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What is the subdifferential of the following norm at the origin

\begin{align} \lVert\eta\rVert_{\mathbf{K}}=(\eta^{\textrm{T}}\mathbf{K}\eta)^{\frac{1}{2}} \end{align} where $\mathbf{K}$ is a positive definite matrix

Question: What is the subdifferential at the origin?

neil
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  • Did you tried something? – Tomás Nov 10 '13 at 11:09
  • @Tomás yes, this norm is a skewed version of $\ell_2$ norm, I did some trial, but not very satisfied. I'll try to answer to it myself first. – neil Nov 10 '13 at 11:16
  • @Tomás I wonder if the answer is ${\eta\mid\lVert\eta\rVert_\mathbf{K}\le\lambda_{\min}}$, $\lambda_\min$ is the smallest eigenvalue of $\mathbf{K}$, but I don't know to prove or disprove – neil Nov 10 '13 at 11:47
  • Today I can not help you. Tomorrow, if you do not get any answer, I will try to help you. – Tomás Nov 10 '13 at 14:20
  • Hi @neil. Is $K$ a symmetric matrix or it is only positive? – Tomás Nov 11 '13 at 11:58
  • @Tomás symmetric and positive definite. I think I have figured it out in the following answer. Is it right? – neil Nov 13 '13 at 02:27

2 Answers2

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This is just a partial answer: We want to find $\mu \in \mathbb{R}^n$ such that $$\langle K\eta,\eta\rangle\geq\langle\mu, \eta\rangle^2,\ \forall\ \eta\in\mathbb{R}^n\tag{1}$$

Let $$f(\eta)=\frac{\langle K\eta,\eta\rangle}{\langle\mu, \eta\rangle^2}$$

Note that $f(\lambda\eta)=f(\eta)$, hence the problem consists in find $\mu$ in such a way that the minimum value of $f$ in $S=\{\eta\in\mathbb{R}^n:\ |\eta|=1\}$ is bigger than or equal to $1$. Define $A$ by $$A=\{\mu\in\mathbb{R}^n\ :\ \langle\mu,\eta\rangle^2\leq\lambda,\ \forall\eta\in S \}$$

where $\lambda$ is the least eigenvalue of $K$. If $\mu \in A$, we have that $\mu$ satisies $(1)$, now the question is: Is the any element outside $A$ which satisfies $(1)$?

Tomás
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\begin{align} \{\frac{\alpha\mathbf{K}\eta}{\lVert\eta\rVert_\mathbf{K}}\mid0\le\alpha\le1,\eta\ne\mathbf{0}\} \end{align}

neil
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