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 !

  • 2
    $\begingroup$ I have seen this problem posted very recently, can not look for it I am on mobile. Try the closed graph theorem $\endgroup$ – 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 $C([0,1],\|\cdot\|_{\infty})$.

  • $\begingroup$ so simple, thank you ! $\endgroup$ – Malik Mar 12 at 2:51

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