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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

Daniele Tampieri
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math student
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    Show that if $h$ has compact support, then $\lvert \int \Phi(y)h(y),dy\rvert \leqslant C\cdot \lVert h\rVert_\infty$. – Daniel Fischer Aug 15 '13 at 20:35
  • Please review the changes I made to the formatting, both to ensure that they are correct and to learn from them. Be sure to read the changes to the source. – dfeuer Aug 15 '13 at 20:55
  • @DanielFischer,in this case i will have to prove that the integral in $R^n$ of $\Phi$ is finite ? – math student Aug 15 '13 at 21:13
  • No. The premise that the support of $h$ is compact is essential. $\Phi$ isn't integrable, but it is locally integrable. So its integral over the support of $h$ is finite. – Daniel Fischer Aug 15 '13 at 21:15

1 Answers1

5

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.

Tomás
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