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Let $F:[0,\infty) \to [0,\infty)$ be a $C^2$ strictly convex function, and let $r_0<r_1$ be positive fixed constants. Let $$a<r_0<r_1<c<b, \tag{1}$$ and let $\lambda \in [0,1]$ satisfy $ \lambda a +(1-\lambda)b=c. $

Set $D(a,b,c)=\lambda F(a)+(1-\lambda)F(b)-F(c) $.

Question: Does there exist a constant $m>0$ (which may depend on $f,r_0,r_1$ but not on $a,b,c$) such that $ D(a,b,c) \ge m\lambda(1-\lambda)(r_1-r_0)^2 $ for any choice of $a,b,c$ satisfying condition $(1)$?

Here is the key point:

If $f'' \ge m$, then $f$ is strongly convex with parameter $m$, so $$ D(a,b,c) \ge \frac{1}{2}m\lambda(1-\lambda)(b-a)^2 \ge \frac{1}{2}m\lambda(1-\lambda)(r_1-r_0)^2 \tag{2} $$ as required. However, in our case, $c$ and $b$ can be arbitrarily large, and $F$ can become "less convex" (closer to being affine) when $x \to \infty$. In other words, if $\lim_{x \to \infty}F''(x)=0$, then the lower bound $(2)$ becomes the trivial bound $$ D(a,b,c) \ge \frac{1}{2} (\inf F'')\lambda(1-\lambda)(b-a)^2=0. $$

So, "naive application" of strong convexity does not apply here as is. However, my intuition is that even if $\lim_{x \to \infty}F''(x)=0$, we should somehow encounter "the strong convexity content" which lies between the fixed $r_0$ and $r_1$ so the "convexity gap" $D(a,b,c)$ should be bounded away from zero.

I thought to express $D(a,b,c)$ as some integral of $F''$ over a domain which contains $[r_0,r_1]$ but so far without success.

Asaf Shachar
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  • I'm not sure I get the question. Define $m=0$ and the result holds by Jensen's inequality. You cannot have $m>0$ because you can consider $f(x) = e^{-x}$ defined for all $x \geq 0$, and note that for $\lambda=1/2$ the gap $D(x,x+2,x+1)=(1/2)e^{-(x+2)} + (1/2)e^{-x} - e^{-(x+1)}$ gets arbitrarily small as $x\rightarrow \infty$. – Michael Aug 20 '20 at 14:54
  • Is $\lambda$ fixed? – Zim Aug 20 '20 at 17:09
  • @Michael Yes, I meant to require $m>0$. Also, note that I specifically assumed that $a<r_0$ where $r_0$ is a fixed constant. In your counterexample $a=x$ is unbounded. – Asaf Shachar Aug 20 '20 at 17:17
  • @zim No, $\lambda$ is in fact uniquely determined by $a,b,c$. That is why I have denoted $D=D(a,b,c)$ and not $D=D(a,b,c,\lambda)$. (We do not need to mention explicitly the dependence of $D$ on $\lambda$ since it's determined by the other three variables $a,b,c$). – Asaf Shachar Aug 20 '20 at 17:19
  • For fixed $[a,b]$, since $f$ is strictly convex, it is also uniformly convex on $[a,b]$. This is noted in Exercise 10.10 in Bauschke & Combettes' book, vol. 2. However, the exercise does not provide the modulus of the uniform convexity, so it is hard to say how the modulus relates to your proposed bound. – Zim Aug 21 '20 at 18:16
  • This is a really interesting question! In looking for a counterexample, I looked at the strictly (but not strongly) convex function $e^{-x}$. I was not able to find a counterexample! Here is the graph I used, for anyone interested: https://www.desmos.com/calculator/swlpftndjk – Zim Aug 21 '20 at 18:43
  • It is an interesting question. I guess that there exists a constant $m > 0$ which depends on $f, r_0, r_1$. – River Li Aug 23 '20 at 04:02

2 Answers2

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If suffices to require that $F$ is strictly convex and differentiable on an interval $I \subset \Bbb R$. (Even the differentiability requirement can be dropped, see the remarks at the end of the answer.)

