# discontinuous Linear operator from a Banach space to a normed vector space

How do I find that example of a discontinuous linear operator A from a Banach space to a normed vector space such that A has a closed graph?

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The energy operator $[Hf](x)=-\frac{1}{2}\frac{d^2}{dx^2} f(x) + \frac{1}{2}x^2f(x)$ is an example. (But, as explained here, it is not defined on the whole domain $L^2(\mathbb R)$. Otherwise, it would contradict the CGT.) – William DeMeo Mar 6 '12 at 6:34
The Closed Graph Theorem is about mappings between Banach spaces. The OP asks about a map from a Banach space into a normed space. – Dirk Mar 6 '12 at 7:47
Let B be some infinite dimensional Banach space with Hamel basis $\{e_\alpha:\alpha\in\mathcal{A}\}$. Fix $\alpha_0\in\mathcal{A}$ and consider linear functional $$f:B\to\mathbb{C}:x\mapsto x_{\alpha_0},$$ where $x_{\alpha_0}$ is a $\alpha_0$-coordinate of $x$ in this basis. – Norbert Mar 6 '12 at 14:49

Take the differentiation operator $D:C^1([0,1]) \to C[0,1]$ given by $Df(x)=f'(x)$ where $C^1([0,1])$ has the norm as a subspace of $C([0,1])$ with the supremum norm.

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Did you mean to equip the domain with the sup norm and the range with the $\|\cdot \|_1$ norm? – William DeMeo Mar 6 '12 at 5:36
What norm is the range? – Jong hee Kim Mar 6 '12 at 6:02
$C^1$ with the sup-norm is not a Banach space. – Dirk Mar 6 '12 at 7:48
My bad. I gave a map from a normed space to a Banach space. – azarel Mar 6 '12 at 14:47

Let $D:=\{z\in\mathbb C, |z|<1\}$ and $E:=\{f\colon D\to\mathbb C, f\mbox{ holomorphic}\}=F$. Let $||f||_E:=\sup_{|z|=2^{—1}}|f(z)|$ and $||f||_F:=\sup_{|z|=2^{—1}}|f'(z)|+|f(0)|$. $E$ is a Banach space, and if we put $A\colon E\to F$ defined by $A(f)=f'$, then $A$ has a closed graph. Indeed, let $\{f_n\}\subset E$ such that $f_n\to f$ in $E$ and $f_n'\to g$ in $F$. Since $f_n''$ converges to $f''$ on $\{z,|z|=2^{-1}\}$ then $f''(z)=g'(z)$ for all $z$ which has modulus $2^{-1}$, and by connectedness of $D$ $f''(z)=g'(z)$ for all $z\in D$, so $f'(z)=g(z)+C$ for some constant $C$. Since $f_n'(0)$ converges to $f'(0)$ and $g(0)$ we get $f'(z)=g(z)$.

But $A$ is not continuous, indeed consider $f_n(z)=2^nz^n$. Then $\lVert f_n\rVert_E=1$ and $f_n'(z)=n2^nz^{n-1}$ so for $n\geq 2$: $\lVert A(f_n)\rVert =2n$.

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This is a bit late, but I would like to ask, how is $E$ a Banach space? Isn't $f_m(z)=\sum_{j=1}^m \left( \frac{4}{3}z\right)^j$ a Cauchy sequence in that norm? – Jose27 Mar 23 '12 at 6:34
@Jose27 You are right, I realize it's not a Banach space (I tried to remember an exercise I've done a long time ago). I think I will be able to find it at my library this afternoon, so I will do the appropriated modification. Thanks for pointing it out. – Davide Giraudo Mar 23 '12 at 9:15