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It is well-known that $l_2$ norm squared, $f(x) = \frac{1}{2}\|x\|_2^2$, is strongly convex with respect to $l_2$ norm. My question is what about other non-Euclidean norm squared, such as $\frac{1}{2}\|x\|^2_1$, $\frac{1}{2}\|x\|^2_{\infty}$. Are they strongly convex w.r.t. itself? ($\frac{1}{2}\|x\|_1^2$ strongly convex w.r.t. $\|\cdot\|_1$, etc.). Here the definition of strong convexity w.r.t. norm $\|\cdot\|$ is given by

$$f(\lambda x+ (1-\lambda)y) \leq \lambda f(x) + (1-\lambda)f(y)-\frac{\lambda(1-\lambda)}{2}\|x-y\|^2 \quad \forall x,y$$

Actually, are they even strictly convex?

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  • $\begingroup$ Isn't this a consequence of the fact that all norms on $\mathbb{R}^n$ are equivalent? $\endgroup$ – madnessweasley Aug 8 '19 at 20:20
  • $\begingroup$ @madnessweasley Strong/Strict convexity are not topological properties; they are metric properties, so they do not remain invariant under equivalence of norms, which only guarantees topological equivalence. $\endgroup$ – uniquesolution Aug 8 '19 at 20:34
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Claim. Assume that a certain norm $\|\cdot\|$ is not strictly convex. That is, there exist points $x,y$ such that $x\neq y$ and $$1=\|x\|=\|y\|=\|\frac{x+y}{2}\|$$ Consider the function $f(x)=\frac{1}{2}\|x\|^2$. Then $f(x)$ is not strongly convex with respect to the norm.

Proof. Choose $x,y$ as above and $\lambda = \frac{1}{2}$, we have $$f(\lambda x+(1-\lambda)y)=\frac{1}{2}\|\frac{x+y}{2}\|^2=\frac{1}{2}$$ and $$(1-\lambda)f(x)+(1-\lambda)f(y)=\frac{1}{2}\cdot\frac{1}{2}\|x\|^2+\frac{1}{2}\cdot\frac{1}{2}\|y\|^2=\frac{1}{2}$$ but $$\frac{\lambda(1-\lambda)}{2}\|x-y\|^2=\frac{1}{8}\|x-y\|^2>0,$$ because $x\neq y$. So the inequality defining strong convexity fails. Q.E.D

Since both the $\|\cdot\|_1$ and the $\|\cdot\|_{\infty}$ are not strictly convex, it follows that the functions in your question are not strongly convex with respect to their respective norms.

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  • $\begingroup$ Thanks! You actually answered both of my questions. $\endgroup$ – Sean Ian Aug 8 '19 at 20:44

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