# Infimum of an integral functional involving a symmetric matrix

I have a symmetric $d \times d$ matrix $A$ and I have the following functional: $$\mathcal J(h) := \int_{B_1(0)} \vert \langle Au,u \rangle\vert \frac{\vert h'(\vert u \vert)\vert}{\vert u \vert} du,$$ where $B_1(0)$ is the unit ball in $\mathbb R^d$ and $h \in C_c^\infty(\mathbb R)$ with $\text{supp}\, h\subset [-1,1]$ and $\int h = 1$. Let me denote by $H$ the set of these functions $h$. I would like to find/estimate $$\inf_{h \in H} \mathcal J(h).$$

Do you have any ideas on how to approach this kind of problems? I have never heard nor read about such a problem. I have been playing with some toy models in $\mathbb R^2$ without success, I do not even manage to guess who is the infimum.

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The infimum is $0$ for $d\ge2$. Indeed, take any $h \in H$. For any real $\ep\in(0,1)$ and all real $r>0$, let $h_\ep(r):=\frac1\ep\,h(\frac r\ep)$, so that $h_\ep\in H$, $\text{supp}\,h_\ep\subset[-\ep,\ep]$, and $$|h'_\ep(r)|=\frac1{\ep^2}\,\Big|h'\Big(\frac r\ep\Big)\Big|\le \frac M{\ep^2}$$ for some real $M>0$ and all real $r>0$, whence $$0\le\mathcal J(h_\ep) \ll\int_0^\ep r^2\frac{|h'_\ep(r)|}{r}\,r^{d-1}\,dr \ll\int_0^\ep r^2\frac{1/\ep^2}{r}\,r^{d-1}\,dr \ll\ep^{d+1-2}\to0$$ as $\ep\downarrow0$.
• Oh nice, thanks for the useful answer! I admit I am a bit puzzled, as I expected the result to depend on the matrix $A$. However seems to be working, nice! Thanks again, +1! – Y.B. May 23 '18 at 20:56
• Sorry, there is still something not clear. It is easy to see that one can re-write the functional $\mathcal J$ as $\int \vert \langle Au, \nabla \tilde{h}(u) \rangle \vert du \ge \vert \int \langle Au, \nabla \tilde{h}(u) \rangle du \vert = \vert \text{tr}\, A\vert$ since $\int u_i \partial_j \tilde{h}(u) du = \delta_{ij}$ (up to a sign, integration by parts). Where am I mistaken? Thanks again. – Y.B. May 23 '18 at 21:15
• @Y.B. : What is $\tilde h$? Please also provide details on the rest of your comment. – Iosif Pinelis May 23 '18 at 21:53
• Sorry for my confusion. So let me try to explain better. We have $J(h) = \int \vert \langle Au, \nabla h(\vert u \vert) \rangle \vert du$. Do you agree on this? Let me define a function $\tilde h$ so that $h(\vert u \vert) = \tilde h(u)$. Then $J(h) = \int \vert \langle Au, \nabla \tilde h(u) \rangle \vert du \ge \vert \int \langle Au, \nabla \tilde h(u) \rangle du \vert$. Am I right? This last guy, by definition, is the sum over i,j of $A_{ji} \cdot \int u_i \cdot \frac{\partial}{\partial u_j}\tilde h(u) du$. Do you agree? Now the last integral can be computed by integration by parts. – Y.B. May 23 '18 at 22:32