# A question about asymptotic affinity and strict convexity with unbounded means

Let $$F:[0,\infty) \to [0,\infty)$$ be a $$C^1$$ strictly convex function.

Let $$\lambda_n \in [0,1],a_n\le c satisfy $$\lambda_n a_n +(1-\lambda_n)b_n=c_n \tag{1}$$ and assume that $$c_n \to ֿ\infty$$. (which implies $$b_n \to ֿ\infty$$). $$c>0$$ is just some constant, to make $$a_n$$ bounded.

Set $$D_n=\lambda_nF(a_n)+(1-\lambda_n)F(b_n)-F\big(c_n\big)$$, and assume that $$\lim_{n \to \infty}D_n=0$$

Question: Does $$\lambda_n \to 0$$?

My intuition is that even if $$F$$ becomes "less convex" (closer to being affine) when $$x \to \infty$$, then we cannot put to much weight on the $$a_n$$-since otherwise we get hit by the "convexity gap" between $$a_n$$ and $$b_n$$ by a non-negligible amount, which should make $$D_n$$ large.

Edit:

This is an attempt to understand Ron P's answer:

We have $$D(a_n,c_n,b_n)=\lambda_n F(a_n)+(1-\lambda_n)F( b_n)-F(c_n)$$, where $$\lambda_n a_n +(1-\lambda_n) b_n=c_n$$.

Similarly, $$D(a,c_n,b_n)=\tilde \lambda_n F(a)+(1-\tilde \lambda_n)F( b_n)-F(c_n)$$, where $$\tilde\lambda_n a +(1-\tilde \lambda_n) b_n=\tilde c_n$$.

Suppose that $$a_n \to a$$. (This implies $$\lambda_n-\tilde \lambda_n \to 0$$). We have

$$D(a_n,c_n,b_n)-D(a,c_n,b_n)=\lambda_n F(a_n)-\tilde \lambda_n F(a)+(\tilde \lambda_n-\lambda_n)F(b_n). \tag{2}$$ The first term tends to zero, since $$F(a_n) \to F(a)$$ and $$\lambda_n-\tilde \lambda_n \to 0$$.

Why does the second term tend to zero? we don't have control over $$F(b_n)$$, right?

• It should be easy to make a strictly convex function with $F(2n)/2+F(0)/2 -F(n)=1/n$, as envelope of a family of lines. (that is a counterexample with $a_n=0, c_n=n, b_n=2n, \lambda_n=1/2$) Aug 18, 2020 at 6:39
• One can also just do it for powers of 2, b_n=2^n Aug 18, 2020 at 7:21
• Dear Pietro, it seems that your suggestion cannot work. You may see Iosif Pinelis's answer below which proves that the answer is in fact positive. Aug 18, 2020 at 18:31
• Yes! I realized it with the tentative counterexample in the answer then deleted. The fact it does not work may be turned into a positive proof Aug 19, 2020 at 6:27
• Is assumed that $F$ is increasing? Aug 20, 2020 at 11:32

Indeed, by rescaling, without loss of generality (wlog) $$c=1$$. To simplify the notations, let $$f:=F$$, $$a:=a_n$$, $$b:=b_n$$, $$c:=c_n$$, $$t:=\lambda_n$$, $$D:=D_n$$. Passing to a subsequence, wlog $$a\to a_*\in[0,1]$$ and $$t\to t_*\in(0,1]$$. Also, wlog $$a+2\le c$$, since $$a\le1$$ and $$c\to\infty$$. Also, wlog $$b>c$$, since wlog $$t>0$$ and $$c>a$$.

By the convexity of $$f$$ and inequalities $$a+1\le a+2\le c$$, $$\begin{equation*} f(a+1)\ge f(c)+\frac{a+1-c}{b-c}\,(f(b)-f(c)).\tag{1} \end{equation*}$$ Using now the convexity of $$f$$ again together with the inequality $$a+2\le c$$ and (1), we have \begin{align*} 0\le d&:=\frac{f(a)+f(a+2)}2-f(a+1) \\ &\le \frac{f(a)}2+\frac12\,\frac{(c-a-2)f(a)+2f(c)}{c-a}-f(a+1) \\ &=\tilde d:=\frac{(c-a-1)f(a)+f(c)}{c-a}-f(a+1) \\ &\le\frac{(c-a-1)f(a)+f(c)+D}{c-a} \\ &\ \ -\Big(f(c)+\frac{a+1-c}{b-c}\,(f(b)-f(c))\Big) \\ &=\tilde D:=\frac{b-a-1}{b-a}\frac Dt\sim\frac D{t_*}\to0, \end{align*} so that $$\begin{equation*} d\to0. \tag{2} \end{equation*}$$ On the other hand, $$\begin{equation*} d\to\frac{f(a_*)+f(a_*+2)}2-f(a_*+1)>0 \end{equation*}$$ by the strict convexity of $$f$$. This contradicts (2). $$\Box$$

Remark: As seen from the above, condition $$c\to\infty$$ can be relaxed to $$\liminf(c-a)>0$$.

