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My question is about the part of proof of the theorem 1 section 2.2.1 of the evans pde(p24):

The only thing that I don't understand is the inequality: $$\int_{B(0, \varepsilon)} \Phi(y) \Delta_{x} f(x-y) d y\leq C\left\|D^{2} f\right\|_{L^{\infty}\left(\mathbb{R}^{n}\right)} \int_{B(0, \varepsilon)}|\Phi(y)| d y \leq\left\{\begin{array}{ll} C \varepsilon^{2}|\log \varepsilon| & (n=2) \\ C \varepsilon^{2} & (n \geq 3) \end{array}\right.$$

$$ f \in C_{\mathrm{c}}^{2}\left(\mathbb{R}^{n}\right),\Phi(x):=\left\{\begin{array}{ll} -\frac{1}{2 \pi} \log |x| & (n=2) \\ \frac{1}{n(n-2) \alpha(n)} \frac{1}{|x|^{n-2}} & (n \geq 3) \end{array}\right.$$

I don't understand the meaning of $C$,it may be connected with the inequality in p22: $$|D \Phi(x)| \leq \frac{C}{|x|^{n-1}},\left|D^{2} \Phi(x)\right| \leq \frac{C}{|x|^{n}} \quad(x \neq 0)$$ Someone can help me to prove detailedly that inequality ?

Thanks in advance.

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Probably, the problem is the notation. In Evans book, $$\|D^2f\|_\infty=\||D^2f|\|_\infty=\left\|\left(\sum_{i,j=1}^n |f_{x_ix_j}|^2\right)^{1/2}\right\|_\infty.$$ Then, we have to obtain right-hand side from the absolute value of the Laplacian. The constant $C$ comes from the equivalence of norms on $\mathbb R^n$:

Since the $L^1$-norm defined by $$\|(v_1,...,v_n)\|_1=|v_1|+\cdots +|v_n|$$ is equivalent to the Euclidean norm defined by $$\|(v_1,...,v_n)\|=(|v_1|^2+\cdots +|v_n|^2)^{1/2},$$ there exists a constant $C>0$ such that $$\begin{align*} |\Delta_x f|&=|f_{x_1x_1}+\cdots +f_{x_nx_n}|\\ &\leq |f_{x_1x_1}|+\cdots +|f_{x_nx_n}|\\ &\leq C (|f_{x_1x_1}|^2+\cdots +|f_{x_nx_n}|^2)^{1/2}\\ &\leq C \left(\sum_{i,j=1}^n|f_{x_ix_j}|^2\right)^{1/2}\\ &=C|D^2 f| \end{align*}$$ and thus $$\||\Delta_x f|\|_{\infty}\leq \|C|D^2 f|\|_{\infty}=C\|D^2 f\|_{\infty}.$$ It follows that $$\begin{align*} \int_{B(0, \varepsilon)} \Phi(y) \Delta_{x} f(x-y)\; d y&\leq \||\Delta_{x} f|\|_{\infty} \int_{B(0, \varepsilon)} |\Phi(y)| \;dy\\ &\leq C\|D^2 f\|_{\infty} \int_{B(0, \varepsilon)} |\Phi(y)| \;dy. \end{align*}$$

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  • $\begingroup$ Thank you very much!!! $\endgroup$ – Johnstein Apr 11 at 12:29

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