7

On page 23 of Lawrence Evans' Partial Differential Equations text (2nd edition) he claims that $$\frac{ f( x + he_i - y) - f( x-y)}{h} \to \frac{ \partial f}{ \partial x_i} ( x-y)$$ uniformly on $\mathbb{R}^n$ as $h \to 0$. So

$$\frac{ \partial u}{ \partial x_i } (x) = \int_{\mathbb{R}^n }\Phi(y) \frac{ \partial f}{ \partial x_i} ( x- y) dy$$ Why does this hold true? I have proved uniform convergence as follows:

Proof: It suffices to show that $$\underbrace{ \frac{ f( x + h e_i - y ) - f( x-y) }{ h} - \frac{ \partial f}{ \partial x_i } ( x-y)}_{g_h(y)} \to 0 $$ uniformly as $h \to 0$. But by the Mean Value Theorem, for each $h$ there exists a $\theta_h$ such that $0 < \theta_h < 1$ and $\frac{ f( x + h e_i - y ) - f( x-y) }{ h} = f_{x_i} ( x+ \theta_h h e_i - y) $. So \begin{align*} g_h(y) &= \frac{ f( x + h e_i - y ) - f( x-y) }{ h} - \frac{ \partial f}{ \partial x_i } ( x-y) \\ &= f_{x_i} ( x+ \theta_h h e_i - y) - f_{x_i} ( x-y) \end{align*} but $f_{x_i}$ is uniformly continuous ($f \in C_c^2$ so $f_{ x_i} \in C_c^1$ and therefore continuous on a compact set, so uniformly continuous). Therefore, regardless of $y$, for small enough $h$ this difference can be made arbitrarily small. So $g_h(y) \to 0$ uniformly, as we intended to show. QED.

But how does uniform convergence allow us to move the derivative through the integral sign?

kam
  • 312

2 Answers2

2

We have

$$\left|\frac{u(x+he_i)-u(x)}{h}- \int_{\mathbb{R}^n}\Phi(y)\frac{\partial f}{\partial x_i}(x-y)\, dy\right|\\= \left|\int_{\mathbb{R}^n}\Phi(y)\frac{f(x+he_i-y)-f(x-y)}{h}\, dy- \int_{\mathbb{R}^n}\Phi(y)\frac{\partial f}{\partial x_i}(x-y)\, dy\right|\\ \leqslant \int_{\mathbb{R}^n}|\Phi(y)|\left|\frac{f(x+he_i-y)-f(x-y)}{h}\, - \frac{\partial f}{\partial x_i}(x-y)\right|\, dy$$

Let $x$ be fixed. Since $f$ has compact support, the terms involving $f$ vanish for all $y$ outside some ball $B_R(0)$ of sufficiently large radius $R$.

Hence,

$$\left|\frac{u(x+he_i)-u(x)}{h}- \int_{\mathbb{R}^n}\Phi(y)\frac{\partial f}{\partial x_i}(x-y)\, dy\right|\\ \leqslant \int_{B_R(0)}|\Phi(y)|\underbrace{\left|\frac{f(x+he_i-y)-f(x-y)}{h}\, - \frac{\partial f}{\partial x_i}(x-y)\right|}_{G(h,x,y)}\, dy$$

Note that the fundamental solution $\Phi$ has a singularity at $y=0$ but is integrable on any bounded neighborhood, so there exists $M > 0$ such that

$$\int_{B_R(0)}|\Phi(y)|< M$$

It has been established that $\displaystyle\frac{f(x+he_i-y)-f(x-y)}{h}$ converges uniformly to the partial derivative of $f$ as $h \to 0$. Hence for any $\epsilon > 0$ there exists $\delta(\epsilon) > 0$ such that if $|h| < \delta(\epsilon)$, then $0 \leqslant G(h,x,y) < \epsilon/M$ for all $x$ and $y$.

Thus, if $|h|< \delta(\epsilon)$, then we have

$$\left|\frac{u(x+he_i)-u(x)}{h}- \int_{\mathbb{R}^n}\Phi(y)\frac{\partial f}{\partial x_i}(x-y)\, dy\right|\leqslant \int_{B_R(0)}|\Phi(y)|\frac{\epsilon}{M}\, dy \leqslant \epsilon$$

Therefore,

$$\frac{\partial u}{\partial x_i} (x) = \lim_{h \to 0}\frac{u(x+he_i)-u(x)}{h}= \int_{\mathbb{R}^n}\Phi(y)\frac{\partial f}{\partial x_i}(x-y)\, dy$$

RRL
  • 93,190
  • Evans omits so many details here that it's hard to get. Of course, it may be that I'm just too weak. Thanks. – Enhao Lan Jun 04 '24 at 11:14
0

Alternatively, one can use the dominated convergence theorem. Note that $$\lim_{h\rightarrow 0}\Phi(y)\frac{f(x-y+he_{i})-f(x-y)}{h} = \Phi(y)\partial_{i}f(x-y)\:\:\text{a.e.}\:\:y\in\mathbb{R}^{n}.$$ Then by the mean value theorem we have $$\Big|\Phi(y)\frac{f(x-y+he_{i})-f(x-y)}{h}\Big| = |\Phi(y)\partial_{i}f(x-y+s(x, y, h)he_{i})|\leq C|\Phi(y)|\chi_{B(0, R)}(y),$$ where $s(x, y, h)\in (0, 1)$ and $R > 0$ is sufficiently large. Here, we have used the fact that $f\in C^{2}_{c}(\mathbb{R}^{n})$. By the coarea formula, \begin{align}\int_{B(0, R)}|\Phi(y)|\:\text{d}y &= \int_{0}^{R}\int_{\partial B(0, r)}|\Phi|\:\text{d}\mathcal{H}^{n-1}\:\text{d}r.\end{align} Since $|\Phi|_{\partial B(0, r)}| = g(r)$ where $$g(r) = \begin{cases}\frac{1}{2\pi}|\log r|, & n = 2\\ \frac{1}{n(n-2)\alpha(n)}\frac{1}{r^{n-2}}, & n\geq 3\end{cases},$$ we get \begin{align}\int_{B(0, R)}|\Phi(y)|\:\text{d}y &= \int_{0}^{R}g(r)\mathcal{H}^{n-1}(\partial B(0, r))\:\text{d}r\\ &= \mathcal{H}^{n-1}(\partial B(0, 1))\int_{0}^{R}g(r)r^{n-1}\:\text{d}r\\ & < \infty.\end{align}

Karthik Kannan
  • 2,000