4
$\begingroup$

I am familiar with the conjugate function of the vector norm, which uses the concept of dual norm and is defined as follows:

$$\|\mathbf{y}\|_p^*=\max_{\mathbf{x}}\left(\mathbf{x}^T\mathbf{y}-\|\mathbf{x}\|_p\right)=\begin{cases}0~~~\|\mathbf{y}\|_q\leq 1 \\\infty ~~~\text{otherwise}\end{cases}$$ where $\frac{1}{p}+\frac{1}{q}=1$ for $p\geq 1$.

My question is:

Is there an equivalent conjugate function for the mixed matrix norm $\|\mathbf{A}\|_{p,q}$ defined for matrix $\mathbf{A}$?

$\|\mathbf{A}\|_{p,q}=\left(\sum_i \|\mathbf{a}_i\|_p^q\right)^{1/q}$ where $\mathbf{a}_i$ is the $i^{\text{th}}$ column of matrix $\mathbf{A}$.

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
7
$\begingroup$

Let $p^*$ and $q^*$ be the conjugate exponents. Some (slightly laborious) algebra shows that the dual-norm is $\|A\|_{p^*,q^*}$. The conjugate function is the indicator function for the (unit) dual-norm ball.

Improve this answer
$\endgroup$
1
  • $\begingroup$ This has also been shown in much more generality in the context of (weighted) Besov (sequence) spaces (characterised by wavelets) - which are a vast generalisation matrix spaces endowed with $\| \cdot \|_{p, q}$ - in the paper On the smoothness and convexity of Besov spaces by Kazimierski. $\endgroup$
    – ViktorStein
    Commented Mar 8, 2022 at 21:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .