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Let $T: D(T) \to L^p(0,1)$ for $D(T) = C[0,1] \subset L^p(0,1)$ and $p \in [1,\infty)$ be defined by

$(Tf)(x) := f(0) x$ for $f \in D(T)$ and $x \in (0,1)$.

Problem: We need to show that $T$ is neither closed nor closable.

My ideas: If I could show that $T$ is not closable then it would be automatically closed. But this seemed to be a bit difficult at first, so I want to try to proove that $T$ is not closed.

This must mean we have to find a sequence $(f_n)_n \subset C[0,1]$ such that $$\lim_{n \to \infty} (f_n, Tf_n) \not\in G(T)$$ where $G(T)$ denotes the graph of $T$. Let us say $f_n \to f$ for $n \to \infty$. So we want to get

$$ f(0)x = Tf(x) \neq g(x) := \lim_{n \to \infty} Tf_n(x) = \lim_{n \to \infty} f_n(0) x $$ for at least an $x \in (0,1)$. But aren't the left and the right hand side the same and I can not see why they should be different?

Could you explain what I am doing wrong? Thank you!

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    $\begingroup$ You have to keep in mind, that $(f_n, Tf_n)$ just has to converge in $L^p$, not pointwise. Try to find a sequence of function converging to the zero function in $L^p$, but $f_n(0)=1$ for all $n$. $\endgroup$ – Severin Schraven Dec 17 '17 at 19:13
  • $\begingroup$ Thanks for the remark, that was a stupid mistake! Can I choose $f_n(x) = (1-nx) \mathbb{1}_{[0,\frac{1}{n}]}$? This should suffice the conditions that you mentioned. $\endgroup$ – Diglett Dec 17 '17 at 20:02
  • $\begingroup$ Indeed, this works. $\endgroup$ – Severin Schraven Dec 18 '17 at 18:54

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