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Let $f = u+iv$ be a holomorphic function on the unit disk $D\subset \mathbb C$ such that $f(0)=0$. I am trying to prove that for each positive integer $k$, there is a constant $c_k$, independent of $r$, such that for all $r < 1$,

$\int_0 ^{2\pi} |v(re^{i\theta})|^{2k} d\theta \leq c_k \int_0 ^{2\pi} |u(re^{i\theta})|^{2k} d\theta$.

So far, I have tried using the fact that $|u|$ and $|v|$ are subharmonic, but this does not seem to produce the desired estimates. I have also tried some manipulations using the Cauchy-Riemann equations and the mean value property, also to no avail. Any suggestions?

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1 Answer 1

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This is a neat trick using Cauchy's theorem for the function $\frac{f(z)^{2k}}z$, i.e., $$ 0 = \int_{|z|=r} \frac{f(z)^{2k}}z \, dz = \int_0^{2\pi} (u(re^{i\theta})+iv(re^{i\theta}))^{2k} \, d\theta. $$ Then multiply it out, take the real part, throw the integral of $v^{2k}$ on one side and get rid of all the negative terms to get the inequality (where for ease of notation $u=u(re^i\theta)$ and $v=v(re^{i\theta})$) $$ \int_0^{2\pi} v^{2k} \, d\theta < \int_0^{2\pi}\left( { 2k \choose 2} u^{2}v^{2k-2} + {2k \choose 6} u^{6} v^{2k-6} + \ldots \right) \, d\theta. $$ Now for $2 \le j \le 2k-2$, Hölder's inequality with $p=\frac{2k}{j}$ and $q=\frac{2k}{2k-j}$ gives $$ \int_0^{2\pi} u^{j} v^{2k-j} \, d\theta \le \left( \int_0^{2\pi} u^{2k} \, d\theta \right)^{\frac{j}{2k}} \left( \int_0^{2\pi} v^{2k} \, d\theta \right)^{\frac{2k-j}{2k}} $$ Note that this inequality is trivially true if $j=2k$. Writing $$ U = \left( \int_0^{2\pi} u^{2k} \, d\theta \right)^{\frac1{2k}} \quad \text{ and } \quad V = \left( \int_0^{2\pi} v^{2k} \, d\theta \right)^{\frac1{2k}} $$ we get $$ V^{2k} < { 2k \choose 2 } U^{2} V^{2k-2} + { 2k \choose 6 } U^{6} V^{2k-6} + \ldots $$ and dividing both sides by $U^{2k}$ and writing $X=V/U$ we have $$ X^{2k} < { 2k \choose 2 } X^{2k-2} + { 2k \choose 6 } X^{2k-6} + \ldots $$ Now the left-hand side is a polynomial of degree $2k$ in $X$, whereas the right-hand side is a polynomial of degree $2k-2$ in $X$, so if $M_k$ denotes the largest solution of the corresponding polynomial equality, then we know that $X < M_k$. Note that $M_k$ depends only on the coefficients of the polynomial, not on $f$ or $r$. To get the claimed inequality, set $c_k = M_k^{2k}$.

I shamelessly stole this proof from Marcel Riesz's original paper Marcel Riesz, Sur les fonctions conjuguées, Math. Z. 27 (1928), no. 1, 218–244 (The proof appears on page 221.)

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