1. ## Tempered distribution.

Hi all,

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

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

2. Hmm, I can't see a thing...

3. 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

4. It is clear that $\displaystyle \frac{1-\chi}{P} \in L^{\infty}$. Take any function $\displaystyle f\in L^1_{loc}$ and $\displaystyle g\in \mathcal{S} (\mathbb{R}^n)$ with $\displaystyle f=O(|x|^r)$ as $\displaystyle |x| \rightarrow \infty$ for some $\displaystyle r\geq 0$ then we have, with $\displaystyle c>0$ and $\displaystyle a>n$:

$\displaystyle \left| \int_{\mathbb{R}^n} fg \right| \leq \int_{|x|<c} |fg| + \int_{|x|>c} |fg| \leq \| f1_{\{ x: |x|<c\} }\| _1 \| g \|_{\infty} + \int_{|x|>c} |x|^{-a}(|x|^{a+r}g) \leq$

$\displaystyle \| f1_{\{ x: |x|<c\} }\| _1 \| g \|_{\infty} + \| |x|^{-a} 1_{\{ x:|x|>c \} }\| _1 \| |x|^{a+r} g\| _{\infty}$

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