# A sufficient condition for convexity: $f$ absolutely continuous on $[a,b]$ and $f'$ increasing a.e. imply $f$ is convex

Problem: I am asked to prove the claim in the title: that if $$f$$ is absolutely continuous on $$[a,b]$$, with $$f'$$ increasing almost everywhere, then $$f$$ is a convex function.

The problem comes with the additional hint that, for $$\xi < \eta$$ and $$\xi,\eta \in [a,b]$$, we have $$\newcommand{\nc}{\newcommand} \nc{\R}{\mathbb{R}} \nc{\AC}{\mathrm{AC}} \nc{\ve}{\varepsilon} \nc{\d}{\delta} \nc{\set}[1]{\left\{#1\right\}} \nc{\br}[1]{\left[ #1 \right]} \nc{\abs}[1]{\left| #1 \right|} \nc{\l}{\lambda} \frac{f(\eta) - f(\xi)}{\eta - \xi} = \int_0^1 f' \Big( (1-\l) \xi + \l \eta \Big) \, \mathrm{d} \l$$

Relevant Definitions: These are mostly in line with Measure & Integral: An Introduction to Real Analysis by Richard Wheeden and Antoni Zygmund.

• We say that $$f : [a,b] \to \R$$ is absolutely continuous (denoted $$f \in \AC[a,b]$$) if, $$\forall \ve > 0$$, $$\exists \d > 0$$, such that, for any collection $$\set{\br{a_i,b_i}}_{i=1}^n$$ of finitely many non-overlapping intervals (i.e. intersect at most on the boundary) $$\sum_i \abs{b_i-a_i} < \d \implies \sum_i \abs{f(b_i) - f(a_i)} < \ve$$

• We say that $$f : (a,b) \to \R$$ is convex if, whenever $$a < \xi < \eta < b$$, and whenever $$\l \in [0,1]$$, then $$f \Big( (1-\l) \xi + \l \eta \Big) \le (1-\l) f(\xi) + \l f(\eta)$$ For whatever reason, Measure & Integral chooses to use an open interval for the domain of $$f$$, and I'm not sure why. I don't think there's a meaningful issue if we extend to closed intervals.

• We say $$f : (a,b) \to \R$$ has a supporting line (at an $$x_0 \in (a,b)$$) if $$\exists m \in \R$$ such that, $$\forall x \in (a,b)$$, $$f(x) \ge f(x_0) + m(x-x_0)$$ (That is, a supporting line through $$x_0$$ has the graph of $$f$$ entirely above it. For differentiable $$f$$, the tangent line is such a support line, but many may exist in general.)

Relevant Results: Not sure how useful these are, but just in case here are some. I'm open to the use of others, though.

• $$f \in \AC[a,b]$$ iff $$f'$$ exists a.e., $$f' \in L[a,b]$$, and $$f(x) = f(a) + \int_a^x f'$$ for every $$x \in [a,b]$$.

• $$f$$ has a supporting line through every point in its domain iff $$f$$ is convex.

• Some sufficient conditions to ensure that $$f$$ is convex include:

• If $$f'$$ exists everywhere (not "almost everywhere") and is increasing, $$f$$ is convex.
• If $$f$$'s right-hand derivative exists everywhere, is increasing, and always finite, then $$f$$ is convex.
• If $$f''(x) \ge 0$$ on $$(a,b)$$, it is convex.

That all said, I'm not sure at all how to handle the argument, or even the relevance of the given hint. I do notice that it is just a reformulation of the fundamental theorem of calculus for $$f$$, as, by letting $$u = (1-\l) \xi + \l \eta$$,

$$\int_0^1 f' \Big( (1-\l) \xi + \l \eta \Big) \, \mathrm{d}\l = \frac 1{\eta-\xi} \int_\xi^\eta f'(u) \, \mathrm{d} u$$

Maybe one could send $$\eta \to \xi$$ in the limit? But handling the right hand side of the hint, if we were to bring the limit in somehow (even if it could be justified) would just tell use the $$f' \equiv 1$$ a.e., which doesn't help.

I know it's not much, but can anyone give me any hints or nudges in the right direction?

• What exactly do you mean by '$f'$ increasing ae.'? Apr 8, 2023 at 7:20
• That $f'$ is increasing "almost everywhere," i.e. except on a set of measure zero. That is, $$x > y \implies f'(x) > f'(y)$$ for all $x,y \in [a,b] - Z$ where $Z \subseteq [a,b]$ is a set of measure zero. Apr 8, 2023 at 7:35

We want to show that $$f(\lambda x + (1-\lambda)y) \le \lambda f(x)+(1-\lambda) f(y)$$ for all $$\lambda \in [0,1]$$ and $$x,y \in [a,b]$$.

Let $$Z \subset [a,b]$$ be the null set for which $$f'$$ is not defined.

