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Let $H^2(\mathbb{D})$ be the space of all functions $f$ holomorphic on the open unit disk $\mathbb{D}$ such that the Hardy norm, given below, is finite: $$||f||_H^2 = \sup_{0<r<1}\frac{1}{2\pi}\int_0^{2\pi} |f(re^{i\theta})|^2 \ \mathrm{d}\theta.$$

I have already shown that the evaluation $f\mapsto f(z) \ (z\in\mathbb{D})$ is continuous with respect to the norm $||\cdot||_H$.

Now let $||\cdot||$ be any other norm with respect to which $H^2(\mathbb{D})$ is a Banach space, and for which the evaluations $$f\mapsto f(1/(n+1)) \ (n=1,2,...)$$ are continuous. How do I prove that $||\cdot||$ is equivalent to $||\cdot||_H$?

I know that, due to a consequence of the Open Mapping Theorem, one need only show one inequality involving these two norms. Moreover, using the Uniform Boundedness Principle, there is a constant $K$ such that, for all $n\in\mathbb{N}, f\in H^2(\mathbb{D})$, we have $$|f(1/(n+1))|\le K<\infty.$$

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  • $\begingroup$ Your uniform boundedness argument is not correct. $\endgroup$
    – zhw.
    Commented Feb 25, 2017 at 4:23
  • $\begingroup$ I've edited my hint into a more complete solution. $\endgroup$
    – zhw.
    Commented May 15, 2017 at 19:12

1 Answer 1

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The key tools here are the closed graph theorem, the open mapping theorem (for Banach spaces), and the identity principle for holomorphic functions.

So suppose $f_m \to f$ in $(H^2,\|\,\|_H)$ and $f_m \to g$ in $(H^2,\|\,\|).$ Because we know point evaluation is continuous on the first space, we have

$$f_m(1/(n+1)) \to f(1/(n+1)), \, n = 1,2,\dots $$

Now we are given the point evaluations at each $1/(n+1)$ are continuous in the second space, so we also have

$$f_m(1/(n+1)) \to g(1/(n+1)), \, n = 1,2,\dots $$

It follows that $f(1/(n+1))= g(1/(n+1)) \, n = 1,2,\dots.$ But $f,g$ are both holomorphic in $\mathbb D$ and $\{1/(n+1): n = 2,3,\dots \}$ is a set with limit point in $\mathbb D.$ By the identity principle, $f=g$ on $\mathbb D.$

By the closed graph theorem, we have shown the identity map

$$I:(H^2,\|\,\|_H)\to(H^2,\|\,\|)$$

is continuous. Now $I$ is obviously a bijection between these two spaces. In particular it is surjective. Thus $I$ is an open map by the open mapping theorem. Thus $I^{-1} = I$ is also continuous. So $I$ is a Banach space isomorphism between $(H^2,\|\,\|_H)$ and $(H^2,\|\,\|).$ This gives the desired equivalence of norms.

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  • $\begingroup$ But we don't know $I$ is continuous until we prove it. We prove it by using the closed graph theorem and the identity principle. $\endgroup$
    – zhw.
    Commented May 15, 2017 at 20:10
  • $\begingroup$ yeah I just realized it. $\endgroup$
    – Kenneth.K
    Commented May 15, 2017 at 20:11

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