# $L^{p} (\mathbb T)$ is complete.

I want to show that:

$(1)$ $L^{p}(\mathbb T)$, where $\mathbb T$ is the circle is complete.

$(2)$ $C(\mathbb T)$ is dense in $L^{p}(\mathbb T).$

Can you help me in (1) for any interval not necessarily the circle?

Thanks.

• There are dozens of books. A good one is Real and Complex analysis, by Walter Rudin. – uniquesolution Mar 6 '17 at 7:31
• I need a simple and detailed book .... Is walter Rudin simple and detailed? – Emptymind Mar 6 '17 at 7:32
• Walter Rudin is one of the great masters of exposition. His three analysis books are classics. They are fully detailed, and their clarity is second to none. However, the subject matter is not very simple, hence I would not say that Real and Complex analysis is a simple book. I am not aware of simple books in functional analysis. – uniquesolution Mar 6 '17 at 7:53

To see that $L^p(S^1)$ is complete note that $S^1=[0,2\pi]/\{0\}\sim\{2\pi\}$ and consider the maps: $$r:L^p(S^1)\to L^p(0,2\pi),\ f\mapsto f\lvert_{(0,2\pi)}\\ i:L^p(0,2\pi)\mapsto L^p(S^1),\ f\mapsto\left(x\mapsto\begin{cases}f(x)&x\neq0 \\0&x=0\end{cases}\right)$$
These maps are linear and actually isometries. Thus since $L^p(0,2\pi)$ is complete so is $L^p(S^1)$. To see that the continuous functions are dense, note that continuous functions with compact support are dense in $L^p(0,2\pi)$ (since you can get step functions from limits of such functions), but these lie in the image of $C(S^1)$ under $r$, so $C(S^1)$ must be dense in $L^p(S^1)$.
• Take an interval of length $2\pi$ and glue the ends together, you get the circle. Thats the basic idea, the induced measure is also the same as usual measure on $S^1$. If you know that $L^p(0,1)$ is complete then the proof that $L^p(0,2\pi)$ is complete is either exactly the same or the map that sends a function $f$ to $x\mapsto \frac {f(2\pi x)}{2\pi}$ is an invertible isometry from $L^p(0,1)$ to $L^p(0,2\pi)$. – s.harp Mar 22 '17 at 12:33
• It is sketched here. The statement "every absolutely convergent series in $X$ converges $\iff$ $X$ is complete" is shown nicely here. – s.harp Mar 22 '17 at 21:17