Let $$\phi(\lambda) = f(\lambda x + (1-\lambda)y)$$, we want to show that $$\phi(\lambda) \le \lambda \phi(1) + (1-\lambda) \phi(0)$$. Note that $$\phi$$ satisfies the same conditions as $$f$$ (you should verify this).

Suppose that $$\lambda \in (0,1)$$ (otherwise the result is immediate). Let $$g_- = \sup_{t \in [x,\lambda) \setminus Z} \phi'(t)$$, $$g_+ = \inf_{t \in (\lambda, y] \setminus Z} \phi'(t)$$. Note that $$g_- \le g_+$$, and pick $$g \in [g_-,g_+]$$.

The fundamental theorem shows that $$\phi(t) = \phi(\lambda) + g (t-\lambda) + \int_\lambda^t (\phi'(s) -g)ds$$.

Hence $$\phi(t) \ge \phi(\lambda) + g (t-\lambda)$$ for $$t \in [0,1]$$.

From this we get, $$\phi(0) \ge \phi(\lambda) + g(-\lambda)$$, $$\phi(1) \ge \phi(\lambda) + g (1-\lambda)$$ and so $$\phi(\lambda) \le \lambda \phi(1) + (1-\lambda) \phi(0)$$.

I do not know the result's name in English (or if it has one to begin with, we French do tend to give anything and everything a name, for better or for worse) but there is a "three slopes lemma" that goes something like this:

Let $$f : I \to \mathbb{R}$$, with $$I$$ an interval of $$\mathbb{R}$$.
Then, $$f$$ is convex if and only if for all $$x_1 < x_2 < x_3$$ in $$I$$, we have: $$\frac{f(x_2) - f(x_1)}{x_2 - x_1} \leq \frac{f(x_3) - f(x_1)}{x_3 - x_1} \leq \frac{f(x_3) - f(x_2)}{x_3 - x_2}$$ Moreover, you only need one inequality out of both for $$f$$ to be convex.

Geometrically it'll make sense if you draw the graph of a convex function, choose any three points and look at the lines connecting them, since those quantities will be the slopes of said lines.

If you can prove this lemma (a demonstration is given here but it's in French), then you should be able to use the indication you were given and arrive at the conclusion rather easily.

EDIT: I thought this would be enough of a "hint or nudge" as asked by the OP but I'll gladly develop my answer.

Since $$f'$$ is a.e. increasing, there exists a negligeable set $$N \subset [a,b]$$ for which $$f'$$ exists and is increasing on $$[a,b] \setminus N$$. Now let $$x_1 < x_2 < x_3$$ in $$[a,b]$$. We want to show that: $$\frac{f(x_2) - f(x_1)}{x_2 - x_1} \leq \frac{f(x_3) - f(x_1)}{x_3 - x_1} \tag{1}$$ and we'll be done thanks to the three slopes lemma.
Thanks to the indication, this is equivalent to showing that: $$\int_0^1 f'\Big((1-\lambda)x_1 + \lambda x_2\Big) \mathrm{d}\lambda \leq \int_0^1 f'\Big((1-\lambda)x_1 + \lambda x_3\Big) \mathrm{d}\lambda \tag{2}$$

Let $$N_1$$ be the set $$[x_1,x_2] \cap N$$ and $$N_2$$ the set $$[x_1, x_3] \cap N$$.
Define the functions $$g_1 : \lambda \in [0,1] \mapsto (1-\lambda)x_1 + \lambda x_2 \in [x_1, x_2]$$ and $$g_2 : \lambda \in [0,1] \mapsto (1-\lambda)x_1 + \lambda x_3 \in [x_1, x_3]$$.
These functions are affine and bijective, thus the sets $$A_1 := g_1^{-1}(N_1)$$ and $$A_2 := g_2^{-1}(N_2)$$ are negligeable due to $$N_1$$ and $$N_2$$ being negligeable.
If we take $$A := A_1 \cup A_2$$, then, for all $$\lambda \in [0,1] \setminus A$$, $$f'\Big((1-\lambda)x_1 + \lambda x_2\Big)$$ and $$f'\Big((1-\lambda)x_1 + \lambda x_3\Big)$$ exist, and since $$f'$$ is increasing on $$[a,b] \setminus N$$: $$f'\Big((1-\lambda)x_1 + \lambda x_2\Big) \leq f'\Big((1-\lambda)x_1 + \lambda x_3\Big)$$ $$A$$ being negligeable, this provides, after integration on $$[0,1]$$: $$\int_0^1 f'\Big((1-\lambda)x_1 + \lambda x_2\Big) \mathrm{d}\lambda \leq \int_0^1 f'\Big((1-\lambda)x_1 + \lambda x_3\Big) \mathrm{d}\lambda$$ Therefore, $$(2)$$ is true, and as such $$(1)$$ is true, thus by the lemma I introduced, $$f$$ is convex.

• I do not mind getting downvoted, however I would at least like to know the reason why? That way I can hopefully change my answer accordingly? Apr 9, 2023 at 11:19