# Norm $||.||$ on $C(X)$ is equivalent to $||.||_{\infty}$ norm if evaluation linear functionals on $(C(X),||.||)$ is continuous.

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.

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

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.