1

Let $U$ be a open set of $\mathbb{R}^n$, C(U) is all continuous functions on U, for example C(0,1), when $U=(0,1)$. And $L^1(U)$ is lp-space where $p=1$.

It was said that $L^1(U)$ is the completion of of $C(U)$ by the norm $\|\cdot\|_{L^1}$. Namely $$ L^1(U)=\overline{C(U)}. $$

But I feel confuse, the function in $C(U)$ my be not integrable, $1/x$ in $(0,1)$ for example. But from $\overline{C(U)}=L^1(U)$ we can infer that $C(U)\subset L^1(U)$. This is a contradiction.

user129161
  • 25
  • Your confusion is warranted, you should either consider the integrable continuous functions, or the compactly supported, continuous functions on $U$. You could even consider the smooth, compactly supported functions on $U$. – Olivier Bégassat Mar 25 '14 at 17:37
  • You can't say $\overline{C(U)}=L^1(U)$. The truth is $\overline{C_c(U)}=L^1(U)$, where $C_c(U)$ is all continuous compactly supported functions on $U$. Of course non-zero constant function is element of $C(\Bbb{R})$ but not in $L^1(\Bbb{R})$ – Hamid Shafie Asl Mar 25 '14 at 17:44

0 Answers0