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I'm reading up on Dirichlet spaces using this document here and, on page two, am stumped by a particular integral inequality. If this is trivial and/or I'm missing something blatant, I apologize.

First we define, for any analytic function $f$ defined on the unit disk $\mathbb{D}$, the quantity $$ D(f) = \frac{1}{\pi} \int_{\mathbb{D}} | f' |^2 dA = \frac{1}{\pi} \int_{\mathbb{D}} | f' |^2 dxdy $$ (The Dirichlet space is the set of all functions $f$ as before such that $D(f)<\infty$. $D(f)$ is only a semi-norm, as $D(c)=0$ for any constant $c$.) Next we define, for any $\zeta \in \partial \mathbb{D}$, the quantity $$ L(f,\zeta) = \int_0^1 |f'(r \zeta)|dr $$ The author now says

If $D(f)<\infty$ we can apply the Cauchy-Schwarz inequality to show that $$ \int_0^{\pi} L(f,e^{i\theta} ) d\theta \leq c D(f) < \infty $$ and so $L(f,e^{i\theta})<\infty$ for almost all $\theta$.

This is what I can not show.

Attempt: The C-S inequality reads $$ \left| \int h(x) \overline{g(x)} dx \right|^2 \leq \int |h(x)|^2dx \int |g(x)|^2dx $$ and applies because, by above, all functions in the Dirichlet space are square integrable. We are trying to show $$ \int_0^{\pi} \int_0^1 |f'(r e^{i\theta})|dr d\theta \leq c \frac{1}{\pi} \int_{\mathbb{D}} | f' |^2 dxdy $$ which, when we transfer to polars, gives $$ \int_0^{2\pi} \int_0^1 |f'(r e^{i\theta})|dr d\theta \leq A \int_0^{2\pi} \int_0^1 |f'(re^{i\theta})|^2rdrd\theta $$ So I defined $h(r) = |f'(r e^{i\theta})| \sqrt{r}$ to match C-S with what we have on the right, and then attempted so pick $g$ so that the left hand sides would match, too. This gives $g(r) = r^{-1/2}$. Subbing all this in we get $$ \left| \int_0^1 |f'(r e^{i\theta})| \sqrt{r} \frac{1}{\sqrt{r}} dx \right|^2 \leq \int_0^1 |f'(r e^{i\theta})|^2 rdr \int_0^1 |r^{-1/2}|^2dx $$ $$ \left| \int_0^1 |f'(r e^{i\theta})| dx \right|^2 \leq \int_0^1 |f'(r e^{i\theta})|^2 rdr \int_0^1 r^{-1}dr $$ When we integrate over $\theta$ we kind of get what we're looking for (the square shouldn't matter as everything on the right above is finite and all we're trying to show is that the left is finite). The problem is the term $$ \int_0^1 r^{-1}dr = + \infty $$ Any thoughts? Apologies again if this is silly.

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I don't see how such an inequality can hold, for example if we replace $f$ by $\alpha f$, $\alpha>0$. –  Davide Giraudo Jul 19 '12 at 15:07
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up vote 3 down vote accepted

Your approach is right, but you should integrate, say, from 1/2 to 1 (instead of from 0 to 1). The part within radius 1/2 is of no concern anyway since the derivative is bounded on compact subsets.

And @Davide is right - the inequality is false as stated, due to wrong homogeneity.

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Of course, that makes sense! I was thinking too hard about trying to prove it using only the method he suggested. Thanks very much! –  James Fennell Jul 19 '12 at 18:14
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