My doubt is about the proof of the theorem 1 section 2.2.1
of the evans pde classic book.
My doubt:
Consider the function
$$\Phi(x) = \begin{cases}
– \frac{1}{2 \pi} \log |x| & \text{if $n= 2$} \\
\frac{1}{n(n-2)\alpha (n)} \frac{1}{|x|^{n-2}} &\text{if $n \geq 3$}
\end{cases},$$
where $x \neq 0$ ($x \in \Bbb R^n$).
Consider
$$u(x) = \int_{\Bbb R^n} \Phi (x-y) f(y) \, dy, $$
where $ f \in C^{2}_{c} (\Bbb R^n)$.
I am trying to prove that
$$\frac{\partial u }{\partial x_i} (x) = \int_{R^n} \Phi (y) \frac{\partial f}{ \partial x_i} (x-y) \, dy.$$
My book does this:
We have
$$u(x) = \int_{\Bbb R^n} \Phi (x-y) f(y) \, dy = \int_{\Bbb R^n} \Phi (y) f(x- y) \ dy.$$
Then for $h \in \Bbb R \setminus \{ 0 \}$ we have
$$ \frac{u(x + h e_i) – u(x)}{h} = \int_{\Bbb R^n} \Phi (y)\: \frac{f(x – y + h e_i) – f(x -y)}{h} \, dy$$
Evans says
$$\frac{f(x + he_i – y ) – f (x-y)}{h} \to \frac{\partial f}{\partial x_i} (x-y)$$
uniformly as $h \to 0$, and thus
$$\frac{\partial u}{\partial x_i} (x) = \int_{\Bbb R^n} \Phi (y) \frac{\partial f}{\partial x_i} (x-y) \, dy. \tag{*}$$
The only thing that I don't understand is the line $(*)$. The uniform convergence I proved. Someone can help me to prove the line $(*)$ ?
Thanks in advance
Best Answer
One way to solve this problem is: Note that
$\left|\int_{\mathbb{R}^n}\Phi(y)\frac{f(x-y+he_i)-f(x-y)}{h}dy-\int_{\mathbb{R}^n}\Phi(y)\frac{\partial f(x-y)}{\partial x_i}\right|\leq \int_{\mathbb{R}^n}|\Phi(y)|\left| \frac{f(x-y+he_i)-f(x-y)}{h}-\frac{\partial f(x-y)}{\partial x_i}\right|$
For small $h$, the support of $\left| \frac{f(x-y+he_i)-f(x-y)}{h}-\frac{\partial f(x-y)}{\partial x_i}\right|$ is contained in a fixed compact set $K$. By using the uniform convergence, we have that for given $\epsilon>0$, there exist $\delta>0$ such that for $|h|<\delta$, $y\in \mathbb{R}^n$ $$\left| \frac{f(x-y+he_i)-f(x-y)}{h}-\frac{\partial f(x-y)}{\partial x_i}\right|<\epsilon $$
If follows that $$\int_{\mathbb{R}^n}|\Phi(y)|\left| \frac{f(x-y+he_i)-f(x-y)}{h}-\frac{\partial f(x-y)}{\partial x_i}\right|\leq \epsilon\int_K|\Phi(y)| $$
Now use the fact $\Phi\in L^1_{loc}(\mathbb{R}^n)$ to conclude.