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Suppose $f$ is strongly convex and twice differentiable on some interval $I$:

(1) $\forall x,y \in I: \forall t \in (0,1) \implies f(tx + (1-t)y)<tf(x)+(1-t)f(y)$

(2) $\forall x \in I \implies f''(x) \in \mathbb{R}$

My question is what exactly can we say about $f''$ on $I$? For example obviously, $f'' \geq 0$, but there's more to it for the values of $x$ when $f''(x) = 0$. For example, $f(x) = x$ is definitely not strongly convex on any interval, but $f'' \geq 0: \forall x \in \mathbb{R}$.

I think a sufficient condition would be if $f''(z) = 0 \implies \forall \delta > 0: \exists y_1,y_2 \in (z-\delta,z+\delta):y_1<z<y_2$ and $f''(y_1) > 0 < f''(y_2)$. Or in other words, $f$ is not linear at $z$. This, combined with $f'' \geq 0$ on $I$ should ensure strong convexity.

I would like to clarify my understanding by knowing a near-miss example(s). For example, what if $f$ is a function such that is convex on $\mathbb{R}$ but, say, $f''(0) = 0$ and $\forall \delta > 0: \exists x_1, x_2 \in (-\delta, \delta): x_1 < 0 < x_2:f''(x_1)=0=f''(x_2)?$ In my head I see something that looks kind of like $f''(x) = \sin(\frac{1}{x}) + 1$, but I would like an example where I can express $f$ in elementary functions if possible?

  • A differentiable function $g$ on an interval $I \subset \mathbb{R}$ is increasing if and only if $g' > 0$ outside a set of empty interior. You can prove this result with the mean value theorem. Apply it with $g = f'$. – halbaroth Oct 31 '20 at 03:13
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    I think most people call this notion "strictly convex." "Strongly convex" refers to something slightly different. – angryavian Oct 31 '20 at 03:17
  • @angryavian Yes you're right I meant strictly convex. – SenZen Oct 31 '20 at 08:27
  • @halbaroth Is {1} a set of empty interior? – SenZen Oct 31 '20 at 08:29
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    A set of $\mathbb{R}$ is of empty interior if and only if it does not contain a non-empty open interval. So the interior of ${ 1 }$ is empty. – halbaroth Oct 31 '20 at 08:33
  • @halbatoth I'm more interested in finding a example where $f$ is strictly increasing on $\mathbb{R}$, but at some point, say $x=0$, $f'$ hits 0 infinitely many times withing any interval containing 0. Something that looks like $f'(x) = 1 + \sin(\frac{1}{x})$ but which I can integrate. – SenZen Oct 31 '20 at 08:35
  • I suggest $g(x)=x^4\sin^2(\frac{1}{x})$. The function $g$ is non-negative and continuous on $\mathbb{R}$. Now integrate $g$ twice. Let $f$ denote the result. You don't need to find a closed-form expression of $f$. If my above suggestion is right, the function $f$ should be strictly convex on $\mathbb{R}$ but $f''$ vanishes (not identically) in any neighbourhood of the origin. – halbaroth Oct 31 '20 at 08:52
  • Thanks unfortunately I havent actually studied integration yet, that's why I want a closed form $g$, so I don't have to rely on something I don't know. Otherwise couldn't I just double integrate $1+ \sin(x)$? – SenZen Oct 31 '20 at 10:29
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    Yes, you can integrate twice your function. It doesn't seem easy to construct an explicit example using only elementary functions. – halbaroth Oct 31 '20 at 11:59

1 Answers1

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Let $I \subset \mathbb{R}$ be a non-empty interval.

Proposition. Let $f : I \to \mathbb{R}$ be a twice differentiable function. Then $f$ is strictly convex on $I$ if and only if $f'' > 0$ outside a set of empty interior.

Proof: I do not assume that $f''$ is non-negative on $I$. But Darboux proved that derivative functions satisfy the conclusion of the intermediate value theorem. Since $f''$ is positive outside a set of empty interior, we deduce that $f''$ is non-negative on $I$. Thus, the function $f$ is convex on $I$. So if $f$ is not strictly convex on $I$, there exists a non-empty interval $J \subset I$ such that $f_{\restriction J}$ is affine. Hence, the function $f''$ is identically zero on $J$, which contradicts our hypothesis. Reciprocally, if $f'' = 0$ on $J$ then $f_{\restriction J}$ is affine and so $f$ is not strictly convex on $I$.

Set $g(x) = x^4 \sin^2(\frac{1}{x})$. Remark that $g$ is non-negative and continuous on $\mathbb{R}$. Consider $f: \mathbb{R} \to \mathbb{R}$ the map defined by $$\forall x \in \mathbb{R},\ f(x) = \int_0^x \Big(\int_0^y g(s)\, ds\Big) dy. $$ The function $f$ is twice differentiable and $f'' = g$. Thus, $f$ is strictly convex on $\mathbb{R}$ according to the above proposition.

halbaroth
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