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Let $X$ be the space of all continuously differentiable functions $f: [0,1] \to \mathbb{R}$ endowed with the norm $\Vert f \Vert = \max_{x \in [0,1]} |f(x)|$. Let $Y$ be the space $C([0,1])$ of all continuous functions $f: [0,1] \to \mathbb{R}$ endowed with the norm $\Vert f \Vert = \max_{x \in [0, 1]} |f(x)|$.

Define a linear mapping $D: X \to Y$ by $D(f) = f'$, where $f'$ is the derivative of $f$.

Prove that $D$ has a closed graph. And why this is not contradicting the closed graph theorem.

The first part of the question is showing that $D$ is unbounded. I have figured it out. Because I plug in the function $t \mapsto t^n$, then $\Vert D \Vert > n$ for any n = 1,2,3,... Thus $\Vert D \Vert = \infty$.

But I don't know how to prove D has a closed graph.

Rondoudou
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Kole
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1 Answers1

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To show $D$ has a closed graph, it is sufficient to show that if $(f_n)_{n\in\Bbb N}\subseteq X$ is a sequence convergent to $f\in X$ and $D(f_n)_{n\in\Bbb N}\to g\in Y$ then $D(f)=g$. However, it is pertinent to note that the topology in $X,Y$ is that of uniform convergence, so we are asking:

If $f_n$ converges uniformly to $f$ and $f_n'$ converges uniformly to $g$, then is $f'=g$?

To which the answer is yes. In fact, $f_n\rightrightarrows f$ is not necessary viz. the post, we only need $f_n(x_0)\to f(x_0)$ at a single point $x_0$. And as per BS Thomson’s comment, we get $f_n\rightrightarrows f$ anyway, which is explained in the post.

FShrike
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  • Note that the link shows that it is enough to assume convergence at a single point, but the conclusion will be that $f_n\to f$ uniformly anyway. – B. S. Thomson Apr 19 '22 at 21:49
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    @B.S.Thomson Yes, it’s probably worth mentioning that in an edit – FShrike Apr 19 '22 at 22:03
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    One more gratuitous comment. The link goes to the correct, most general version of this result. Everyone should know that. But this version has a very strong extra assumption: all the functions ${f_n'}$ are continuous and converge uniformly on $[0,1]$ to a continuous function $g$. So it is much easier to prove. Note that $$f(x)-f(0) = \lim_n [f_n(x)-f_n(0)] = \lim_n \int_0^x f_n'(t),dt =\int_0^x g(t),dt.$$ – B. S. Thomson Apr 19 '22 at 23:06