On convergence of convex-concave functions

Question

Let $(f_n)$ be a sequence of twice differentiable functions on $\mathbb R$ such that for each $n$ there exists some $x_n\in\mathbb{R}$ such that:

• $f_n$ is strictly convex on $(-\infty,x_n)$,
• $f_n$ is strictly concave on $(x_n, +\infty)$.

Suppose also that $f_n$ uniformly converges to a twice differentiable function $f$.

It was conjectured that then the sequence $(x_n)$ will be convergent.

This conjecture was disproved.

The OP then asked in a comment whether the additional condition that $f''$ not vanish on any nonempty interval can help.

Below it will be shown that the answer is still negative, even with the latter additional condition.

In another answer, it will be shown how conditions on the $f_n$'s or on $f$ can be modified to make the sequence $(x_n)$ convergent.

Answer 1

Let $b$ be the continuously differentiable bump function defined by the formula \begin{equation} b(u):=60u_+^2(1-u)_+^2 \end{equation} for real $u$, where $u_+:=\max(0,u)$. So, $b$ is supported on the interval $[0,1]$. For real $x$, let \begin{equation} B(x):=-1+\int_{-\infty}^x b(u)\,du =\begin{cases} -1 & \text{ if }x\le 0, \\ 1 &\text{ if } x>1, \\ 12 x^5-30 x^4+20 x^3-1 & \text{otherwise.} \end{cases} \end{equation} Then $B$ is a twice continuously differentiable nondecreasing function such that $|B|\le1$.

In what follows, assume that $n\ge3$. Then $x^2-B(x/n)\ge0$ for all real $x$. So, we can define \begin{equation} f_n(x):=\sqrt{x^2-B(x/n)}. \end{equation} Let also \begin{equation} f(x):=\sqrt{x^2+1}; \end{equation} here and in what follows, $x$ is a real number, sometimes subject to additional conditions. Note that $f$ is a smooth function and $f''$ does not vanish on any nonempty interval.

If $x\le0$, then $f_n(x)=f(x)$.

If $0<x=o(n)$, then $B(x/n)\to-1$ and hence $f_n(x)\to f(x)$ (as $n\to\infty$) uniformly over all $x$ such that $0<x=o(n)$.

If $n=O(x)$, then $x\to\infty$. Therefore and because $|B|\le1$, we have $f_n(x)\to f(x)$ uniformly over all $x$ such that $n=O(x)$.

Thus, $f_n(x)\to f(x)$ uniformly in all real $x$.

Next, $f_n$ is twice continuously differentiable, $f''_n(x)>0$ if $x\le0$, $f''_n(x)<0$ if $x\ge n$, and $f_n(x)>0$ and \begin{equation} f''_n(x)=\frac{(r(u)-1/n^2)q(u)}{f_n(x)^3} \end{equation} if $x\in(0,n)$, where $u:=x/n\in(0,1)$, \begin{equation} r(u):=\frac{p(u)}{q(u)}, \end{equation} \begin{equation} p(u):=1 - 20 u^3 + 90 u^4 - 72 u^5, \end{equation} \begin{equation} q(u):=60 (1-u) u \left(9 u^6-27 u^5+25 u^4-5 u^3-2 u+1\right)>0. \end{equation} So, the sign of $f''_n(x)$ for $x\in(0,n)$ is the same as the sign of $r(u)-1/n^2$.

Note that $r'(u)<0$ for $u\in(u_*,1)$, where $u_*:=\frac{3 +\sqrt3}6\in(0,1)$.
Also, $$\min\{r(u)\colon u\in(0,u_*]\}=\frac2{205}\, (45 - 16\sqrt3)=0.168\ldots>1/n^2$$ (since $n\ge3$). Also, $r(1-)=-\infty$. So, there is some $u_n\in(u_*,1)\subset(0,1)$ such that $r(u)-1/n^2>0$ for $u\in(0,u_n)$ and $r(u)-1/n^2<0$ for $u\in(u_n,1)$. Letting now \begin{equation} x_n:=nu_n, \end{equation} we see that $f''_n(x)>0$ if $x<x_n$ and $f''_n(x)<0$ if $x>x_n$.

