For any $f\in L^1(S^1)$ and let $y\in\mathbb{R}$, let $f_y$ be defined as $f_y(x)=f(x+y)$. Prove that $\lim_{y\to 0 } ||f_y-f||=0.$

Question

Here is an exercise from Mckean's book "Fourier Series and Integrals".

---

Problem:

For any $f\in L^1(S^1)$ and let $y\in\mathbb{R}$, let $f_y$ be defined as $f_y(x)=f(x+y)$. Prove that $$\lim_{y\to 0 } ||f_y-f||=0.$$

---

Solution:

There was a hint for this exercise which told me first to prove this for the space $C(S^1)$ (the space of continuous function on the circle). We have that the norm on that space is given by $$||f||_{\infty}=\max_{0\leq x< 1} |f(x)|.$$ In our case, we want to consider $$\lim_{y\to 0}||f_y-f||_{\infty}=\lim_{y\to 0}\max_{0\leq x< 1} |f(x+y)-f(x)|=$$ $$\max_{0\leq x< 1}\lim_{y\to 0} |f(x+y)-f(x)|=\max_{0\leq x< 1}|f(x)-f(x)|=0.$$

I am not sure if I am allowed to interchange the order of $\max$ and $\lim$ like I did, but it felt like a natural step for me (and I don't know what could have went wrong in that step). The second last step is motivated by the continuity of $f$.

The last step to conclude this exercise is to use that $C(S^1)$ is dense in $L^1(S^1)$ (according to the last part of the hint). I know what it means to be dense, but I don't know how to apply it in such a case to prove that the other limit must also be $0$.

---

Questions:

1. Was my manipulations of the limit correct?

2. How can I apply that $C^1(S^1)$ is dense in $L^1(S^1)$ to conclude that $$\lim_{y\to 0 } ||f_y-f||=0?$$

Answer 1

You cannot exchange the limit and the max like that; you don't even know if the limit exists at that point! Besides, you didn't use continuity, which is essential. Note, also that you functions evaluate in $S^1$, so you cannot say $0\leq x\leq 1$. The usual assumption, that the book seems to be following, is to think of $S^1$ as $[0,2\pi]$ (as $S^1$ is not closed under addition).

What you have is that, since $f$ is continuous on $[0,2\pi]$, which is compact, it is uniformly continuous. So given $\varepsilon>0$, there exists $\delta>0$ such that $|y|<\delta$ implies $|f(x+y)-f(x)|<\varepsilon$ for all $x$; that is, $\|f_y-f\|_\infty<\varepsilon$.

Now if $f\in L^1(S^1)$ and $\varepsilon>0$ is given, there exists $g\in C(S^1)$ with $\|f-g\|_1<\varepsilon/3$. Then $$\tag{1} \|f_y-f\|_1\leq\|f_y-g_y\|_1+\|g_y-g\|_1+\|g-f\|_1<\frac{2\varepsilon}3+\|g_y-g\|_1. $$ If $y$ is small enough, then $\|g_y-g\|_\infty<\varepsilon/6\pi$. Then $$ \|g_y-g\|_1=\int_0^{2\pi}|g_y-g|\leq2\pi\|g_y-g\|_\infty<\frac{\varepsilon}3. $$ Now $(1)$ has become, if $y$ is small enough, $$ \|f_y-f\|_1<\varepsilon $$

Answer 2

I'm rather sure that the intention of the hint you are referring to asks you to prove that, for continuous $g$, $$\lim_{y\rightarrow 0} ||g_y- g||_{L^1} = 0$$

For, if that is true and then $f\in L^1$, you can find to given $\varepsilon >0$ a funktion $g^\varepsilon\in C(S)$ such that $$||f-g^\varepsilon||_{L^1} <\varepsilon$$ and simultaneously $$||f_y-g_y^\varepsilon||_{L^1} < \varepsilon$$ (why?) so that then $$||f-f_y||_{L^1} = ||f-g^\varepsilon + g^\varepsilon - g_y^\varepsilon + g_y^\varepsilon -f_y||_{L^1}$$ can be estimated easily using the triangle inequality.