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:
$$||f||_H^2 = \sup_{0 I have already shown that the evaluation $f\mapsto f(z) \ (z\in\mathbb{D})$ is continuous with respect to the norm $||\cdot||_H$. Now let $||\cdot||$ be any other norm with respect to which $H^2(\mathbb{D})$ is a Banach space, and for which the evaluations $$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.$$
Prove the equivalence of norms on the Hardy space $H^2(\mathbb{D})$.
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0Your uniform boundedness argument is not correct. – 2017-02-25
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0I've edited my hint into a more complete solution. – 2017-05-15
1 Answers
The key tools here are the closed graph theorem, the open mapping theorem (for Banach spaces), and the identity principle for holomorphic functions.
So suppose $f_m \to f$ in $(H^2,\|\,\|_H)$ and $f_m \to g$ in $(H^2,\|\,\|).$ Because we know point evaluation is continuous on the first space, we have
$$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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0But we don't know $I$ is continuous until we prove it. We prove it by using the closed graph theorem and the identity principle. – 2017-05-15
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0yeah I just realized it. – 2017-05-15