So, all the conditions on $f$ and the $f_n$'s hold.

However, $x_n\ge nu_*\to\infty$. $\quad\Box$

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For an illustration, here are the graphs $\{(x,f_3(x))\colon-3\le x\le7\}$ (above) and $\{(u,r(u))\colon0<u<1\}$ (below):

Answer 2

Here it will be shown how conditions on the $f_n$'s or on $f$ can be modified to make the sequence $(x_n)$ convergent.

First of all, the condition that the $f_n$'s be twice differentiable is of no help; so, this condition will be dropped.

Suppose that for each $n$ there exists some $x_n\in\mathbb{R}$ such that:

• $f_n$ is semi-strongly convex on $(-\infty,x_n)$ in the sense that $f_n$ is convex on $(-\infty,x_n)$ and $f_n(x)/|x|\to\infty$ as $x\to-\infty$;

• $f_n$ is semi-strongly concave on $(x_n,\infty)$ in the sense that $f_n$ is concave on $(x_n,\infty)$ and $f_n(x)/x\to-\infty$ as $x\to\infty$.

Note that, if $f_n$ is strongly convex on $(-\infty,x_n)$, then $f_n$ is semi-strongly convex on $(-\infty,x_n)$; similarly, if $-f_n$ is strongly convex on $(x_n,\infty)$, then $f_n$ is semi-strongly concave on $(x_n,\infty)$.

Suppose also that $(f_n)$ uniformly converges to a function $f$, which is not affine on any nonempty interval.

Then the sequence $(x_n)$ is convergent.

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Indeed, suppose first that the sequence $(x_n)$ is unbounded. Then, using a left-right symmetry and passing to a subsequence, without loss of generality (wlog) assume that $x_n\to\infty$. It follows that the limit function $f$ is convex on $\mathbb R$. So, $f(x)\ge a+bx$ for some real $a,b$ and all real $x$. Therefore and in view of semi-strongly concavity of $f_n$ on $(x_n,\infty)$, for each $n$ we have $f_n(x)-f(x)\to-\infty$ as $x\to\infty$, which contradicts the uniform convergence of $(f_n)$ to $f$.

So, the sequence $(x_n)$ is bounded. Take any subsequence $(x_{n_k})$ of the sequence $(x_n)$ converging to a limit $x_*$. Then $f$ is convex on $(-\infty,x_*)$ and concave on $(x_*,\infty)$. Since $f$ is not affine on any nonempty interval, the point $x_*$ is uniquely determined by $f$. So, any converging subsequence of the sequence $(x_n)$ converges to the same limit $x_*$. Since the sequence $(x_n)$ is bounded, we conclude that $x_n\to x_*$, as claimed.

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This reasoning shows that the following is true as well:

Suppose that for each $n$ there exists some $x_n\in\mathbb{R}$ such that:

• $f_n$ is convex on $(-\infty,x_n)$;

• $f_n$ is concave on $(x_n,\infty)$.

Suppose also that $(f_n)$ converges pointwise (not necessarily uniformly) to a function $f$, which is not affine on any nonempty interval, not convex on $\mathbb R$, and not concave on $\mathbb R$.

Then the sequence $(x_n)$ is convergent.

Details on the latter assertion, requested by Gaetano: Suppose first that the sequence $(x_n)$ is unbounded. Then, using a left-right symmetry and passing to a subsequence, wlog assume that $x_n\to\infty$. It follows that the limit function $f$ is convex on $\mathbb R$, which contradicts the assumption that $f$ is not convex on $\mathbb R$.

So, the sequence $(x_n)$ is bounded. Take any subsequence $(x_{n_k})$ of the sequence $(x_n)$ converging to a limit $x_*$. Then $f$ is convex on $(-\infty,x_*)$ and concave on $(x_*,\infty)$. Since $f$ is not affine on any nonempty interval, the point $x_*$ is uniquely determined by $f$. So, any converging subsequence of the sequence $(x_n)$ converges to the same limit $x_*$. Since the sequence $(x_n)$ is bounded, we conclude that $x_n\to x_*$, as claimed.