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I found the following definition in this answer $$ \partial f(x) := \{x^* \in X^* \mid f(x') \ge f(x) + \langle x^*, x'-x\rangle\;\forall x' \in X\} $$

Can I define this $$ \langle A, B\rangle\ = Re(tr(B^H A))$$ and rewrite this definition

$\partial\|A\| := \{S\in \mathbb R^{m \times n} \quad \big| \quad \|B\|\geq\|A\| + tr((B-A)^T S),\forall B \in \mathbb R^{m \times n}\}$

in Characterization of the Subdifferential of Some Matrix Norms by G. A. Watson to

$\partial\|A\| := \{S\in \mathbb C^{m \times n} \quad \big| \quad \|B\|\geq\|A\| + Re(tr((B-A)^H S)),\forall B \in \mathbb C^{m \times n}\}$

I tried to find any paper where this was written in this way but found nothing. Any idea whether what I did here is correct?

I used this to find an element of $\partial\|A\|_2$ of a rank 1 hermitian matrix.

Med B
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  • The expression $\langle A, B\rangle\ = Re(tr(B^H A))$ is not an inner product so the 'definitions' above are not subdifferentials. The function $f(z) = |z|$ is not complex differentiable, so expecting $|\cdot|$ to have a subdifferential is probably unrealistic. You may want to consider $\mathbb{C}^n$ as two copies of $\mathbb{R}^n$ and deal with the real case. – copper.hat May 28 '22 at 05:07
  • [link] (https://www.sciencedirect.com/science/article/pii/0024379593902097?via%3Dihub) here I found the "inner product" (it is based on a Hermitian product and respects all properties for inner product). – Med B May 28 '22 at 07:53
  • It cannot be linear over $\mathbb{C}$ in either argument hence cannot be an inner product. – copper.hat May 28 '22 at 20:50
  • $\langle A, B\rangle\ = Re(tr(B^H A))$ gives back real numbers so for linearity wouldn't you use a read number r?

    if so $\langle A+rC, B\rangle\ = Re(tr(B^H (A+rC)))=Re(tr(B^H A+r B^HC)))=Re(tr(B^H A))+Re(tr(r B^HC)))=Re(tr(B^H A))+rRe(tr( B^HC)))$

     – Med B May 29 '22 at 07:29
  • the variables A and B are complex matricies but the function maps back to $\mathbb R$. https://en.wikipedia.org/wiki/Inner_product_space#Definition I checked the 3 points in here: Conjugate symmetry, Linearity in the first argument, Positive-definiteness. – Med B May 29 '22 at 07:38

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