The above solution might look somewhat mysterious. In fact, the idea is a rather simple geometric one. For any real $$A,B,C$$ such as $$A\le B\le C$$, let the "gain" $$g(A,B;C)$$ denote the distance between the point on the graph of the convex function $$f$$ with abscissa $$C$$ and the point with the same abscissa on the chord connecting the points on the graph of $$f$$ with abscissas $$A$$ and $$B$$.

So (see the picture below), $$D=g(a,b;c)$$ and $$\tilde d=g(a,c;a+1)$$, where $$\tilde d$$ is as defined in the above multi-line display. In that display, it was shown that $$\tilde d\le\tilde D$$, which is clear from the picture. Also, if $$t$$ is bounded away from $$0$$ -- that is, if $$c/b$$ is bounded away from $$1$$, then, as it is clear from the picture by looking at the similar triangles, we have $$\tilde D\asymp D\to0$$; cf. the last line of the above multi-line display. This and the inequality $$\tilde d\le\tilde D$$ imply $$\tilde d\to0$$.

By the convexity of $$f$$, for any fixed real $$A,C$$ such as $$A\le C$$, the gain $$g(A,B;C)$$ is nondecreasing in $$B\in[C,\infty)$$ (here you may want to draw another picture). Therefore and because $$a+2\le c$$, we have $$d=g(a,a+2;a+1)\le g(a,c;a+1)=\tilde d$$, so that $$d\le\tilde d$$, which was shown in the first three lines of the above multi-line display.

This is the geometric explanation of (1) and the above multi-line display. • wow! This is an amazing answer, really. I liked a lot your geometric explanation and picture-very clear. (and indeed it was a bit mysterious before you added them). Just to be sure- in your third (last) inequality you have added the non-negative term $\frac{D}{c-a}$? I am also not sure how did you deduce that this was the "right thing" to add in order to get $\tilde D$. Did you also use the visual intuition from the picture here, or did you just take the difference and then wrote it "backwards"? Aug 19, 2020 at 12:10

First let's reformulate the question. For $$0\leq a\leq c\leq b$$, let $$\lambda=\lambda(a,c,b)\in[0,1]$$ be the number such that $$c=\lambda a + (1-\lambda)b$$, and for $$f\colon \mathbb R_+\to\mathbb R$$ define $$D_f(a,c,b)= \lambda f(a)+(1-\lambda)f(b)-f(c).$$

Lemma 1. Let $$f\colon \mathbb R_+\to\mathbb R$$ be strictly convex and continuously differentiable. Let $$0\leq a_n\leq c_n\leq b_n$$ be sequences such that $$a_n$$ is bounded, $$c_n-a_n$$ is bounded away from 0, and $$\limsup \lambda(a_n,c_n,b_n)>0$$. Then, $$\limsup D_f(a_n,c_n,b_n)>0$$.

We first apply a sequence of reduction steps that allow us to assume wlog that $$a_n=0$$, $$c_n\geq 1$$, , for all $$n$$, and $$\liminf\lambda(a_n,c_n,b_n)>0$$. If you trust that that is possible, you may skip directly to Lemma 5 below.

By taking a sub-sequence $$n'$$ on which $$\liminf \lambda(a_{n'},c_{n'},b_{n'})>0$$, Lemma 1 follows from Lemma 2.

Lemma 2. Let $$f\colon \mathbb R_+\to\mathbb R$$ be strictly convex and continuously differentiable. Let $$0\leq a_n\leq c_n\leq b_n$$ be sequences such that $$a_n$$ is bounded, $$c_n-a_n$$ is bounded away from 0, and $$\liminf \lambda(a_n,c_n,b_n)>0$$. Then, $$\limsup D_f(a_n,c_n,b_n)>0$$.

By further taking a sub-sequence $$n'$$ on which both $$a_{n'}$$ converges, Lemma 2 follows from Lemma 3.

Lemma 3. Let $$f\colon \mathbb R_+\to\mathbb R$$ be strictly convex and continuously differentiable. Let $$0\leq a_n\leq c_n\leq b_n$$ be sequences such that $$a_n\to a$$, $$c_n-a_n$$ is bounded away from 0, and $$\liminf \lambda(a_n,c_n,b_n)>0$$. Then, $$\limsup D_f(a_n,c_n,b_n)>0$$.

