields $f'=g,$ > > as required. > > > **The completeness assumption on $Y$ can't be dropped.** > - Let $(X, \|\cdot\|)$ be an [[dimension|infinite-dimensional]] [[norm|normed]] [[vector space|space]], and suppose $\{ x_{\alpha} \}_{\alpha \in A}$ is a [[basis|Hamel basis]] for $X$ and that $\|x_{\alpha}\|=1$ for all $\alpha \in A$. We define a new norm $\left|\!\left|\!\left|\,\cdot\,\right|\!\right|\!\right|$ on $X$ as $\left|\!\left|\!\left|x\right|\!\right|\!\right|=\sum_{\alpha \in A} |\lambda_{\alpha}| \text{ for } x=\sum_{\alpha \in A}\lambda_{\alpha} x_{\alpha}.$ > > Then $(X, \left|\!\left|\!\left|\, \cdot \,\right|\!\right|\!\right|)$ is not [[separable space|separable]] when $(X, \|\cdot\|)$ is a [[Banach space]], and from this we can construct examples for which the [[identity map]] $\begin{align} > T:(X, \|\cdot\|) &\to (X, \left|\!\left|\!\left| \, \cdot \, \right|\!\right|\!\right|) \\ > x & \mapsto x > \end{align}$ > is [[closed linear operator|closed]] but not [[operator norm|bounded]]. > > > [!proof]- Proof. > > *Verification that $\left|\!\left|\!\left|\, \cdot \, \right|\!\right|\!\right|$ is a [[norm]].* > > > > Let $x, y \in X$; $x=\sum_{\alpha \in A} \lambda_{\alpha} x_{\alpha}$, $y=\sum_{\alpha \in A} \xi_{\alpha}x_{\alpha}$. Let $c>0$. > > > > *(Triangle inequality)* We have $\begin{align} > > \left|\!\left|\!\left|x+y\right|\!\right|\!\right| &= \left|\!\left|\!\left| \sum_{\alpha \in A} (\lambda_{\alpha}+\xi_{\alpha}) x_{\alpha} \right|\!\right|\!\right| \\ > > &= \sum_{\alpha \in A} |\lambda_{\alpha}+ \xi_{\alpha}| \\ > > & \leq \sum_{\alpha \in A} |\lambda_{\alpha}| + |\xi_{\alpha}| \\ > > &= \sum_{\alpha \in A} |\lambda_{\alpha}| + \sum_{\alpha \in A} |\xi_{\alpha}| \\ > > &= \left|\!\left|\!\left|x\right|\!\right|\!\right|+\left|\!\left|\!\left|y\right|\!\right|\!\right|. > > \end{align}$ > > > > > > *(Homogeneity)* We have $\begin{align} > > \left|\!\left|\!\left|c x\right|\!\right|\!\right|&= \sum_{\alpha \in A} |c \lambda_{\alpha}| \\ > > &= |c| \sum_{\alpha \in A} |\lambda_{\alpha}| \\ > > &= | c| \left|\!\left|\!\left|x\right|\!\right|\!\right|. > > \end{align}$ > > > > > > *(Positive definiteness)* Clearly $\left|\!\left|\!\left|x\right|\!\right|\!\right|=0$ whenever $x=0$. Conversely, suppose $\left|\!\left|\!\left|x\right|\!\right|\!\right|=0$, i.e., $\sum_{\alpha \in A} |\lambda_{\alpha}|=0.$ A sum of nonnegative numbers equals zero iff each summand is zero; thus $x=0$. > > > > > > > > *$(X, \left|\!\left|\!\left|\, \cdot \,\right|\!\right|\!\right|)$ is not [[separable space|separable]] when $(X, \|\cdot\|)$ is a [[Banach space]].* > > > > > > ~~Consider $(X, \|\cdot\|)=\ell^{\infty}$. Suppose $(X, \left|\!\left|\!\left|\,\cdot\,\right|\!\right|\!\right|)$ is [[separable space|separable]]: there exists a [[sequence]] $(x_{n})$ such that for all $x \in X$ and $r>0$, $\left|\!\left|\!\left|x-x_{n}\right|\!\right|\!\right|<r$ for some $n \in \mathbb{N}$. Since $\|\cdot\| \leq \left|\!\left|\!\left|\, \cdot \,\right|\!\right|\!\right|$, this implies $(X, \|\cdot\|)$ is [[separable space|separable]]. But $\ell^{\infty}$ is not separable; this is a contradiction. ~~ > > > > By [[Hamel basis of infinite-dimensional Banach space is uncountable]], $\{ x_{\alpha} \}$ is [[uncountably infinite|uncountable]]. Observe $\left|\!\left|\!\left|x_{\alpha}-x_{\beta }\right|\!\right|\!\right|=2$ for $\alpha \neq \beta$. Thus $\{ x_{\alpha} \}$ is a $2$-[[separable space|separated]] subset of $(X, \left|\!\left|\!\left|\, \cdot \,\right|\!\right|\!\right|)$. Recall that [[separable space|in a separable metric space, every]] $r$-[[separable space|separated subset is countable]]. Hence $(X, \left|\!\left|\!\left|\, \cdot \,\right|\!\right|\!\right|)$ must not be separable. > > > > > > > > > > > > *Construction of a closed unbounded linear operator $X \to Y$ when only $X$ is Banach.* > > > > First that for all $\|x\| =\|\sum_{\alpha \in A}\lambda_{\alpha} x_{\alpha}\|\leq \sum_{\alpha \in A} |\lambda_{\alpha}|\overbrace{ \|x_{\alpha}\| }^{ =1 }=\left|\!