I am working on exercise whose first part is as follows:
Let $f\in L^{1}(\mathbb{S}^{1})$. For any $k\in\mathbb{N}$, we define $f_{k}(x):=f(kx)$. Note that $f_{k}\in L^{1}(\mathbb{S}^{1})$. Does $f_{k}$ have a limit in $L^{1}(\mathbb{S}^{1})$ under $L^{1}$ norm?
I have some attempts as follows:
Firstly, using the change of variable $y:=kx$, we have $$\int_{-\pi}^{\pi}f_{k}(x)dx=\int_{-\pi}^{\pi}f(kx)dx=\int_{-k\pi}^{k\pi}f(y)\dfrac{1}{k}dy=\dfrac{1}{k}\int_{-k\pi}^{k\pi}f(y)dy.$$
Now, since $f(y)$ is a $2\pi-$perodic function, and $[-k\pi,k\pi]$ contains $k$ copies of $2\pi$, we must have $$\dfrac{1}{k}\int_{-k\pi}^{k\pi}f(y)dy=\dfrac{1}{k}\cdot k\int_{-\pi}^{\pi}f(y)dy=\int_{-\pi}^{\pi}f(y)dy.$$
This implies that $$\lim_{k\rightarrow\infty}\int_{-\pi}^{\pi}f_{k}(x)dx=\int_{-\pi}^{\pi}f(y)dy.$$
However, I am not sure if this implies that $\|f_{k}-f\|_{L_{1}}\longrightarrow 0$. What should I do to step further?
Thank you!
Let, for a continuous function $g$ defined on $S^1 \sim [0,1)$, $S^n_g(x)=\frac{1}{n}\sum_{k=0}^{n-1}{g(x+k/n)}$.
Then $f_ng$ and $fS^n_g$ have the same integral.
Assume $f$ bounded. Assume that $f_n$ converges in $L^1$ (to some $u$) and let us show that $f$ is constant. Up to subtracting a constant, we may assume that $f$ has integral zero. As $\|f_n\|_{L^1}=\|f\|_{L^1}$ goes to $\|u\|_{L^1}$, all we need to show is that $f_n$ goes weakly to zero (because it implies $u=0$ thus $\|u\|_{L^1}=0$).
Now, let $g$ be a continuous function on $S^1$. It follows from uniform continuity that $S^n_g$ converges uniformly to a constant, so the integral of $f_ng$ converges to a multiple of the integral of $f$ so goes to zero.
If $g$ is simply $L^{\infty}$, let $h$ be continuous, then $$|\int{f_ng}| \leq |\int{f_n(g-h)}|+|\int{f_nh}| \leq \|f\|_{\infty}\|g-h\|_{L^1}+o(1).$$ Therefore by standard arguments $\int{f_ng} \rightarrow 0$ and we are done.
If $f$ is unbounded, apply the result to its arctangent (because $\arctan$ is Lipschitz continuous, bounded, and odd).
Edit: a shorter argument. Let, for $g \in L^1$, $a_k(g)$ be the $k$-th Fourier coefficient of $g$. It is well-known that $|a_k(g)| \leq C\|g\|_{L^1}$ for some universal constant $C$ and that for any $g$, $a_k(g)$ goes to zero as $k$ goes to infinity.
Assume that $f_{n_k}$ converges to $g$. Then for every nonzero $p \in \mathbb{Z}$, as $k \rightarrow \infty$, $$|a_p(f)|=|a_{pn_k}{f_{n_k}}| \leq |a_{pn_k}(g)|+|a_{pn_k}(f_{n_k}-g)| \leq o(1)+C\|f_{n_k}-g\|_{L^1}=o(1).$$
So each nonconstant Fourier coefficient of $f$ is zero. So $f$ is constant.