# Is there a strictly convex function on a finite domain that is not strongly convex?

A strictly convex function on $$D\subseteq\mathbb{R}$$ is called strongly convex if there exists $$\beta>0$$ such that if $$\theta \in [0,1], x\neq x'$$ $$f(\theta x + (1-\theta) x') \le \theta f(x) + (1-\theta) f(x') - \frac{\beta}{2}\theta(1-\theta)(x-x')^2.$$ I understand that the exponential function is a convex function that is not strongly convex. I am trying to come up with such an example on a finite domain.

My intuitive understanding is that a strongly convex function cannot be arbitrarily close to constant anywhere on the domain, which is what happens with the exponential function, as $$x$$ tends to negative infinity. Can it happen if $$x$$ does not tend to either infinity?

I would greatly appreciate any comment or suggestions. Thank you.

• Wouldn't $f(x) = x$ be a much simpler example of a convex but not strongly convex function? Apr 13, 2020 at 8:46
• @ClementYung Thank you! I think that's a convex function but not strictly convex. I am wondering if there's a strictly convex function that satisfies the conditions, or if it can be disproved. Apr 13, 2020 at 16:02

The function $$f \colon [0, 1] \to \mathbb R$$, $$x \mapsto x^4$$ is strictly convex, but not strongly.
If is strictly convex: we have $$f'(x) = 4 x^3$$ and thus $$\big( f'(x) - f'(y) \big) ( x - y) = 4 (x^3 - y^3)(x - y) > 0$$ for all $$x, y \in [0, 1]$$ with $$x \ne y$$.
If it were strongly convex, then there is an $$m > 0$$ such that $$(f'(x) - f'(y))(x - y) \ge m (x - y)^2$$ for all $$x, y \in [0, 1]$$. Supposing without loss of generality $$x > y$$ we can divide this inequality by $$x - y$$ to obtain $$4 (x^3 - y^3) \ge m (x - y)$$ i.e. $$4 (x^2 + x y + y^2) \ge m$$, but this is not true.