# 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:

$$\lVert A \rVert_2^2 \leq \lVert A \rVert_1 \lVert A \rVert_\infty$$

• Rather than \parallel, use \lVert and \rVert for the norms, that gives proper spacing when rendered. Feb 18, 2015 at 13:11

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 $$\|A\|_1 =\|A^H\|_\infty$$.
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 $$0 and $$\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. In other words, $$\|A\|_p\leq (\|A\|_1)^{\tfrac1p}(\|A\|_\infty)^{1-\tfrac1p}$$ The particular choice $$p=2$$ gives the result in the OP.