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Assume that $Q$ is a positive definite matrix, is it true to say that the function $f(v)=v^TQv$ is strongly convex with respect to the norm $||u||=\sqrt{u^TQu}$?

Thanks

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  • $\begingroup$ I think yes, as matter of eigen values determined by Q $\endgroup$ – Cardinal Jul 20 '15 at 14:52
  • $\begingroup$ Yes, That's my intuition too, but I can't prove it $\endgroup$ – Raba Poco Jul 20 '15 at 14:54
  • $\begingroup$ Why ? Just try with Hessian Matrix $\endgroup$ – Cardinal Jul 20 '15 at 14:56
  • $\begingroup$ The Hessian will work only for the euclidean norm no? $\endgroup$ – Raba Poco Jul 20 '15 at 14:58
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    $\begingroup$ Apply a change in basis, and this amounts to the question of the convexity of $f(x) = x^Tx$ with respect to the usual norm, $\|x\| = \sqrt{x^Tx}$. $\endgroup$ – Ben Grossmann Jul 20 '15 at 18:05
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Let $\lambda \in [0,1]$, $v_1, v_2 \in \mathbb{R}^n$, and $0 < Q \in \mathbb{R}^{n \times n}$. Denote ${\bar \lambda} := 1 - \lambda$ and $\|v\|^2 := v^T Q v$. Then, \begin{align} f(\lambda v_1 + \bar \lambda v_2) &= \lambda^2 \|v_1\|^2 + {\bar \lambda}^2 \|v_2\|^2 + 2\lambda \cdot {\bar \lambda} \cdot v_1^T Q v_2 \\ &= \lambda\, \|v_1\|^2 + {\bar \lambda} \, \|v_2\|^2 - \lambda \cdot \bar \lambda \cdot \Big ( \|v_1\|^2 + \|v_2\|^2 - 2 v_1^T Q v_2 \Big ) \\ &= \lambda\; f(v_1) + {\bar \lambda} \; f(v_2) - \lambda \cdot \bar \lambda \cdot \|v_1 - v_2\|^2. \end{align} Therefore, $f(v) = \|v\|^2 = v^T Q v$ is strongly convex (with respect to $\|v\| = \sqrt{v^T Q v}$).

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