Does $f(t)+sf\left(\frac{r-t}{s}\right)$ have local extrema at $r/(s+1)$ and $-r/(s-1)$ for $s > 1$, $r \neq 0$.

Question

Let $f:\mathbb{R}\mapsto [-1,1]$ be a nondecreasing function, such that

1. $f(-t) = -f(t)$, i.e. it's odd.
2. It is convex on $(-\infty, 0]$ and concave on $[0, \infty)$

Is it then true that for $s > 1, r\neq 0$ the function $$ g(t)=f(t)+sf\left(\frac{r-t}{s}\right), $$ has two local extrema at precisely $t_1 = r/(s+1)$ and $t_2 = -r/(s-1)$? Also, for $r=0$, then $t_1=t_2$ is a saddle point.

I've observed this behaviour while playing around with sigmoid functions, such as $\tanh(t)$, $\operatorname{erf}(t)$, $\DeclareMathOperator{\sign}{sign}\sign(t)$ and even $$ f(t) = \max(-1,\min(1,t)). $$ If need be, we may enforce additional conditions, such as it being strictly increasing, differentiable, etc.

What I can infer is that $g$ is concave on $[0,r]$ if $r > 0$, so it must contain a local maximum in this interval, and numerical analysis points to it being $t_1$. By numerical inspection, it also appears to be convex on some interval $[c,0]$, but I can't seem to prove that directly. A similar, opposite observation can be made when $r < 0$.

Answer 1

First note that $$\frac{r-t_1}{s}=\frac{r-r/(s+1)}{s}=\frac{sr}{s(s+1)}=t_1$$ and $$ \frac{r-t_2}{s}=\frac{r+r/(s-1)}{s}=\frac{sr}{s(s-1)}=-t_2. $$ Let us also assume that $r>0,s>1$.

Now, because $f$ is concave on $[0, \infty)$ we have for $t\in [-t_1, r-t_1]$ \begin{align} g(t_1)&=(1+s)f(t_1)=(1+s)f((t_1+t)/(s+1) + s(t_1-t/s)/(s+1))\geq\\ &\geq\frac{(1+s)(f(t_1+t)+sf(t_1-t/s))}{s+1}=f(t_1+t)+sf(t_1-t/s)=\\&=f(t_1+t)+sf((r-t_1 - t)/s)=g(t_1+t). \end{align} Furthermore for $t>0$ and since $f(0)=0$ \begin{align} g(r)-g(r+t)&=f(r)+sf(0)-f(r+t)-sf(-t/s)=f(r)+sf(t/s)-f(r+t)\geq\\ &\geq f(r)+ s(f(t)+f(0))/s - f(r+t)=f(r)+f(t)-f(r-t)\geq\\ &\geq\frac{r}{r+t}f(r+t)+\frac{t}{r+t}f(r+t)+f(0)-f(r-t)=0, \end{align} so in fact $g(t_1)\geq g(t)$ for all $t\in[0,\infty)$.

On the other hand, because $f$ is convex on $(-\infty,0]$ we have for $t\in(-\infty,-st_2]$ that $$ f(t_2+t/s)=f\left(\frac{t_2 + t + (s-1)t_2}{s} \right)\leq \frac{f(t_2+t)+(s-1)f(t_2)}{s} $$ or $$ f(t_2+t) - f(t_2) \geq sf(t_2+t/s) - sf(t_2). $$ Therefore \begin{align} g(t_2+t)-g(t_2) &= f(t_2 + t)+sf(-t_2 - t/s) - f(t_2) - sf(-t_2)=\\ &= f(t_2+t)-f(t_2) - sf(t_2+t/s)+sf(t_2)\geq\\ &\geq sf(t_2+t/s) - sf(t_2) -sf(t_2+t/s)+sf(t_2) = 0. \end{align}

The case $r < 0$ is handled in a similar way, except the roles of $t_1$ and $t_2$ are reversed, and we switch the direction of the inequalities.