inequality between matrix norms A is a $n\times k $ matrix.
I have to show that $\|A\|_2\leq \sqrt{\|A\|_1\cdot \|A\|_\infty}$. 
I know that $\|A\|_2^2 = \rho(A^H\cdot A)\leq \|A^H \cdot A\| $ for every $\| \cdot  \|$ submultiplicative matrix norm, but I don't know how to conclude.
Any idea? 
 A: As you noted that you now the inequality $\|A\|_2^2=\rho(A^HA)$ then it is assumed that you work with the standard induced matrix norms.
The definition is as follows:
Let $B\in \mathbb R^{n\times k}$ and let $\|.\|$ is some vector norm in $\mathbb R^k$. Then the induced matrix norm, denoted again with $\|.\|$ is:
$$\|B\|=\sup\limits_{\vec x\in \mathbb R^k}{\frac{\|B\vec x\|}{\|\vec x\|}}$$
So we have:
$$\|A\|_2^2=\rho(A^HA)\leq \|A^HA\|_\infty\leq \|A^H\|_\infty\cdot \|A\|_\infty =\|A\|_1\cdot \|A\|_\infty$$.
The first inequality is known for you, as you said, and it is shown like this:
Let $\vec x$ be an eigenvector of $B$ for the eigenvalue $\lambda$, where $\max\limits_{i}|\lambda_i|=|\lambda|$. Then for arbitrary matrix norm $\|.\|$, subordinate to the vector norm $\|.\|$, we have $\|B\|=\max\limits_{\vec y\neq \vec 0} \frac{\|B\vec y\|}{\|\vec y\|}\ge \frac{\|B\vec x\|}{\|\vec x\|}=\frac{\|\lambda \vec x\|}{\|\vec x\|}=|\lambda|$ 
The second inequality is because the $\|.\|_\infty$ is induced norm from the vector $\infty$ - norm, so it is sub-multiplicative.
The last equality is because $\|A^H\|_\infty$ is the maximum absolute row sum of the matrix $A^H$, which is equal to the maximum absolute column sum of $A$ which is equal to the $\|A\|_1$ .
A: or use  $\|A\|_2^2 = \rho(A^H\cdot A)\leq \|A^H \cdot A\| _\infty\leq\|A^H\|_\infty \cdot \|A\|_\infty \leq \|A\|_1.\|A\|_\infty$
Because $\|A^H\|_\infty=\|A\|_1 $ 
A: $\|A\|_{2}^2\leq trac(A^H\cdot A)\leq{\Vert A^H\cdot A\Vert_1\leq  \| A\|_1\|A\| _\infty}$. 
for more information see $D.46$ and $D.52$ of abstract harmonic analysis hewitt&ross pages 706 and 709
