Understanding the $1$-norm of $L^1(\mathbb{R}^d)$

Asked Apr 21 '21 at 00:00

Active Apr 21 '21 at 00:00

Viewed 46 times

I have a quick question about a proof I was reading. The proof is basically asking about the continuity of the translation operator in the $L^1$-norm. Here is a snippet of the proof

I am not sure how their integral over $\mathbb{R}^d$ turned into just an integral over $E$ (first line to second). It seems to me that this is not too clear. Unless I am missing something very simple...

Thanks in advance!

asked Apr 21 '21 at 00:00

BingerBiter

Think about what the support of $F - F_h$ is – SescoMath Apr 21 '21 at 00:11
@PhysMath: Thank you for your comment. But wouldn't the support of $F-F_h$ be $E \Delta (E+h)$? – BingerBiter Apr 21 '21 at 00:12
The proof is wrong. Where did you find it? – Kavi Rama Murthy Apr 21 '21 at 00:13
Yeah I believe you're right, but i think you can upper bound by all of $E$ plus the set $E+h$ which is just $E$ shifted by $h$. (So actually i think there a small error in the proof but the same idea works). – SescoMath Apr 21 '21 at 00:14
@KaviRamaMurthy: I found a link while studying and this problem was interesting to me but I could not follow that line. Do you have any recommendations on how to prove this when $F \in C_c(\mathbb{R}^d)$? – BingerBiter Apr 21 '21 at 00:18
@PhysMath: Thank you, yes I agree that maybe not all hope is lost! – BingerBiter Apr 21 '21 at 00:18

Understanding the $1$-norm of $L^1(\mathbb{R}^d)$

0 Answers0