$f:[0,1]\to \mathbb R$ is differentiable, prove that $f$ must be either linear or $|f(1)-f(0)|<|f'(t)|$ for some $t\in (0,1)$

Question

A function $f:[0,1]\to \mathbb R$ is such that $f$ is differentiable in it's domain. Prove that $f$ , is either linear in $x$ or $$|f(1)-f(0)|<|f'(t)|$$ for some $t\in (0,1)$

My attempt:

By Rolle theorem, we know that there exist atleast one $t\in (0,1)$ where the equality holds $$|f(1)-f(0)|=|f'(t)|$$Now , if such equality would have hold for all $t$ in the domain, then clearly $f$ is linear as the slope remains constant.

Now, I try to show that opposite inequality holds false i.e. :

$$f(1)-f(0)>f'(t)$$

By integrating both sides, we get that : $$\int_0^1f(1)-f(0) \mathrm dt>\int_0^1 f'(t) \mathrm dt$$

$$f(1)-f(0)>f(1)-f(0)$$ which is a contradiction !

Problems:

1. I had integrated without taking absolute value as asked in question and hence proof is incomplete.

2. This was the last question in my test and hence was toughest. I believe that my (incomplete) proof is not right in itself because it's too simple.

Can you help me to solve it in correct way ?

Answer 1

First prove that the statement holds for $f(0) = f(1)$ (this should be straightforward). Then assume that $f(1) \neq f(0)$ and set $g(x) = f(x) - (f(1) - f(0))x$, such that $g(1) = g(0)$. Therefore, either $g$ is constant or there is $t_0 \in (0,1)$ such that $$0 < |g'(t_0)| = |f'(t_0) - (f(1) - f(0))|$$ Can you conclude from there ?

Answer 2

Hint:

The function $f$ is linear if and only if $f(x) = (1-x) f(0) + x f(1)$ for all $x \in [0,1]$. (This says, for instance, that $f(1/4)$ is $1/4$ of the way between $f(0)$ and $f(1)$.)

So if $f$ is not linear, there exists an $x$ such that $f(x) \ne (1-x) f(0) + x f(1)$. Look at what the mean value theorem tells you about $f'(t_1)$ for some $0 < t_1 < x$, and for $f'(t_2)$ for some $x < t_2 < 1$. Now show that in either case, one of these two cases gives you a $t$ with $|f(1)-f(0)| < |f'(t)|$.

You could break it down according to whether $f(x) < (1-x) f(0) + x f(1)$ or $f(x) > (1-x) f(0) + x f(1)$. If the absolute values bother you, you can break it down further into the cases $f(1) - f(0) \ge 0$ and $f(1) - f(0) < 0$.

You can see the idea by drawing a picture: sketch a straight line L0 from $(0, f(0))$ to $(1, f(1))$. If any point $(x, f(x))$ of the graph of $f$ is not on this line, then draw line segments L1, L2 from $(0, f(0))$ to $(x, f(x))$ to $(1, f(1))$. You'll see that either L1 or L2 must have a greater slope than L0 (and the other one must have a lesser slope), and by the mean value theorem, there is a point where the derivative of $f$ equals this slope.

Answer 3

This answer uses the FTC on $f'$.

Do you know Cauchy-Schwarz inequality?

It states that, $$\left|\int u(x)v(x)\mathrm d x\right| \le \left(\int u(x)^2 \mathrm d x\right)^{\frac12}\left(\int v(x)^2 \mathrm d x\right)^{\frac12}$$ With equality only and only if $u = \lambda v$ with $\lambda$ a constant.

If $f$ is not linear then $f'$ is not constant. Let $$M = \sup_{t\in (0, 1)} \left|f'(t)\right|$$

• If $M$ is a maximum $\left(\exists t_0 \in (0, 1),\, M = \left|f'\left(t_0\right)\right|\right)$. Apply Cauchy-Schwarz for $u = 1$ and $v = f'$, So

$$\left|f(1) - f(0)\right| = \left|\int_0^1 f'(x) \mathrm d x\right| < \left(\int_0^1 f'(x)^2\right)^{\frac12} \le \left(f'\left(t_0\right)^2\right)^{\frac12} = \left|f'(t_0)\right|$$

• If $M$ is not a maximum $$\left|f(1) - f(0)\right| \le \int_{0}^1 \left|f'(t)\right|\mathrm d t < M$$ and by the definition of the supremum, there is $t_0\in (0, 1)$, $$\left|f(1) - f(0)\right| < \left|f'\left(t_0\right)\right| < M$$