Inequality for holomorphic functions 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?
 A: 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.)
