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Let $A$ be a square Matrix and $||\cdot ||_p$ the induced Matrixnorm for $1 \leq p \leq \infty$. Is it true that $$||A||_p\leq \max(||A||_1,||A||_{\infty})?$$ For $p=2$ the answer is yes because $||A||_2^2\leq ||A||_1||A||_{\infty}$.

The motivation is that I have a family of matrices where I can bound the infinity and 1-Norm but I dont know how to bound the $p$-Norm.

Thx for any help in advance

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closed as off-topic by Bill Johnson, David Roberts, Willie Wong, Andrey Rekalo, David White Jun 27 '13 at 18:22

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up vote 4 down vote accepted

Yes, by the Riesz-Thorin theorem.

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Thank you very much – user35593 Jun 25 '13 at 20:05

For the Shatten $p$-norm (i.e. $\|A\|_p=tr_n(|A|^p)^{1/p}$) it is true. This follows from the positivity of the trace: $tr_n(|A|^p)^{1/p}=tr_n(\sqrt{A^*A}^{p})^{1/p}=tr_n((A^*A)^{\frac{p-1}{2}}(A^*A)^{\frac{1}{2}})^{1/p}\leq (\|A\|_{\infty}^{p-1} tr_n(|A|))^{1/p}\leq{\max(\|A\|_1,\|A\|_{\infty})}$

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It is not clear to me what the OP means by $\|\cdot\|_p$. – Bill Johnson Jun 25 '13 at 19:03
"Induced matrix norm" normally means the $\ell_p^n \to \ell_p^n$ operator norm. – Mark Meckes Jun 25 '13 at 19:34
yes I meant the operator norm – user35593 Jun 25 '13 at 19:53

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