Intersection of two convex functions such that their derivatives are also the same at the intersections

Question

I have two functions $T_1(x)$ and $T_2(x)$ defined on the interval $[0,1]$ such that $T_1(0) = T_2(0) = 0$ and $T_1(1) = T_2(1) = 1$. Furthermore, both of these are strictly monotone functions and are convex in the given interval. Also, they are differentiable at least twice. My aim is to prove these two functions are exactly equal. I have the following additional constraints on them.

1. $T_1'(0) = T_2'(0)$ and $T_1'(1) = T_2'(1)$
2. They intersect at $2$ points in (0,1).

Intuitively, I think that the functions should be the same given the constraints and I think only the first condition is sufficient to prove this. However, I am not able to get a proof for the same even with both these conditions. I tried a variety of things but they did not lead me anywhere. Any help on this would be highly appreciated. Thanks!

Answer 1

Here is a rather tedious counterexample.

Define $f(x) = {1 \over 3} x$ for $x \in [0, { 1\over 3}]$, and $f(x) = 3x-2$ for $x \in [{8 \over 9}, 1]$.

Note that $f({1 \over 3}) = {1 \over 9}$, $f'({1 \over 3}) = {1 \over 3}$, $f''({1 \over 3}) = 0$, and $f({8 \over 9}) = {2 \over 3}$, $f'({8 \over 9}) = 3$, $f''({8 \over 9}) = 0$.

In particular, if we let $a={1 \over 3}, b={8 \over 9}$, then we have $f'(a) < f'(b), f''(a) = f''(b) = 0$.

We can construct many convex functions $\phi$ on $[a,b]$ such that $\phi(a) = f(a), \phi(b) = f(b) $, $\phi'(a) = f'(a), \phi'(b) = f'(b) $, $\phi''(a) = \phi''(b) = 0$. Then if we define $f(x) = \phi(x)$ for $x \in [a,b]$ we have many functions that satisfy the criteria.

To show one such construction, it will be easier to work in terms of $g=\phi'$. Then the constraints are $g(a)=f'(a), g(b)=f'(b)$, $g'(a)=g'(b) = 0$, $g'(x) \ge 0$ for $x \in [a,b]$ and $\int_a^b g(t)dt = f(b)-f(a)$.

To see one way of doing this, choose $t_t,t_2$ such that $a < t_1 < t_2 < b$ and define $g(x) = f'(a)$ for $x \in [a,t_1]$, $g(x) = f'(b)$ for $x \in [t_2,b]$ and $g(x) = {f'(a)+f'(b) \over 2} + {f'(a) -f'(b) \over 2} \cos ({x - t_2 \over t_2-t_1}\pi)$ for $x \in [t_1,t_2]$. It is easy to confirm that all of the constraints except the integral one are satisfied.

To show that there are multiple values of $(t_1,t_2)$ that satisfy the integral constraint, note that $\int_a^b g(t)dt= f'(a) ({t_1+t_2 \over 2} -a) + f'(b) (b-{t_1+t_2 \over 2})$. Since $f'(a)(b-a) < {f(b)-f(a)} < f'(b)(b-a)$, we see that we can choose a value of ${t_1+t_2 \over 2}$ such that the integral constraint is satisfied.

Now we can choose $t_1-\delta,t_2+\delta$ for small $\delta$ to get different functions.

Answer 2

This is not true. Set $T_1(x) = x^2$, consider a smooth function $g$ which has compact support in $(0,1)$ and set $T_2(x) := T_1(x) + \epsilon \, g(x)$ for $\epsilon$ small enough.