Fourier coefficients and series of functions We define $f_n\in C(\mathbb{T})$ by
$$f_1=f, \quad f_{n+1}=f_{n}*f$$
where $f$ is a function s.t
$$\frac{1}{2\pi}\int^{\pi}_{-\pi}|f(x)|^2 dx \leq 1$$
Now we define
$$I_n=\frac{1}{2\pi}\int^{\pi}_{-\pi}|f_n(x)|^2 dx$$
and we want to prove that either $I_n\rightarrow 0$ or $I_n\rightarrow 1$. Using Parsevals equality, I get that
\begin{align}
I_n &= \frac{1}{2\pi}\int^{\pi}_{-\pi}|f_n(x)|^2 dx \\
&=\sum_{k\in\mathbb{Z}}|\tilde f_n(k)|^2 \\
&=\sum_{k\in\mathbb{Z}}|\tilde f_1(k)|^{2n} \\
&\leq \left(\sum_{k\in\mathbb{Z}}|\tilde f(k)|^2 \right)^n
\end{align}
where I used $\tilde f(k)$ to denote the $k$th fourier coefficient, and that $\tilde{(f_1*f_2)}(k) = \tilde f_1(k) \tilde f_2(k)$.
Using this, if
$$\frac{1}{2\pi}\int^{\pi}_{-\pi}|f(x)|^2 dx < 1$$
then showing $I_n\rightarrow 0$ is quite easy. However, I'm having troubles handling the case where
$$\frac{1}{2\pi}\int^{\pi}_{-\pi}|f(x)|^2 dx = 1$$
I tried proving by induction that for each $n\in\mathbb{N}$, $I_n=1$. For the case $n=1$ it's obvious from the assumption, and by assuming for $n$ I want to conclude that it happens for $n+1$. But, how do I go from
$$\sum_{k\in\mathbb{Z}}|\tilde f_n(k)|^2 \rightarrow \sum_{k\in\mathbb{Z}}|\tilde f_{n+1}(k)|^2$$
? Is it just using Cauchy's convolution between $\sum_{k\in\mathbb{Z}}|\tilde f_{n}(k)|^2$ and $\sum_{k\in\mathbb{Z}}|\tilde f_1(k)|^2$?
 A: As per OP, we consider the case $I_1=1$ or $\sum_{k \in \mathbb Z} |a_k|^2=1$, where $a_k=\hat f_1(k)$
There are two cases; if there is $m, |a_m|=1$ then $a_k=0, k \ne m$ and clearly all $I_n=1$ so done!
Assume now $|a_k|<1, k \in \mathbb Z$; then pick a fixed $m$ for which $0<|a_m|<1$; then since $\sum_{k \ne m} |a_k|^2=A<1$ we get $\sum_{k \ne m} |a_k|^{2n} \le A^{2n} \to 0$, while of course $|a_m|^{2n} \to 0$ also, so we are done too!
A: Note that $\hat f_n(k)=\hat f_1(k)^n$, so
$\newcommand{\pint}{\frac1{2\pi}\int_{-\pi}^\pi}$ $$I_n=\sum_k|\hat f_1(k)|^{2n}.$$ In particular $$\sum|\hat f_1(k)|^2=I_1\le1,$$so $|\hat f_1(k)|\le1$ for all $k$.


Case 1: $f=0$.


I'll leave this case to you...


Case II: There exist two distinct values of $k$ with $\hat f_1(k)\ne0$.


In this case it follows that $|\hat f_1(k)|<1$ for every $k$, and since $\hat f_1(k)\to0$ there exists $c<1$ with $|\hat f_1(k)|\le c$ for all $k$. Hence $$I_n\le c^{2(n-1)}I_1\to0.$$


Case III: There exists $k_0$ such that $\hat f(k)=0$ for all $k\ne k_0$.


In this case $$I_n=|\hat f(k_0)|^{2n},$$which tends to $0$ or$1$.
