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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.

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  • $\begingroup$ Wouldn't $f(x) = x$ be a much simpler example of a convex but not strongly convex function? $\endgroup$ Apr 13, 2020 at 8:46
  • $\begingroup$ @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. $\endgroup$
    – alicec
    Apr 13, 2020 at 16:02

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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.

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