# closed operator in Hilbert spaces

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.

• 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

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 $(x,y)\in G(T)$