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$f(x)=\frac{1}{2}x^\intercal W x$, 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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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 $W$.

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