\left|\!\left|x\right|\!\right|\!\right|.$ > > Thus $\|\cdot\|\leq \left|\!\left|\!\left|\cdot\right|\!\right|\!\right|$. > > > > > > Consider the [[identity map]] $\begin{align} > > T:(X, \|\cdot\|) &\to (X, \left|\!\left|\!\left| \, \cdot \, \right|\!\right|\!\right|) \\ > > x & \mapsto x > > \end{align}.$ > > *$T$ is closed.* The [[graph]] of $T$ is the [[diagonal]] $\text{Gr }T= \{ (x ,x ): x \in X\}$ endowed with the [[direct sum of Banach spaces|norm]] $\max \{ \|\cdot\|, \left|\!\left|\!\left|\, \cdot \,\right|\!\right|\!\right| \}$. Let $(x_1, x_{1}), (x_{2}, x_{2}),\dots$ be a [[sequence]] in $\text{Gr }T$ converging to some $(x_{\infty}, x_{\infty}') \in X \times X$. We have to show $x_{\infty}=x_{\infty}'$. Fix $\varepsilon>0$. Choose $N$ large enough that $\left|\!\left|\!\left|x_{n}-x_{\infty}'\right|\!\right|\!\right|<\varepsilon$ for all $n \geq N$. Then also $\|x_{n}-x_{\infty}'\|<\varepsilon$ for all $n \geq N$ (recalling $\|\cdot\| \leq \left|\!\left|\!\left| \, \cdot \,\right|\!\right|\!\right|$), meaning $x_{n} \to x_{\infty}'$ wrt $\|\cdot\|$. Since limits in a given [[metric space]] are unique and $x_{n} \to x_{\infty}$ wrt $\|\cdot\|$ by assumption, we conclude $x_{\infty}=x_{\infty}'$ as desired. > > > > *$T$ is unbounded.* Let $(X,\|\cdot\|)$ be [[separable space|separable]], say, $(X, \|\cdot\|)=\ell^{2}$. If $T$ were bounded, then there would exist $C \in [0, \infty)$ such that $\left|\!\left|\!\left|x\right|\!\right|\!\right|\leq C \|x\|$ for all $x \in X$. Combining with our earlier inequality, this gives $\|\cdot\| \leq \left|\!\left|\!\left|\, \cdot \, \right|\!\right|\!\right| \leq C \|\cdot\|$, showing the two norms to be [[norm|equivalent]]. Equivalent norms induce the same topology and separability is a topological invariant, thus $(X, \left|\!\left|\!\left|\, \cdot \, \right|\!\right|\!\right|)$ must be separable. But part (b) showed this is not the case. > > ^proof-2 > > > [!proof]- Proof. ([[closed graph theorem]]) > This is a corollary of the [[open mapping theorem]] (or more precisely the [[open mapping theorem|bounded inverse theorem]]; in [[the principle results of functional analysis (PROFA)]] it is shown that these three results are all in fact equivalent). > > First suppose $T$ is a [[operator norm|bounded linear map]]. Suppose $(f_{1},Tf_{1}), (f_{2},Tf_{2})$ is a [[sequence]] in $\text{graph}(T)$ [[sequence|converging]] to $(f,g) \in V \times W$. Thus $\lim f_{k}=f, \lim Tf_{k}=g.$ > Since $T$ [[the sequential continuity lemma|is]] [[continuous]], $\lim Tf_{k}=Tf$, hence $g=Tf$. Thus $(f,g) \in \text{graph}(T)$. [[limit point|Thus]] $\text{graph}(T)$ is [[closed set|closed]]. > > Conversely, suppose $\text{graph}(T)$ is a [[closed set|closed]] [[linear subspace]] of $V \times W$. Thus $\text{graph}(T)$ is a [[Banach space]] with the [[subspace topology|with the]] [[norm]] inherited from $V \times W$ (by [[complete|closed subspace of complete metric space is complete]]). Let $\pi:\text{graph}(T) \to V$ denote the [[projection]] $(f, Tf) \mapsto f$, note that $\pi$ is [[operator norm|bounded]] since $\|\pi (f, Tf)\|=\|f\| \leq \max \{ \|f\|, \|Tf\| \}=\|(f, Tf)\|$ > for all $f \in V$. $\pi$ has [[inverse map]] given by $f \xmapsto{\iota} (f,Tf)$; by the [[open mapping theorem|bounded inverse theorem]] we know $\iota=\pi ^{-1}$ is a [[operator norm|bounded linear map]]. We then have $\begin{align} > \|Tf\| &\leq \max \{ \|f\|, \|Tf\| \} \\ > &= \|(f, Tf)\| \\ > &= \|\iota f\| \\ > & \leq \|\iota \| \| f\| > \end{align}$ > for all $f \in V$. This implies $\|T\| \leq \|\iota\|<\infty$, from which it follows that $\|T\|<\infty$ and so $T$ is bounded. ---- #### [^1]: That $(p_{n})$ is assumed to converge to a polynomial doesn't contradict the [[Weierstrauss Approximation Theorem]] because the limit is taken *in $X$*. ----- #### 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 > ```