4
$\begingroup$

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.

$\endgroup$
3
  • 5
    $\begingroup$ There are dozens of books. A good one is Real and Complex analysis, by Walter Rudin. $\endgroup$ Mar 6, 2017 at 7:31
  • $\begingroup$ I need a simple and detailed book .... Is walter Rudin simple and detailed? $\endgroup$
    – Emptymind
    Mar 6, 2017 at 7:32
  • 2
    $\begingroup$ 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. $\endgroup$ Mar 6, 2017 at 7:53

1 Answer 1

3
+50
$\begingroup$

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 $L^p(S^1)$.

A simple book is in my opinion Kolmogorov and Fomin.

$\endgroup$
7
  • $\begingroup$ thank you .....what is the title of the book? these are the authors . $\endgroup$
    – Emptymind
    Mar 6, 2017 at 9:26
  • 2
    $\begingroup$ Elements of the Theory of Functions and Functional Analysis $\endgroup$
    – s.harp
    Mar 6, 2017 at 9:43
  • $\begingroup$ @Keen-ameteur .... Could u explain this in more detail for me? $\endgroup$
    – Emptymind
    Mar 9, 2017 at 15:20
  • 1
    $\begingroup$ 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 $L^p(0,2\pi)$. $\endgroup$
    – s.harp
    Mar 22, 2017 at 12:33
  • 1
    $\begingroup$ It is sketched here. The statement "every absolutely convergent series in $X$ converges $\iff$ $X$ is complete" is shown nicely here. $\endgroup$
    – s.harp
    Mar 22, 2017 at 21:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.