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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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    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.2012-03-06

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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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    @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.2012-03-23
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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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    My bad. I gave a map from a normed space to a Banach space.2012-03-06