Given a bounded linear operator $T:L^2([0,1])\to L^2([0,1])$ which maps $C^0([0,1])$ functions to $C^0([0,1])$ functions, I'm trying to show that $T$ is bounded as an operator on $C^0([0,1])$ in the sup norm $||f||_{\infty} = \sup_{x\in [0,1]} |f(x)|$.
Clearly if $f\in C^0([0,1])$, then $||f||_{\infty}$ exists, is finite, and is actually attained at some point in the interval by $f$ (as $f$ will be continuous on a closed interval). Similar for $||Tf||_{\infty}$. However, I'm not sure how I can use the above and the fact that
$$||Tf||_2 \leq C||f||_2, \quad C \in\mathbb{R}^+$$
to show that
$$||Tf||_{\infty} \leq \tilde{C}||f||_{\infty}, \quad \tilde{C} \in \mathbb{R}^+$$
Does anyone have any advice?
Best Answer
Let $\{f_k\}$ be a sequence of functions in $C^0([0,1])$ which converges in the uniform norm to a function $f$. By hypothesis each $Tf_k$ belongs to $C^0([0,1])$. Assume in addition that $Tf_k$ converges in the uniform norm to a function $g$.
Since $$\|Tf_k - Tf\|_2 \le C \|f_k - f\|_2 \le C \|f_k - f\|_\infty$$ you have $Tf_k \to Tf$ in $L^2([0,1])$. This leads to the following situation:
By hypothesis $Tf$ is continuous so you conclude that $Tf = g$ in $C^0([0,1])$.
The boundedness of $T$ follows from the closed graph theorem.