For any fixed $$\epsilon>0$$, the functions $$\lambda(a,c,b)$$ is continuous in $$a$$ uniformly in $$c$$ and $$b$$ over the domain $$\epsilon\leq a +\epsilon\leq c\leq b$$; therefore, under the assumptions of Lemma 3, $$0<\liminf\lambda(a_n,c_n,b_n)=\liminf\lambda(\lim a_n,c_n,b_n)$$. Furthermore, for $$\lim a_n small enough, we have $$\liminf\lambda(a,c_n,b_n)>0$$. Since $$D_f(a,c,b)$$ is decreasing in $$a$$, $$\limsup D_f(a_n,c_n,b_n)\geq\limsup D_f(a,c_n,b_n)$$. Therefore, Lemma 3 follows from Lemma 4.

Lemma 4. Let $$f\colon \mathbb R_+\to\mathbb R$$ be strictly convex and continuously differentiable. Let $$0\leq a\leq c_n\leq b_n$$ be sequences such that $$c_n-a$$ is bounded away from 0, and $$\liminf \lambda(a,c_n,b_n)>0$$. Then, $$\limsup D_f(a,c_n,b_n)>0$$.

Let $$T\colon \mathbb R\to\mathbb R$$ be the affine transformation that maps $$a$$ to $$0$$ and $$\inf c_n$$ to $$1$$. Replacing $$f$$ by $$F=f\circ T^{-1}$$, and $$a,c_n,b_n$$ by $$T(a),T(c_n),T(b_n)$$ respectively, Lemma 4 follows from Lemma 5.

Lemma 5. Let $$F\colon \mathbb R_+\to\mathbb R$$ be strictly convex and continuously differentiable. Let $$1\leq c_n\leq b_n$$ be sequences such that $$\liminf \lambda(0,c_n,b_n)>0$$. Then, $$\limsup D_F(0,c_n,b_n)>0$$.

Proof of Lemma 5. We assume wlog that $$F(0)=0$$ and denote $$\lambda_n=\lambda(0,c_n,b_n)$$ and $$D_n=D_F(0,c_n,b_n)$$.

Define a function $$G\colon [1,\infty)\times (1,\infty)\to \mathbb R$$ by $$G(x,y)=\tfrac 1 y F(xy)-F(x).$$

Claim 6. $$G$$ is positive and increasing in both $$x$$ and $$y$$.

Proof of Claim 6. Since $$F$$ is strictly convex, $$F(0)=0$$, and $$x = 1/y(xy)+(1-1/y)0$$, $$G(xy)>0$$. Since $$F'$$ is increasing, we have $$\frac {d}{dx}G(xy)=F'(xy)-F'(x)>0$$, so $$G$$ increases in $$x$$. Since $$F'$$ is increasing and $$G(x,y)=1/y\int_0^yF'(xt)x\,dt - F(x)$$, $$G$$ increases in $$y$$, completing the proof of Claim 6.

Suppose there is $$\lambda_0>0$$ such that $$\lambda_n\geq \lambda_0$$ for all $$n$$. Then, $$D_n = G(c_n,1/(1-\lambda_n))\geq G(1,1/(1-\lambda_0)>0, \quad\text{for all n.}$$ QED

• @AsafShachar you're right. I should elaborate about this point. I will do it soon. The idea is to replace $a_n$ with its limit in the expression of $D_n$. By continuity, such a modification does not change whether it converges to zero or not. Aug 21, 2020 at 7:09
• @AsafShachar I've elaborated on "re-scaling." Note that the answer does not assume neither that $F$ is non-negative, nor that $c_n\to\infty$, rather just that it is bounded away from $a_n$. Aug 21, 2020 at 12:39
• Thanks for the elaboration. I am still somewhat troubled tough. It seems to me that the most crucial non-trivial step is in passing from Lemma 3 to Lemma 4 (from $a_n$ to $\lim a_n$). I understand why $\liminf\lambda(a_n,c_n,b_n)=\liminf\lambda(\lim a_n,c_n,b_n)$ (because the difference between the 'lambdas' tend to zero.) However, I don't see why $\limsup D_f(a_n,c_n,b_n)= \limsup D_f(\lim a_n,c_n,b_n)$. I think it reduces to $\lim_{n \to \infty }( \lambda(a_n,c_n,b_n)-\lambda(\lim a_n,c_n,b_n))F(b_n)=0,$ but I don't see why this limit should be zero. Aug 24, 2020 at 11:39
• I have edited the question to make my misunderstanding clear. Thanks. Aug 24, 2020 at 11:39
• @AsafShachar, you're right. I've corrected the reduction from Lemma 3 to Lemma 4. Instead of using the continuity of $D_f$ in $a$, I now use monotonicity. Aug 25, 2020 at 8:54