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Got stuck in this problem:

Let us denote by $X$ the linear space $C^1([0,1])$ equipped with the norm of $C^0([0,1])$ and consider the following statement:

"The differentiation operator $L:X \rightarrow C^0([0,1]): u \mapsto u'$ is linear and closed. By the Closed Graph Theorem, the operator is then continuous"

Show that the conclusion is wrong and find the mistake in the argument

To show that the conclusion is wrong it should be enough to show that the operator is not bounded (as it is obviously linear and, as it is linear, it is continuous iff it is bounded), which is not hard.

So I believe the mistake is that $L$ is not closed, but how can I prove it? I was trying to find an example of a closed set in $X$ whose image is not closed but I'm not sure whether this is a good approach...

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Manlio
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    You want to say "$L$ has closed graph", not "$L$ is closed". But this isn't the real mistake. The mistake is that $X$ is not a Banach space. – David Mitra Jan 02 '15 at 11:35
  • Sorry, it seems "closed operator" indeed means "has closed graph". – David Mitra Jan 02 '15 at 11:55
  • @DavidMitra, in my courses and notes, I have tried to remember to write "graph-closed" rather than just "closed", because the latter is not explanatory... :) – paul garrett Jun 11 '20 at 20:04

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$L$ is closed; that is, $L$ has closed graph: If $u_n$ converges uniformly to $u$ and $Lu_n=u_n'$ converges uniformly to $f$, it follows that $Lu=u'=f$. See this, for example.

The mistake in the argument is that you can't appeal to the Closed Graph Theorem, as the (needed) hypotheses that the domain space be complete is not satisfied by your $X$. (For instance, there is a sequence of polynomials converging uniformly to the function $x\mapsto |x-1/2|$ on $[0,1]$ by the Weierstrass Approximation Theorem.)

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David Mitra
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    Oh yes, thanks! I was stuck because I believed that the hypotheses that X is a Banach space was not mandatory. Looking on the Bathia's textbook (Notes on functional analysis) it looks like the differentiation is the typical example of a closed operator which is not continuous and the problem is exactly the non-closedness of the domain. Thanks. – Manlio Jan 02 '15 at 15:40