Show that $ \lVert A \rVert_2^2 \leq \lVert A \rVert _1 \lVert A \rVert _ \infty $ With the definition of $ \lVert A \rVert_2$ and $\lVert A \rVert_1$ and $\lVert A \rVert_ \infty$ that is:
\begin{gather}
\lVert A\rVert_1 = \max_{j} \sum_{i=1}^m \lvert a_{ij}\rvert\\
\lVert A\rVert_2 = \sqrt{\rho(A^HA)}\\
\lVert A\rVert_\infty = \max_{i} \sum_{j=1}^n \lvert a_{ij}\rvert
\end{gather}
prove that:
$$\lVert A \rVert_2^2 \leq  \lVert A \rVert_1 \lVert A \rVert_\infty$$
 A: Let $\|\cdot\|$ be any matrix norm induced by a vector norm. Then we have $$\|A\|_2^2= \rho(AA^H) \leq \|AA^H\| \leq \|A\|\|A^H\|.$$
Here the first inequality follows from a "famous theorem" (see e.g. Proposition 4.4) and the second inequality follows from the fact that $\|\cdot\|_\infty$ is a matrix norm induced by a vector norm and thus is submultiplicative. Finally note that $\|A\|_1 =\|A^H\|_\infty$.
A: For any $A\in\mathbb{M}_n(\mathbb{C})$ and for any $1\leq p\leq\infty$, denote by $\|A\|_p:=\sup_{|x|_p=1}|Ax|_p$, where $|x|^p_p=\sum^n_{k=1}|x_k|^p$ if $1\leq p<\infty$ and $|x|_\infty=\max_{1\leq k\leq n}|x_k|$.
For any $x\in\mathbb{C}^n$ and $1\leq p_0,p_1\leq\infty$
$$\|Ax\|_{p_0}\leq \|A\|_{p_0}|x|_{p_0}, \qquad \|Ax\|_{p_1}\leq \|A\|_{p_1}|x|_{p_1}$$
By the Riesz-thorin interpolation theorem for any $0<t<1$ and $\frac{1}{p_t}=\frac{1-t}{p_0}+\frac{t}{p_1}$,  $A$ (as an operator from $(\mathbb{C}^n,|\;|_{p_t})$ to itself) satisfies
$$|A x|_{p_t}\leq (\|A\|_{p_0})^{1-t}(\|A\|_{p_1})^t|x|_{p_t}$$
In particular, for  $p_0=1$ and $p_1=\infty$ we have that
$$|A x|_p\leq (\|A\|_1)^{\tfrac1p}(\|A\|_\infty)^{1-\tfrac1p}|x|_p$$
for any $1<p<\infty$. In other words,
$$\|A\|_p\leq (\|A\|_1)^{\tfrac1p}(\|A\|_\infty)^{1-\tfrac1p}$$
The particular choice $p=2$ gives the result in the OP.
