I have some question on deriving Poisson's equation on Evan's book. we define $$u=\int_{\mathbb{R}^n}\phi(x-y)f(y)dy,$$ where $\phi$ is the fundamental solution of Laplace equation, and $f\in C_c^2(\mathbb{R}^n)$. What I am confused is that he says $$\frac{f(x+he_i-y)-f(x-y)}{h}\rightarrow \frac{\partial f}{\partial x_i}(x-y)$$ uniformly on $\mathbb{R}^n$ when $h\rightarrow 0$. I could just see pointwise convergence here, how uniformly convergence comes?
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3Possible duplicate of theorem 1 chapter 2 - Evans PDE – Sahiba Arora Oct 23 '19 at 09:23
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Let's use the mean value theorem twice.
First: $$ f(z + he_i) - f(z) = h \frac{\partial f}{\partial x_i}(z + h'e_i) $$ for some $h' \in (0,h)$.
Second: $$ \frac{\partial f}{\partial x_i}(z + h'e_i) - \frac{\partial f}{\partial x_i}(z) = h' \frac{\partial^2 f}{\partial^2 x_i}(z + h''e_i) $$ for some $h'' \in (0, h')$.
Now, since $\frac{\partial^2 f}{\partial^2 x_i}$ has compact support by assumption, there is some $M > 0$ such that $$\sup_z \left|\frac{\partial^2 f}{\partial^2 x_i}(z)\right| \leq M.$$
Putting this together (you filling in the couple steps between the inequality from our discussion above): $$\sup_z\left| \frac{f(z+he_i) - f(z)}{h} - \frac{\partial f}{\partial x_i}(z) \right| \leq h' \sup_z \left|\frac{\partial^2 f}{\partial^2 x_i}(z)\right| \leq h'M $$
Tom
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4why does $\frac{\partial^2 f}{\partial^2 x_i}$ have compact support? I thought only $f$ has compact support? – cooselunt Feb 08 '21 at 01:46