With some conditions on $f$, resolve the claim $|f(x)-f(y)|\le|x-y|$ for all $x,y\in[0,1]$.

Question

Here's the problem . . .

Prove or disprove:$\;$If $f:[0,1]\to\mathbb{R}$ is a twice differentiable function such that

• $f(0)=f(1)=0$.
• $|f''(x)|\le 2$ for all $x\in[0,1]$.

then $|f(x)-f(y)|\le|x-y|$ for all $x,y\in[0,1]$.

Some context . . .

This problem was posted yesterdayby @Alex (user 1258161)

https://math.stackexchange.com/questions/4829557/

but was closed for lack of context, and then deleted by the post author.

I found the problem interesting so I tried it just for fun.

Considering the possibility that the claim might be false, I first tried to construct a counterexample. Not finding one, I tried to prove the claim, but so far without success. Perhaps I'm missing something simple, a commonly used approach or a standard trick.

In any case, as shown below, I made some progress, and perhaps my partial results could be of some use in the quest for a complete solution.

Attempting to prove the claim, here's what I tried . . .

Assume the hypothesis and suppose $|f(u)-f(v)| > |u-v|$ for some $u,v\in[0,1]$.

Our goal is to derive a contradiction.

We can assume

• $u < v$, else replace $f(x)$ by $f(1-x)$.
• $f(u) > f(v)$, else replace $f(x)$ by $-f(x)$.

Thus we have $f(u)-f(v) > v-u$.

By the MVT, there exists $a\in(u,v)$ such that $f(u)-f(v)=f'(a)(u-v)$, hence $f'(a) < -1$.

By the MVT, for arbitrary $x\in[0,1]{\setminus}\{a\}$, there exists $b$ with $\min(x,a) < b < \max(x,a)$ such that $f'(x)-f'(a)=f''(b)(x-a)$, so $f'(x) < 1$.

Thus we have $f'(x) < 1$ for all $x\in[0,1]$.

Then by the MVT

• Since $f(0)=0$, it follows that $f(x) < x$ for all $x\in(0,1]$.
• Since $f(1)=0$, it follows that $f(x) > x-1$ for all $x\in[0,1)$.

By Rolle's Theorem, there exists $c\in(0,1)$ such that $f'(c)=0$.

By the MVT, for arbitrary $x\in[0,1]{\setminus}\{c\}$, there exists $d$ with $\min(x,c) < d < \max(x,c)$ such that $f'(x)-f'(c)=f''(d)(x-c)$, so $f'(x) > -2$.

Thus we have $f'(x) > -2$ for all $x\in[0,1]$.

Then by the MVT

• Since $f(0)=0$, it follows that $f(x) > -2x$ for all $x\in(0,1]$.
• Since $f(1)=0$, it follows that $f(x) < 2-2x$ for all $x\in[0,1)$.

Then we have

• $f(x) < \min(x,2-2x)$ for all $x\in(0,1)$, hence $f(x) < 2/3$ for all $x\in[0,1]$.
• $f(x) > \max(-2x,x-1)$ for all $x\in(0,1)$, hence $f(x) > -2/3$ for all $x\in[0,1]$.

That's my progress so far.

How to resolve the problem?

Answer 1

By $\lvert f''(x) \rvert \leq 2$ and the MVT, we have $\lvert f'(x) - f'(y) \rvert \leq 2 \lvert x - y \rvert.$ Now, suppose $\exists c \in (0, 1)$ such that $f'(c) > 1$ for contradiction. Then, for all $t \in (0, 1)$, $$f'(c) - f'(t) \leq 2 \lvert t - c \rvert \implies f'(t) \geq f'(c) - 2 \lvert t - c \rvert > 1 - 2 \lvert t - c \rvert,$$ implying $$\begin{split} f(1) - f(0) &= \int_0^1 f'(t) dt \\ &> \int_0^1 (1 - 2 \lvert t - c \rvert) dt \\ &= \int_0^c (1+2(t-c)) dt + \int_c^1 (1-2(t-c)) dt \\ &= [t + (t-c)^2]_0^c + [t-(t-c)^2]_c^1\\ & = c - c^2 + (1-(1-c)^2) - c \\ &= -2c^2+2c \\ &= 2c(1-c)\\ &> 0, \end{split}$$ contradicting $f(0) = f(1)$. Therefore, $\forall c \in (0, 1), f'(c) \leq 1$. Similarly, $\forall c \in (0, 1), f'(c) \geq -1$. In other words, $\forall c \in (0, 1),\lvert f'(c) \rvert \leq 1$, and by the MVT, we have the desired inequality.

Answer 2

Edit: Direct integration can be used to show that $|f'(x)|<1$ for all $x\in(0,1)$.

By hypothesis, for all $0\leq t\leq 1$ $$ -2\leq f''(t)\leq 2$$ Integration over any interval with endpoints $0\leq u,v\leq 1$ yields $$ -2|v-u|\leq f'(v)-f'(u)\leq 2|v-u|$$ Fix $0<u<1$. Then, for all $0\leq v\leq 1$ $$f'(u)-2|v-u|)\leq f'(v)\leq f'(u)+2|v-u|$$ Integrating over $[0,1]$ yields $$f(u)-2\int^1_0|v-u|\,dv\leq 0\leq f'(u)+2\int^1_0|v-u|\,dv$$ that is $$f'(u)-1 + 2(u-u^2)\leq 0\leq f'(u)+1-2(u-u^2)$$ As $u-u^2>0$, $$f'(u)-1<f'(u)-1 + 2(u-u^2)\leq 0\leq f'(u)+1-2(u-u^2)< f'(u)+1$$ Hence $$-1< f'(u)< 1, \qquad 0<u<1$$ This shows that for any $0\leq x<y\leq 1$ $$|f(y)-f(x)|<|y-x|$$

Answer 3

As a special case of the mean value theorem for divided differences, we have:

Lemma. For any distinct $x_0, x_1, x_2 \in [0, 1]$ there exists $\xi \in [0, 1]$ such that

$$ \frac{ \frac {f(x_{2})-f(x_{1})}{x_{2}-x_{1}} - \frac {f(x_{1})-f(x_{0})}{x_{1}-x_{0}}}{x_{2}-x_{0}} = \frac{f''(\xi)}{2!}. $$

Plugging $(x_0, x_1, x_2) = (0, x, y)$ and $(1, x, y)$ respectively, the lemma tells that there exist $\xi_0, \xi_1 \in [0, 1]$ satisfying

\begin{align*} yf(x) - x f(y) &= xy(x - y) \frac{f''(\xi_0)}{2!}, \\ (1-y) f(x) - (1-x) f(y) &= -(1-x)(1-y)(x-y) \frac{f''(\xi_1)}{2!}. \end{align*}

From this and using the bound $|f''(t)| \leq 2$, $t \in [0, 1]$, we get

\begin{align*} |f(x) - f(y)| &\leq |yf(x) - x f(y)| + |(1-y) f(x) - (1-x) f(y)| \\ &\leq (xy + (1-x)(1-y)) |x - y|. \end{align*}

Now the desired conclusion follows from the continuity of $f$ and the fact that $xy + (1-x)(1-y) \leq 1$ for all $x, y \in [0, 1]$.