Jensen's inequality for frobenius norm When I was going through a proof, I saw the following step:
$$\frac{1}{p}\operatorname{trace}(\Sigma) \leq \frac{1}{\sqrt{p}} \|\Sigma\|$$
where, $p$ is the number of variables, $\sigma$ is the population covariance matrix and $\|\cdot\|$ stands for Frobenius norm.
They stated that this from jensen's inequality. Can anyone explain me how they got this result.
Thanks alot
 A: It follows from (3) here with $\phi(t)=t^2$ which implies that for $x=[x_1,\ldots,x_p]$, we have
$$
\left(\frac{\sum_i x_i}{p}\right)^2\leq \sum_i\frac{x_i^2}{p}.
$$
You just need to replace $x$ with the vector of eigenvalues of $\Sigma$ to get the result.
EDIT: Let $\lambda_1,\ldots,\lambda_p$ be the positive eigenvalues of $\Sigma$ (since it is a covariance matrix, I suppose it is symmetric and positive definite; it works also for semi-definite matrices). Then
$$
\mathrm{trace}(\Sigma)=\sum_{i=1}^p\lambda_i, \qquad
\|\Sigma\|=\left(\sum_{i=1}^p\lambda_i^2\right)^{1/2}.
$$
So putting $\lambda$'s instead of $x$'s of the inequality above gives
$$
\left(\frac{\sum_i\lambda_i}{p}\right)^2
\leq
\sum_i\frac{\lambda_i^2}{p}
\quad\Leftrightarrow\quad
\frac{1}{p^2}[\mathrm{trace}(\Sigma)]^2\leq\frac{1}{p}\|\Sigma\|^2.
$$
Then take the square root.
EDIT: The Jensen's inequality (3) on the linked page states that
$$
\phi\left(\frac{\sum_i x_i}{n}\right)\leq \frac{\sum_i \phi(x_i)}{n}
$$
for $n$ numbers $x_1,\ldots,x_n$ in the domain of a convex function $\phi$. Just put $\phi(t)=t^2$ (which is convex).
