1

I'm trying to understand some notation from Evans' PDEs. Specifically, this question asked about the following inequality,

$$\left|\int_{B_\epsilon(0)}\Phi(y)\Delta f(x-y)dy\right| \leq C||D^2f||_{L^\infty}\int_{B_\epsilon(0)}|\Phi(y)|dy$$

Here $D^2f$ represents the Hessian of $f$ and $\Phi$ is the fundamental solution to the Laplace equation. Since the Laplacian is the trace of the Hessian, the inequality amounts to showing,

$$ |\Delta f(x-y)| = |tr(D^2f)| \leq C||D^2f||_{L^\infty} $$

for some constant $C$. But what does the $L^\infty$ norm of a matrix mean here?

J W
  • 2,351

1 Answers1

2

Someone could probably post a better answer than this, but since there is no answer so far I will put down what I have time for.

This is probably the matrix infinity norm: https://en.m.wikipedia.org/wiki/Matrix_norm

The trace can be bounded by the sum of the absolute values of the matrix elements, which is a kind of one norm, and since all norms on a finite-dimensional space are equivalent, https://en.m.wikipedia.org/wiki/Norm_(mathematics) this can be changed into the infinity norm.

Evans may have intended something more elegant, but if you just want to prove the last inequality in your question, that would do it.

Zach Boyd
  • 968