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Let $X$ be compact Hausdorff space. Let $||.||$ be a norm on $C(X)$ which makes it into a Banach space and also assume that the linear functional $\lambda_x$ given by $\lambda_x(f)=f(x)$ is continuous for each $x$. Show that there exist positive constants $A, B$ such that $$A||f||_{\infty}\le ||f|| \le B||f||_{\infty}.$$

I was trying to show that $f\to f$ is a bounded linear map from $(C(X),||.||)$ to $C(X)$ equipped with the sup norm. The map $f\to f$ is clearly one to one and onto, therefore open mapping theorem will give me the desired result.

In order to show the continuity of the above map, I observe that $$|f(x)|=|\lambda_x(f)|\le ||\lambda_x|| ||f||.$$ I would want to supremum over all $x$ in both sides so that I can get the sup norm in LHS. But my issue is that I do not know a uniform bound on Norms of functionals $\lambda_x$. I do not know how to by pass this, I have a feeling that Uniform boundedness principle can be useful to get a uniform bound on the Norms of these functionals but I am unable to carry out the task. Any help in this regard or any other way to attack this problem would be appreciated.

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  • $\begingroup$ Banach-Steinhaus Theorem should solve your problem. $\endgroup$
    – Yesfun Yeh
    Aug 17, 2019 at 17:18
  • $\begingroup$ Ahhh! I see it now. Thanks! I will soon post the solution. $\endgroup$ Aug 18, 2019 at 3:00

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Since OP never posted the solution (as promised in the comments!) I will add the answer:

We consider the identity operator $I:(C(X),\|\cdot\|)\to(C(X),\|\cdot\|_\infty)$ and we want to show that it is bounded. Since both spaces are Banach, we apply the closed graph theorem. Let $(f_n)\subset C(X)$ so that $f_n\to0$ in $\|\cdot\|$. We want to show that $f_n\to0$ in $\|\cdot\|_\infty$ and by the closed graph theorem we can assume that $f_n\to f$ in $\|\cdot\|_\infty$ for some function $f\in C(X)$. All we need to do is show that $f=0$. Let $x\in X$. Then $|f(x)|=|\lambda_x(f)|=\lim_{n\to\infty}|\lambda_x(f_n)|$. But $f_n\to 0$ in $\|\cdot\|$ and $\lambda_x$ is continuous for $\|\cdot\|$, so $\lambda_x(f_n)\to\lambda_x(0)=0$. This shows that $f(x)=0$ and since $x$ was arbitrary we have that $f=0$.

We also want to show that the identity operator $I:(C(X),\|\cdot\|_\infty)\to(C(X),\|\cdot\|)$ is bounded. Again, we apply the closed graph theorem. Let $f_n\to 0$ in $\|\cdot\|_\infty$ and we want to show that $f_n\to0$ in $\|\cdot\|$. By the closed graph theorem we can assume that $f_n\to f$ in $\|\cdot\|$ for some $f\in C(X)$. All we need to do is show that $f=0$. By the previous result we also have that $f_n\to f$ in $\|\cdot\|_\infty$, so since both $0$ and $f$ are uniform limits of $(f_n)$ we have that $f=0$. This concludes the proof.

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  • $\begingroup$ Thanks for adding the answer. I completely forgot it. $\endgroup$ Feb 6, 2021 at 21:50
  • $\begingroup$ @WhoKnowsWho No problem! $\endgroup$ Feb 6, 2021 at 21:50

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