# Tempered distribution.

• May 15th 2011, 07:20 AM
measureman
Tempered distribution.
Hi all,

I am trying to follow the attached proof of the elliptic regularity theorem.

http://s2.postimage.org/4vvpb25vd/ertproof.jpg

Just wondering if someone could explain to me where we go from (1 - chi)/P <= delta to saying (1 - chi)/P is a tempered distribution. I do not see why this would be true.

Thanks
• May 15th 2011, 09:34 AM
Jose27
Hmm, I can't see a thing...
• May 15th 2011, 09:35 AM
measureman
Quote:

Originally Posted by Jose27
Hmm, I can't see a thing...

It shows as an image for me. The direct link is http://s2.postimage.org/4vvpb25vd/ertproof.jpg
• May 15th 2011, 10:44 AM
Jose27
It is clear that $\frac{1-\chi}{P} \in L^{\infty}$. Take any function $f\in L^1_{loc}$ and $g\in \mathcal{S} (\mathbb{R}^n)$ with $f=O(|x|^r)$ as $|x| \rightarrow \infty$ for some $r\geq 0$ then we have, with $c>0$ and $a>n$:

$\left| \int_{\mathbb{R}^n} fg \right| \leq \int_{|x|c} |fg| \leq \| f1_{\{ x: |x|c} |x|^{-a}(|x|^{a+r}g) \leq$

$\| f1_{\{ x: |x|c \} }\| _1 \| |x|^{a+r} g\| _{\infty}$

This implies that the natural functional defined by $f$ on $\mathcal{S}$ is indeed a distribution (Just take a sequence tending to $0$ on $\mathcal{S}$ and use this estimate). Now, for your case take $r=0$.