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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:

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

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

2 Answers 2

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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 $\|A\|_1 =\|A^H\|_\infty$.

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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.

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