For $a, b \in I$ with $a < b$ and $c = \lambda a + (1 - \lambda) b$ with $0 \le \lambda \le 1$ we can write $$ D(a, b, c) = \lambda F(a)+(1-\lambda)F(b)-F(c) \\ = \lambda \bigl \{ F(a) - F(c) - (a-c)F'(c) \bigr\} + (1- \lambda) \bigl \{F(b) - F(c) - (b-c)F'(c)\bigr\} \, . $$

This suggests to introduce $$ H(u, v) = F(u) - F(v) - (u-v) F'(v) $$ for $u, v \in I$. $H$ has the following properties:

  1. $H(u, v) > 0$ if $u \ne v$.
  2. $H(u_1, v) > H(u_2, v)$ if $u_1 < u_2 \le v$, i.e. $H(u,v)$ is decreasing in $u$ as long as $u \le v$.
  3. $H(u, v_1) < H(u, v_2)$ if $u \le v_1 < v_2$, i.e. $H(u, v)$ is increasing in $v$ as long as $u \le v$.

Property (1) is a direct consequence of the strict convexity: $F(u)$ is larger than the corresponding value of the tangent line at $x=v$.

For property (2) we assume $u_1 < u_2 \le v$ and compute $$ H(u_1, v) - H(u_2, v) = F(u_1) - F(u_2) - (u_1 - u_2) F'(v) \\ \ge F(u_1) - F(u_2) - (u_1 - u_2) F'(u_2) = H(u_1, u_2) > 0 \, . $$ Here we used that $F'$ is increasing.

For property (3) we assume $u \le v_1 < v_2$ and compute $$ H(u,v_1) - H(u, v_2) = -F(v_1) - (u-v_1)F'(v_1) + F(v_2) + (u-v_2) F'(v_2) \\ \le -F(v_1) - (u-v_1)F'(v_2) + F(v_2) + (u-v_2) F'(v_2) \\ = -H(v_1, v_2) < 0 \, . $$

With these tools, estimating $D(a, b, c)$ from below becomes easy. If $a \le r_0 < r_1 \le c < b$ then $$ D(a, b, c) = \lambda H(a, c) + (1-\lambda)H(b,c) \\ \ge \lambda H(a, c) \ge \lambda H(r_0, r_1) \ge \lambda(1- \lambda) H(r_0, r_1) \\ = m \lambda(1-\lambda) (r_1-r_0)^2 $$ with $m$ defined as $$ m = \frac{H(r_0, r_1)}{(r_1-r_0)^2} = \frac{F(r_0) - F(r_1) - (r_0 - r_1) F'(r_1)}{(r_1-r_0)^2} > 0 \, . $$

Remarks:

  • The assumption that $F$ is only defined on $[0, \infty)$ with values in $[0, \infty)$ was not used in the proof.
  • The differentiability requirement can also be dropped. A convex function has one-sided derivatives in every inner point of the interval. The above proof still works if we replace $F'$ by the right (or left) derivative.
Martin R
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  • What if $F''(x) = 0$ for some $r_0 < x < r_1$? – River Li Aug 23 '20 at 13:09
  • @RiverLi: Actually, since $m$ is allowed to depend $r_0$ and $r_1$, we can define $m$ as $\frac{1}{(r_1-r_0)^2} \int_{r_0}^{r_1} (t-r_0) F''(t), dt$. The integral cannot be zero. – Martin R Aug 23 '20 at 14:56
  • Yes, you are right. – River Li Aug 23 '20 at 15:13
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    Your solution is nice. – River Li Aug 23 '20 at 15:24
  • @RiverLi: Thanks! I have simplified it further, the result is now similar to what you obtained. Could it be that a square is missing in your definition of $m$? (I apologize if there are overlaps between our solutions but I had already prepared part of this.) – Martin R Aug 23 '20 at 16:20
  • Yes, you are right. – River Li Aug 24 '20 at 00:01
  • @MartinR Thank, this is a very interesting answer! you in fact proved that the $1-\lambda$ factor is not needed. This is a bit surprising! It means that when $D(a,b,c)$ becomes small, then $\lambda$ becomes close to $0$ (it cannot get close to $1$ instead). – Asaf Shachar Aug 24 '20 at 12:56
  • @AsafShachar: It is perhaps not so surprising: $\lambda$ close to one would mean that $c$ is close to $a$, and that is prevented by the restriction $a < r_0 < r_1 < c$. – Martin R Aug 24 '20 at 13:35
  • Yes, I agree with that. However $\lambda$ could a-priori get close to $1$ while $b$ tends to infinity at a suitable rate so $(1-\lambda)b \simeq c-a$. By your bound, this actually cannot happen. (This was in fact my target all along. You can see here - https://mathoverflow.net/questions/369446/a-question-about-asymptotic-affinity-and-strict-convexity-with-unbounded-means- for different proofs of that claim. I am happy that the strong form of your claim implies this other corollary as well). – Asaf Shachar Aug 24 '20 at 13:41
  • @AsafShachar: You are welcome! I worked a while on this (starting with trying to construct a counterexample :) because I found it an interesting problem. I wasn't aware of the question on MO, but there could be a similarity in the arguments: $H(u, v)$ has the geometrical meaning of the vertical distance between the graph of $F$ and the tangent line at $(v, F(v)$, taken at $x=u$. – Martin R Aug 24 '20 at 13:57
  • @AsafShachar: Using my terminology, your MO question could be solved with the quantitative estimate $D(a_n, b_n, c_n) \ge \lambda_n H(c, c+1)$ if $a_n < c < c+1 < b_n$ (which might be what the other answers already say). – Martin R Aug 24 '20 at 14:59
  • Yes, this was exactly my point:) I thought that the proofs given at MO were too complicated, so I asked this question in an attempt to find a simpler proof to a "sub-problem". – Asaf Shachar Aug 24 '20 at 15:00
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Alternative solution

Let us prove that the best constant $m$ is $$m = \frac{F(r_0) - F(r_1) - F'(r_1)(r_0 - r_1)}{(r_1 - r_0)^2} > 0.$$ (Note: Actually, it is equal to $\frac{1}{(r_1 - r_0)^2}\int_{r_0}^{r_1} (x- r_0) F''(x) \mathrm{d} x$ which is positive since $F(x)$ is strictly convex.)

First, we rephrase the problem as follows:

Let $F : [0, \infty) \to [0, \infty)$ be a $\mathrm{C}^2$ strictly convex function. Let $0 < r_0 < r_1$ be fixed constants. Does there exist a constant $m > 0$ such that $$\lambda F(a) + (1 - \lambda)F(b) - F(\lambda a + (1 - \lambda)b) \ge m \lambda (1 - \lambda) (r_1 - r_0)^2$$ for any real numbers $a, b, \lambda$ satisfying $$0 < \lambda < 1, \quad 0 \le a < r_0 < r_1 < \lambda a + (1 - \lambda) b\ ?$$

Second, we have \begin{align} &\inf_{0 < \lambda < 1,\ 0 \le a < r_0 < r_1 < \lambda a + (1 - \lambda) b} \frac{\lambda F(a) + (1 - \lambda)F(b) - F(\lambda a + (1 - \lambda)b)}{\lambda (1 - \lambda)}\\ =\ & \inf_{0 < \lambda < 1, \ 0 \le a < r_0} \left(\inf_{b > \frac{r_1 - \lambda a}{1 - \lambda}} \frac{\lambda F(a) + (1 - \lambda)F(b) - F(\lambda a + (1 - \lambda)b)}{\lambda (1 - \lambda)}\right)\\ =\ & \inf_{0 < \lambda < 1, \ 0 \le a < r_0} \frac{\lambda F(a) + (1 - \lambda)F(\frac{r_1 - \lambda a}{1 - \lambda}) - F(r_1)}{\lambda (1 - \lambda)} \tag{1}\\ =\ & \inf_{0 < \lambda < 1} \left(\inf_{0 \le a < r_0} \frac{\lambda F(a) + (1 - \lambda)F(\frac{r_1 - \lambda a}{1 - \lambda}) - F(r_1)}{\lambda (1 - \lambda)} \right)\\ =\ & \inf_{0 < \lambda < 1} \frac{\lambda F(r_0) + (1 - \lambda)F(\frac{r_1 - \lambda r_0}{1 - \lambda}) - F(r_1)}{\lambda (1 - \lambda)} \tag{2}\\ =\ & \inf_{y > r_1} \frac{(y - r_0)(y - r_1)F(r_0) + (r_1 - r_0)(y - r_0)F(y) - (y - r_0)^2F(r_1)}{(r_1 - r_0)(y - r_1)} \tag{3}\\ =\ & \lim_{y \to r_1} \frac{(y - r_0)(y - r_1)F(r_0) + (r_1 - r_0)(y - r_0)F(y) - (y - r_0)^2F(r_1)}{(r_1 - r_0)(y - r_1)} \tag{4}\\ =\ & F(r_0) - F(r_1) - F'(r_1)(r_0 - r_1). \tag{5} \end{align} Explanations:
(1): By letting $f(b) = (1 - \lambda)F(b) - F(\lambda a + (1 - \lambda)b)$, we have $f'(b) = (1 - \lambda)F'(b) - (1 - \lambda) F'(\lambda a + (1 - \lambda)b) \ge 0$ (note: $F'(x)$ is non-decreasing) and thus $f(b)$ is non-decreasing on $[b, \infty)$.
(2): By letting $g(a) = \lambda F(a) + (1 - \lambda)F(\frac{r_1 - \lambda a}{1 - \lambda})$, we have $g'(a) = \lambda F'(a) - \lambda F'(\frac{r_1 - \lambda a}{1 - \lambda}) \le 0$ (note: $F'(x)$ is non-decreasing) and thus $g(a)$ is non-increasing on $[0, r_0)$.
(3): Use the substitution $y = \frac{r_1 - \lambda r_0}{1 - \lambda}$.
(4): Use the following fact (the proof is given at the end):
Fact 1: Let $$g(y) \triangleq \frac{(y - r_0)(y - r_1)F(r_0) + (r_1 - r_0)(y - r_0)F(y) - (y - r_0)^2F(r_1)}{(r_1 - r_0)(y - r_1)}.$$ Then $g'(y) \ge 0$ on $(r_1, \infty)$.
(5) Apply L'Hopital rule.

We are done.

$\phantom{2}$

Proof of Fact 1: We have, for $y > r_1$, \begin{align} (r_1 - r_0)(y - r_1)^2g'(y) &= (y - r_1)^2F(r_0) - (r_1 - r_0)^2F(y)\\ &\quad + (r_1 - r_0)(y - r_0)(y - r_1)F'(y)\\ &\quad + (-2(y - r_0)(y - r_1) + (y - r_0)^2)F(r_1) \\ &= (y - r_1)^2F(r_0) - (r_1 - r_0)^2( F(y) - F(r_1) ) \\ &\quad - (r_1 - r_0)^2F(r_1) + (r_1 - r_0)(y - r_0)(y - r_1)F'(y)\\ &\quad + (-2(y - r_0)(y - r_1) + (y - r_0)^2)F(r_1)\\ &\ge (y - r_1)^2F(r_0) - (r_1 - r_0)^2(y - r_1)F'(y) \\ &\quad - (r_1 - r_0)^2F(r_1) + (r_1 - r_0)(y - r_0)(y - r_1)F'(y)\\ &\quad + (-2(y - r_0)(y - r_1) + (y - r_0)^2)F(r_1)\\ &= (y - r_1)^2F(r_0) - (y - r_1)^2F(r_1) + (r_1 - r_0)(y - r_1)^2F'(y) \\ &\ge (y - r_1)^2F(r_0) - (y - r_1)^2F(r_1) + (r_1 - r_0)(y - r_1)^2F'(r_1)\\ &= (y - r_1)^2[F(r_0) - F(r_1) - F'(r_1)(r_0 - r_1)]\\ &\ge 0 \end{align} where we have used $(y - r_1)F'(y) \ge F(y) - F(r_1)$ and $F(r_0) - F(r_1) - F'(r_1)(r_0 - r_1) \ge 0$ and $F'(y) \ge F'(r_1)$ (Note: $F(x) \ge F(y) + F'(y)(x-y)$ for convex functions; $F'(x)$ is non-decreasing.). We are done.

River Li
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