For $f\in [0,1]^n$ with $\sum_i f_i = 1$, define the self-convolution $g=f*f$ with $$g_i = \sum_{j,k \mid j+k \equiv i \bmod n} f_j \cdot f_k$$ and define the L2-norm as $$\|f\| = \sqrt{\sum_i f_i^2}.$$

My question is it true that $$\|g\|\le \|f\|^2 \text{?}$$

I arrived here by translating into the frequency domain where the convolution becomes multiplication as far as I understand. I also tried to get here with some facts about convolutions and Lp norms https://en.wikipedia.org/wiki/Young%27s_convolution_inequality but I got a much weaker bound.

EDIT: Sorry, my bound is definitely wrong. The L1-norm of $f,g$ is $1$, so the L2-norm will be at least $1/\sqrt{n}$. My real question: what bounds can we give for $||g||$ in terms of $||f||$?

I'd also be interested if someone can point out the problem with my original hypothesis. Here was my thought process: Let $F$ denote the fourier transform of $f$, and let $\circ$ denote point-wise multiplication. A basic fact of fourier analysis is $F\circ F = f*f$, and also $F,f$ have the same L2-norm. Now we compute $$||F\circ F||^2 = \sum_i F_i^4 \le \left(\sum_i F_i^2\right)^2$$ which seems to imply $$||f*f||\le ||f||^2.$$ Where have I gone wrong?

Also, it seems that maybe stronger bounds can be achieved for prime $n$. If so, please assume that $n$ is prime.