Standard deviation of coefficients of a polynomial: Is it true that $\sigma_{f+g} \leq \sigma_f + \sigma_g$?

Question

For a fixed $N$, let $f=f_0+f_1X+\cdots+f_{N-1}X^{N-1}$ and $g=g_0+g_1X+\cdots+g_{N-1}X^{N-1}$ be two polynomials in $\mathbb{Z}[X]$.

Consider the standard deviation of the coefficients of $f$, $g$ and $f+g$, defined by $\sigma_f$, $\sigma_g$ and $\sigma_{f+g}$ respectively.

We have $\sigma_f = \frac{1}{\sqrt{N}}\sqrt{\sum_{i=0}^{N-1}{(f_i-\bar{f})^2}}$ where $\bar{f}$ is the mean of the coefficients of $f$.

Is it true that $\sigma_{f+g} \leq \sigma_f + \sigma_g$ ?

When I tried to prove this, assuming that this is true, I developed squares but I get an extra factor $2ab$ that is ruining the proof. And this proposition seems to be true, so there must be a way of proving it.

Can someone help? Thanks in advance.

Answer 1

This is an application of the triangle inequality for the $2-$norm (or Euclidean norm, if you prefer). Define the vectors \begin{align} \mathbf{f}&=(f_0,f_1,\ldots,f_{N-1}), \\ \mathbf{\overline{f}}&=(\overline{f},\overline{f},\ldots,\overline{f}), \\ \mathbf{g}&=(g_0,g_1,\ldots,g_{N-1}), \\ \mathbf{\overline{g}}&=(\overline{g},\overline{g},\ldots,\overline{g}), \end{align} with all vectors of length $N$. Now the $2-$norm of a vector is defined by $$\|\mathbf f\|_2=\sqrt{\sum_{i=0}^{N-1}f_i^2 }$$ So, in your notation, $$\sigma_f=\frac{\|\mathbf f-\mathbf{\overline{f}}\|_2}{\sqrt{N}}.$$ Now, using the triangle inequality, we have \begin{align} \sigma_{f+g} &=\frac{\|(\mathbf f+\mathbf g)-(\mathbf{\overline{f}}+\mathbf{\overline{g}})\|_2}{\sqrt{N}}\\ &=\frac{\|(\mathbf f-\mathbf{\overline{f}})+(\mathbf g-\mathbf{\overline{g}})\|_2}{\sqrt{N}}\\ &\leq \frac{\|\mathbf f-\mathbf{\overline{f}}\|_2}{\sqrt{N}}+\frac{\|\mathbf g-\mathbf{\overline{g}}\|_2}{\sqrt{N}}\\ &=\sigma_f+\sigma_g. \end{align}