# Proof of a discrete isoperimetric inequality

The following inequality appears in the proof of certain isoperimetric-type inequalities for analytic functions in two dimensions:

$$\sum_{m=0}^{\infty}\frac{|c_m|^2}{m+1} \leq \pi \left(\sum_{m=0}^{\infty}|a_m|^2 \right)^2,$$ where $$c_m=a_0a_m+a_1 a_{m-1}+ \dots +a_ma_0.$$

It sounds like a basic inequality, but I wasn't able to prove it. Does this inequality have a name? Where I can find a proof or how can one prove it?

• Where did you find it? – Iosif Pinelis Apr 5 at 21:46

Suppose you have a power series with coefficients $$a_n$$ $$f(z):= \sum_{k=1}^\infty a_k z^k .$$ Then the coefficients of $$f^2$$ are exactly $$c_n$$. Also if we denote by $$\odot$$ the Hadamard multiplication of powerseries (coefficient-wise or equivalently convolution of the boundary values), $$c_k^2$$ are the coefficients of $$f^2\odot f^2$$. We apply first Hardy's inequality (Theory of Hp spaces, Duren, Corollary of Theorem 3.15) and then Young's inequality for convolution \begin{align*} \sum_{k=1}^\infty \frac{|c_k|^2}{k+1} & \leq \pi \Vert f^2 \odot f^2 \Vert_{H^1} \\ & \leq \pi \Vert f^2 \Vert_{H^1} \Vert f^2 \Vert_{H^1} \\ &= \pi \Vert f \Vert_{H^2} ^4 \\ & = \pi \Big( \sum_{k=0}^\infty |a_k|^2 \Big)^2 \end{align*}. If your inequality is true I'm missing a constant of $$\pi$$.
• You are right. The inequality was missing a $\pi$ which is fixed. Thanks. – MathLearner Apr 6 at 1:27