$f(x)=\frac{1}{2}x^\intercal W x$, where $W$ is positive definite, not necessarily the identity matrix. $\nabla^2 f(x)=W$. Is $f(x)$ strongly convex with the parameter $m={}?$
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https://math.stackexchange.com/q/1367642 This may help – David M. Jun 08 '18 at 01:29
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Strong convexity of $f$ with parameter $m$ is equivalent to $\nabla^2 f(x) \succeq m I$, so in your case, $f$ is strongly convex with parameter $m$ being the smallest eigenvalue of $W$.
angryavian
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