# Showing an linear operator is bounded

Let $$T:L^2([0,1])\to L^2([0,1])$$ be a bounded linear operator such that $$T$$ maps $$C([0,1],||\cdot||_\infty)\to C([0,1],||\cdot||_\infty)$$. Show that $$T$$ is bounded on $$C([0,1],||\cdot||_\infty)$$.

I thought of using the closed graph theorem but I'm a bit confused with how to start. Thanks !

• I have seen this problem posted very recently, can not look for it I am on mobile. Try the closed graph theorem – clark Mar 10 at 18:21

Let $$(f_n)$$ be a sequence in $$C([0,1],\|\cdot\|_{\infty})$$ such that $$(f_n,Tf_n)\to (f,g)$$ in $$C([0,1],\|\cdot\|_{\infty})\times C([0,1],\|\cdot\|_{\infty})$$. Then $$\|f_n-f\|_{\infty}\to 0$$ and $$\|Tf_n-g\|_{\infty}\to 0$$. This implies that $$\|f_n-f\|_2\to 0$$ and $$\|Tf_n-g\|_2\to 0$$.
Since $$T$$ is bounded from $$L^2([0,1])$$ to $$L^2([0,1])$$, $$\|f_n-f\|_2\to 0$$ implies that $$\|Tf_n-Tf\|_2\to 0$$, therefore $$Tf=g$$. Hence, $$(f_n,Tf_n)\to (f,Tf)$$ in $$C([0,1],\|\cdot\|_{\infty})\times C([0,1],\|\cdot\|_{\infty})$$. Then, since $$C([0,1],\|\cdot\|_{\infty})$$ is a Banach space, the closed graph theorem implies that $$T$$ is bounded from $$C([0,1],\|\cdot\|_{\infty})$$ to $$C([0,1],\|\cdot\|_{\infty})$$.