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Observe $C[0,1]$ and for $1\leq p<\infty$ the norm $\|f\|_p=\left(\int_0^1 |f(t)|^p\, dt\right)^{1/p}$. Let $T: C[0,1]\to C[0,1]$ be an arbitrary linear operator. Show, that when it exists a $1\leq p<\infty$ such that $T: (C[0,1],\|\cdot\|_p)\to (C[0,1],\|\cdot\|_p)$ is continuous, then is $T: (C[0,1],\|\cdot\|_\infty)\to (C[0,1],\|\cdot\|_\infty)$ continuous.

I do not really know how to start here. Thanks in advance for any help.

Cornman
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  • The closed graph theorem should do the trick. – gerw Dec 18 '17 at 14:38
  • So I have to show, that $G(T)$ is closed in $(C[0,1], C[0,1])$ both with the $|\cdot|_\infty$ norm? – Cornman Dec 18 '17 at 14:52
  • Yes, this is what you have to show. – gerw Dec 18 '17 at 19:13
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    Do you mind giving a full answer, or a guide, on how to do this? I do not know how I can show that $(C[0,1], C[0,1])$ is closed. A full answer will be rewarded with 200 reputations as a bounty. – Cornman Dec 18 '17 at 22:28

1 Answers1

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For convenience, we set $C_\infty := (C([0,1]), \|\cdot\|_\infty)$ and $C_p := (C([0,1]), \|\cdot\|_p)$. Since $C_\infty$ is a Banach space, we can apply the closed graph theorem to show the boundedness of $T : C_\infty \to C_\infty$.

Therefore, given a sequence $\{f_n\} \subset C([0,1])$ with $f_n \to f$ in $C_\infty$ and $T f_n \to g$ in $C_\infty$, we have to show that $g = Tf$. Since the measure of $[0,1]$ is finite, $f_n \to f$ in $C_\infty$ implies $f_n \to f$ in $C_p$. The continuity of $T$ in $C_p$ gives $T f_n \to T f$ in $C_p$. Similarly, $T f_n \to g$ in $C_\infty$ implies $T f_n \to g$ in $C_p$. Since the limit in $C_p$ is unique, this gives $T f = g$.

Hence, the graph of $T$ is closed in $C_\infty \times C_\infty$ and $T$ is continuous by the closed graph theorem.

gerw
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