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show-that-the-operator-norm-is-submultiplicative.

We all know that Cauchy-Schwartz inequality works for vector inner product. Also, it works for the matrix innner product. Specifically, the following is the relation between matrix inner product and matrix norms $$\tag{1} |<A,B>|^2\le\|A\|^2\|B\|^2 $$ when $A=cB$, the equality holds. The definition of inner product for matrices is $$\tag{2} <A,B>=\text{vec}(A)^H\text{vec}(B)=\text{tr}(A^HB) $$

If the LHS of (1) is replaced by induced matrix norms and matrix product, does the Cauchy Schwartz inequality still work, like the following induced $L_2$-norm(also known as, spectral norm) $$\tag{3} \|AB\|_2^2\le \|A\|_2^2 \|B\|_2^2 $$ where $\|A\|_2=\sup_{u\neq0}\frac{\|Au\|_2}{\|u\|_2}$. Since $\|A\|_2=\max\{\text{eig(A)}\}=\lambda_{\text{max}}(A)$, if (3) holds, it would imply the following inequality $$\tag{4} \lambda_{\text{max}}(AB)^2\le \lambda_{\text{max}}(A)^2\lambda_{\text{max}}(B)^2 $$

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    $\begingroup$ Please provide the definition of that norm, and some thoughts on why it might/might not work $\endgroup$ Commented Oct 20, 2021 at 15:38
  • $\begingroup$ This is not the CS inequality, but the result holds. $\endgroup$ Commented Oct 20, 2021 at 15:46
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    $\begingroup$ This is known as submultiplicativity. $\endgroup$
    – While I Am
    Commented Oct 20, 2021 at 16:06

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Those types of norms are called submultiplicative. Not every norm is submultiplicative, but every operator norm is.

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