Uniformly convex and strictly convex

Question

I have the following definitions of uniformly convex and strongly convex

Let $f:R^n \to R$ be smooth.

(1) $f$ is uniformly convex if there exists $\theta > 0 $ such that $$\Sigma_{i,j}f_{x_i x_j} \xi_i \xi_j \geq \theta |\xi|^2 \tag1$$ for every $x,\xi \in R^n$.

(2) $f$ is strictly convex if $$\Sigma_{i,j}f_{x_i x_j} \xi_i \xi_j > 0 \tag2$$ for every $x \in R^n$ and every $0 \neq \xi \in R^n $.

I can see that uniformly convex implies strictly convex, but what is an example of a strictly convex function that is not uniformly convex? Also, is there a stronger notion than uniformly convex?

Interesting choice of terminology. In my experience, the first definition is strong convexity and the second is strict convexity. — Michael Grant, Apr 25 '15 at 00:59
I got the first definition from Evan's Partial Differential equations and as for the second definitinon I mistakenly put down strongly instead of strictly. I just fixed it — mononono, Apr 25 '15 at 03:06
Ah, of course, strong convexity is a special case of uniform convexity. If by $|\xi|$ above you mean the Euclidean norm, then you have strong convexity. — Michael Grant, Apr 25 '15 at 03:22

score 4 · Accepted Answer · answered Apr 25 '15 at 12:43

$f(x)=e^x$ is strictly convex but not strongly or uniformly convex. The scalar definition of uniform convexity reduces to $f''(x)>\theta$. For any fixed $\theta>0$, this inequality fails for any $x\le\log(\theta)$.

Indeed, consider any smooth, positive function that approaches zero asymptotically; integrate twice (from, say, the origin) and you have another counterexample.

If you need a multivariate example then just consider $f(x)=\sum_i e^{x_i}.$

score 4 · Answer 2 · answered Apr 25 '15 at 13:29

4

My favorite example is $$ f(x) =\sqrt{1+|x|^2}. $$

The Hessian is always positive definite, but degenerates for $|x|\to\infty$.

answered Apr 25 '15 at 13:29

Dirk

11,680

Uniformly convex and strictly convex

2 Answers2