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If we extend $f_i$ for all $i\in\mathbb Z$ to be $n$-periodic, we may simply write $g_i=\sum_{j=1}^n f_jf_{i-j}$. By Cauchy-Schwarz, $|g_i|\le \sum_{j=1}^n|f_j|^2$ and $\|g\|\le \sqrt n \|f\|^2$. This is sharp because for $f_i=1/n$ for all $1\le i\le n$ we get the equality.

By Cauchy-Schwarz, $|g_i|\le \sum_{j=1}^n|f_j|^2$ and $\|g\|\le \sqrt n \|f\|^2$. This is sharp because for $f_i=1/n$ for all $1\le i\le n$ we get the equality.

If we extend $f_i$ for all $i\in\mathbb Z$ to be $n$-periodic, we may simply write $g_i=\sum_{j=1}^n f_jf_{i-j}$. By Cauchy-Schwarz, $|g_i|\le \sum_{j=1}^n|f_j|^2$ and $\|g\|\le \sqrt n \|f\|^2$. This is sharp because for $f_i=1$ for all $1\le i\le n$ we get the equality.

If we extend $f_i$ for all $i\in\mathbb Z$ to be $n$-periodic, we may simply write $g_i=\sum_{j=1}^n f_jf_{i-j}$. By Cauchy-Schwarz, $|g_i|\le \sum_{j=1}^n|f_j|^2$ and $\|g\|\le \sqrt n \|f\|^2$. This is sharp because for $f_i=1/n$ for all $1\le i\le n$ we get the equality.

If we extend $f_i$ for all $i\in\mathbb Z$ to be $n$-periodic, we may simply write $g_i=\sum_{j=1}^n f_jf_{i-j}$. By Cauchy-Schwarz, $|g_i|\le \sum_{j=1}^n|f_j|^2$ and $\|g\|\le \sqrt n \|f\|^2$. This is sharp because for $g_i=1$ for all $1\le i\le n$ we get the equality.

If we extend $f_i$ for all $i\in\mathbb Z$ to be $n$-periodic, we may simply write $g_i=\sum_{j=1}^n f_jf_{i-j}$. By Cauchy-Schwarz, $|g_i|\le \sum_{j=1}^n|f_j|^2$ and $\|g\|\le \sqrt n \|f\|^2$. This is sharp because for $f_i=1$ for all $1\le i\le n$ we get the equality.