Harmonic functions and Holder's inequality In the book Real and Complex Analysis by Rudin, it is given that by applying Holder's inequality to the $u(re^{i\theta})=\frac{1}{2\pi}\int_{-\pi}^\pi 
P_r(\theta-t)f(t)dt$ we get $|u(re^{i\theta})|^p\leq\frac{1}{2\pi}\int_{-\pi}^\pi 
P_r(\theta-t)|f(t)|^pdt$. I am unable to fill in the details. Any help would be appreciable. 
 A: Assume $1/p+1/q=1$ for $1 < p < \infty$. Because $P$ is positive and $\frac{1}{2\pi}P_{r}(\theta-t)$ integrates to $1$ in $t$,
$$
        \left|\int_{0}^{2\pi}\frac{1}{2\pi}P_{r}(\theta-t)f(t)\,dt\right|
    \le \int_{0}^{2\pi}\frac{1}{2\pi}P_{r}(\theta-t)|f(t)|\,dt \\
        \le \int_{0}^{2\pi}\left(\frac{1}{2\pi}P_{r}(\theta-t)\right)^{1/q}
                 \left[\left(\frac{1}{2\pi}P_{r}(\theta-t)\right)^{1/p}|f(t)|\right]\,dt \\
    \le \left(\int_{0}^{2\pi}\frac{1}{2\pi}P_{r}(\theta-t)\,dt\right)^{1/q}
        \left(\int_{0}^{2\pi}\frac{1}{2\pi}P_{r}(\theta-t)|f(t)|^{p}\right)^{1/p}\\
          = \left(\int_{0}^{2\pi}\frac{1}{2\pi}P_{r}(\theta-t)|f(t)|^{p}\right)^{1/p}
$$
Applying a power of $p$ then gives what you want.
A: Another (related) way to proceed is to use that
$$d\mu = \frac{1}{2\pi} P_r(\theta - t) \, dt$$
is a probability measure (again since $P$ is positive and integrates to $1$ with respect to $dt$). Hence Jensen's inequality implies for $p \geq 1$
$$|u(re^{i\theta})|^p \le \left( \int_{-\pi}^\pi \! |f(t)| \, d\mu(t) \right)^p \le \int_{-\pi}^\pi \! |f(t)|^p \, d\mu(t)$$
which is what you want.
