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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?

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  • $\begingroup$ Where did you find it? $\endgroup$ – Iosif Pinelis Apr 5 at 21:46
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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 $\pi$.

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  • $\begingroup$ You are right. The inequality was missing a $\pi$ which is fixed. Thanks. $\endgroup$ – MathLearner Apr 6 at 1:27

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