1. ## Inequality of norm

$||f||_1 = (\int\int_D |f(z)|^2dxdy)^{\frac{1}{2}}$, $||f||_2 = (sup_{0\leq r<1}\frac{1}{2\pi} \int_0^{2\pi} |f(re^{i\varphi})|^2dxdy)^{\frac{1}{2}}$. $H_i=\{f \in \mathcal{H}(D) : ||f||_i < \infty\}$. $(D=\{z: |z| <1\})$.

Prove: For $f \in H_2$: $||f||_1 \leq \sqrt{\pi}||f||_2$.

2. Originally Posted by veljko
$||f||_1 = (\int\int_D |f(z)|^2dxdy)^{\frac{1}{2}}$, $||f||_2 = (sup_{0\leq r<1}\frac{1}{2\pi} \int_0^{2\pi} |f(re^{i\varphi})|^2{\color{red}d\varphi})^{\frac{ 1}{2}}$. $H_i=\{f \in \mathcal{H}(D) : ||f||_i < \infty\}$. $(D=\{z: |z| <1\})$.

Prove: For $f \in H_2$: $||f||_1 \leq \sqrt{\pi}||f||_2$.
Notice that the integral defining the second norm is a single integral with respect to $\varphi$, not a double integral in x and y.

Squaring both sides, you are trying to prove that $\iint_D |f(z)|^2dxdy \leqslant \sup_{0\leqslant r<1}\frac1{2\pi}\int_0^{2\pi} |f(re^{i\varphi})|^2d\varphi$.

Make a change of variables from Cartesian to polar coordinates for the first of those integrals, giving $\|f\|_1^2 = \int_0^1\!\!\!\int_0^{2\pi}|f(re^{i\varphi}|^2d\va rphi\,rdr \leqslant \int_0^1\biggl(\sup_{0\leqslant r<1}\int_0^{2\pi} |f(re^{i\varphi})|^2d\varphi\biggr)rdr$. The expression inside the parentheses is independent of r, so $\|f\|_1^2\leqslant \sup_{0\leqslant r<1}\int_0^{2\pi} |f(re^{i\varphi})|^2d\varphi\int_0^1r\,dr = \sup_{0\leqslant r<1}\int_0^{2\pi} |f(re^{i\varphi})|^2d\varphi\Bigl[\tfrac12r^2\Bigr]_0^1 = \pi\|f\|_2^2$. Take square roots to get the desired conclusion.