0

I have to show the following: for $f \in L^1(S^1)$ Show that the map $f(x) \to f_y = f(x+y)$ is continuous in the distance of $L^1(S^1)$. i.e.
$\lim_{y \to 0} ||f_y-f||_1 = 0$

I am supposed to use that $||f_y-f||_1 \le ||f_y-f||_\infty$ and that $L^\infty$ is dense in $L^1$.
I do not see how one should do this properly.

I treid to do the following and would apperciate if you can say me whether this is right.
let $ g \in L^\infty$ and continuous
then we know that $g(x+y) \to g(x)$ as $ y \to 0$.
therefore also $\sup_x g(x+y)\to g(x)$ as $y \to 0$, what means that $g(x+y)\to g(x)$ in $L^\infty$ and therefore also in $L^1$.
take $f \in L^1$ and take $\epsilon > 0$ find $g \in L^\infty$ such that $||g-f||_{L^1} < \epsilon / 3$
then $||f-f_y||_{L^1}=||f-g -( f_y -g_y)+g-g_y||_{L^1} \le|| f-g-(f_y-g_y)||_{L^1}+||g-g_y||_{L^1}$
where the first part is $\le 2||f-g||_{L^\infty}$ and the second $\le ||g-g_y||_{L^\infty} \le \epsilon/3$ for $y$ sufficiantly close to $x$.
so all in all $||f-f_y||_{L^1} \le \epsilon$
is this correct?

lukas
  • 45
  • 8
  • I don't see how the density of $L^\infty$ helps. You could (i) prove the result for continuous $f$ and then (ii) show how the general case follows. – David C. Ullrich Feb 10 '16 at 15:05
  • I don't see exactly. why is it easier to proofe it for continuous functions? – lukas Feb 10 '16 at 22:19

1 Answers1

0

Hint: Show this for functions with compact support first, and generalize it after that.

John B
  • 16,854
Börge
  • 1,207
  • 1
    He's talking about functions on the circle; every function has compact support. Compact support per se doesn't help. If you meant continuous functions with compact support that would work. – David C. Ullrich Feb 10 '16 at 15:04
  • I still cannot see how one should proof this properly. – lukas Feb 11 '16 at 10:50