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Assume $A\in\mathbb{R}^{m\times n}$ is a $m$ by $n$ matrix. Let $|\cdot|$ denote the Frobenius norm of matrix. Define function $f:\mathbb{R}^{m\times n}\to\mathbb{R}$ as $f(A):=|A^TA|$. Is $f$ a convex function?

Intuitively, I think this function is a composition of a norm and something with quadratic structure, and thus should be convex.

To prove this, $f(tA+(1-t)B)=|t^2A^TA+t(1-t)(A^TB+B^TA)+(1-t)^2B^TB|\le t^2|A^TA|+(1-t)^2|B^TB|+t(1-t)|A^TB+B^TA|$.

If $|A^TB+B^TA|\le|A^TA|+|B^TB|$, then we get the convexity. Is this inequality true?

Rodrigo de Azevedo
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Syoung
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    For your last question, think about it in the reversed way. Since f is intuitively convex, then the inequality is correct. Observe that $A^T A$ is real symmetric, and positive semidefinite. – GNUSupporter 8964民主女神 地下教會 May 07 '19 at 01:45
  • @GNUSupporter8964民主女神地下教會: The problem with this approach is that "intuitively convex functions" may fail to be convex... – gerw May 17 '19 at 06:48

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We have $$ |A^\top B|^2 = \operatorname{trace}(B^\top A A^\top B) = \operatorname{trace}(B B^\top A A^\top) = (B B^\top, A A^\top)_F \le |B B^\top| \, |A A^\top| = |B^\top B| \, |A^\top A|.$$

From this, we get $$|A^\top B + B^\top A| \le |A^\top B| + |B^\top A| \le 2 \, |B^\top B|^{1/2} \, |A^\top A|^{1/2} \le |A^\top A| + |B^\top B|.$$

gerw
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