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If $X$ be a closed subspace of $L^2([0,1])$ and each element in $X$ belongs to $L^{\infty}([0,1])$, So $X$ is closed in $L^{\infty}([0,1])$ and $\lVert f \rVert_2 \leq \lVert f \rVert_{\infty} $. Therefore X is Banach space. If I define a map $T: X \rightarrow L^{\infty}([0,1]$, I have to show that this is closed. I am not sure that it is continuous and I was trying to use the definition of closed operator but the problem is I have two different norms.I don't know how to use the definition of closed operators in case of two different norms.

Tien
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  • Do You know the closed graph theorem? it implies that the graph of a continuous operator $T$ is closed in the product $X\times L^{\infty}([0,1])$ equipped with the product topology, induced e.g. by the norm $||.||2+||.||{\infty}$. – Peter Melech Jun 22 '17 at 10:00
  • Yes,I want to use closed graph theorem to prove T bounded but first I have to show it is closed. – Tien Jun 22 '17 at 10:06
  • What is Your definition of closed if not that the graph is closed? – Peter Melech Jun 22 '17 at 10:07
  • I edit the question. If I am not sure about continuity then I can't use closed graph theorem directly – Tien Jun 22 '17 at 10:09
  • If $T$ is not continuous it is not closed, again by the closed graph theorem. – Peter Melech Jun 22 '17 at 10:12
  • Thats the problem How can I show the continuity ? – Tien Jun 22 '17 at 10:16
  • Is$T$ defined by $Tf=f$ ? – Peter Melech Jun 22 '17 at 10:18
  • I don't know how to define T as there are two different norms.The only hint i have is takeT as a identity map of X into $L^{\infty}( [0,1])$. – Tien Jun 22 '17 at 10:28
  • That´s exactly $Tf=f$ and You have $X$ closed in $L^{\infty}([0,1])$ and thus the graph of $T$ namely $X\times X$ is closed in $X\times L^{\infty}([0,1])$ equipped with the norm $||.||2+||.||{\infty}$ and so by the closed graph theorem $T$ is continuous – Peter Melech Jun 22 '17 at 10:31
  • sorry i meant $graph(T)={(x,x).x\in X}$ closed in ... – Peter Melech Jun 22 '17 at 10:46
  • You mean if T: $X \rightarrow L^{\infty}([0,1])$ is an operator such that the graph $G(T)={(x,x) \in X\times X \mid Tx=x}$ is closed in $ X \times L^{\infty}([0,1]$. – Tien Jun 22 '17 at 10:46
  • Yes because $X$ is closed wrt $||.||_2$ and by assumption closed in $L^{\infty}$ too – Peter Melech Jun 22 '17 at 10:48
  • But don't you think first i have to show that $G(T)={ (x,x) \in X \times X \mid Tx=x}$ is closed and then i can relate this with closed in $X \times L^{\infty}$. – Tien Jun 22 '17 at 10:56
  • since You know $X$ is closed wrt $||.||2$ and $||.||{\infty}$ it follows immediately that the diagonnal $G(T)={(x,x):x\in X}$ is closed wrt $||.||2+||.||{\infty}$ – Peter Melech Jun 22 '17 at 11:04
  • OK not immediately, see my answer – Peter Melech Jun 22 '17 at 11:26

1 Answers1

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Consider a sequence $\{(x_n,x_n)\}$ in $G(T)$ convergent to $(x,y)$ i.e. $||(x_n,x_n)-(x,y)||=||x_n-x||_2+||x_n-y||_{\infty}\rightarrow 0,n\rightarrow \infty$. It follows $||x_n-x||_2\rightarrow 0$ and $||x_n-y||_{\infty}\rightarrow 0$ for $n\rightarrow \infty$. You get $x,y\in X$ and

$||x-y||_2\leq ||x-x_n||_2+||x_n-x_m||_2+||x_m-y||_2\leq||x-x_n||_2+||x_n-x_m||_2+||x_m-y||_{\infty}\rightarrow 0,n,m\rightarrow \infty$

hence $x=y$ and thus $(x,y)\in G(T)$

Peter Melech
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