TITLE: Doubt about a passage on page 24 of the Evans Partial Differential Equations book
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: 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 $$\vert \Delta f \vert \leqslant \sum_{i=1}^n \vert D_{ii}f \vert \leqslant n \| D^2 f \|_{L^\infty(\mathbb{R}^n)}.$$