Doubt about a passage on page 24 of the Evans Partial Differential Equations book

Question

Let $\Phi$ the fundamental solution of $\Delta v=0$, i.e. $$ \Phi(x)=\left\{\begin{array}{ccc}-\frac{1}{2\pi}\log|x|& \mbox{if}& n=2\\ \frac{1}{n(n-1)\alpha(n)}\frac{1}{|x|^n-2}& \mbox{if}& n\geq 3\\ \end{array} \right. $$ here $\alpha(n)$ denote the volume of unite ball in $\mathbb{R}^n$. In page 23 of Evans Partial Differential Equations book Theorem 1 ( Solving Poisson's equation ) we have for $f\in C_c^{2}(\mathbb{R}^n)$ and $$ u(x)=\int_{\mathbb{R}^n}\Phi(x-y)f(y)\mathrm{d}y=\int_{\mathbb{R}^n}\Phi(y)f(x-y)\mathrm{d}y $$ the following conclusions: $$ \begin{array}{crl} \mathrm{(i)} & u &\in C^2(\mathbb{R}^n) \\ \mathrm{(ii)}& -\Delta u &= f \end{array} $$ In step 2 of the proof of this theorem, we have the following inequality $$ \left|\int_{B(0,\epsilon)} \Phi(y)\Delta_x f(x-y)\mathrm{d} y\right| \leq C\cdot \|D^2f\|_{L^\infty(\mathbb{R}^n)}\int_{B(0,\epsilon)} |\Phi(y)|\mathrm{d} y $$

Question. How can I prove that $$ \left|\int_{B(0,\epsilon)} \Phi(y)\Delta_x f(x-y)\mathrm{d} y\right| \leq C\cdot \|D^2f\|_{L^\infty(\mathbb{R}^n)}\int_{B(0,\epsilon)} |\Phi(y)|\mathrm{d} y\quad ? $$

My attempt: We have \begin{align} \left|\int_{B(0,\epsilon)} \Phi(y)\Delta_x f(x-y)\mathrm{d} y\right| \leq & \int_{B(0,\epsilon)} |\Phi(y)\Delta_x f(x-y)|\mathrm{d} y \\ &\leq \mathop{\mathrm{sup \;ess}}|\Delta_x f(x-y)|\int_{B(0,\epsilon)} |\Phi(y)|\mathrm{d} y \end{align} This seems to indicate that I have to prove that $$ \mathop{\mathrm{sup \;ess}}|\Delta_x f(x-y)| \leq \mathop{\mathrm{sup \;ess}}|D^2_x f(x-y)|_{\mathscr{L}(\mathbb{R}^n,\mathbb{R}^n)} $$ where $|\;\cdot \;|_{\mathscr{L}(\mathbb{R}^n,\mathbb{R}^n)}$ is a matrix norm.

Answer 1

To answer this question, you should check which matrix norm Evan's is using. The truth of the result and proof depend on this. You can find a list of matrix norms on wikipedia.

I think that he is using the Frobenius norm, in which case the result follows because $|\Delta_{x} f(x-y)| = |TrD^{2}_{x}f(x-y)| = \sqrt{\sum_{i=1}^{n} (D_{x_{i}}^{2}f(x-y))^{2}} \leq \sqrt{\sum_{i=1,j=1}^{n} (D_{x_{i}}D_{x_{j}}f(x-y))^{2}} = |D^{2}_{x}f(x-y)|$

If the norm were say the spectral norm, then the result would be false because the trace is the sum of the eigenvalues and the spectral norm is only the largest eigenvalue.

Hope this helps. The norm that he is using is probably in Evan's notation appendix.

Answer 2

Evan's defines$$\vert D^2 f \vert = \bigg (\sum_{i,j=1}^n (D_{ij}f)^2 \bigg)^{\frac 1 2} $$ and $$ \| D^2 f \|_{L^\infty(\mathbb{R}^n)} = \| \vert D^2 f \vert \|_{L^\infty(\mathbb{R}^n)} = \mathrm{ess}\sup_{x \in \mathbb{R}^n} \bigg (\sum_{i,j=1}^n (D_{ij}f)^2 \bigg)^{\frac 1 2}.$$ Then for all $x \in \mathbb{R}^n$, $D_{ij} f (x) \leqslant \vert D^2 f \vert \leqslant \| D^2 f \|_{L^\infty(\mathbb{R}^n)}$. Thus, by triangle inequality $$\vert \Delta f \vert \leqslant \sum_{i=1}^n \vert D_{ii}f \vert \leqslant n \| D^2 f \|_{L^\infty(\mathbb{R}^n)}.$$