If $\lvert f(x)-f(y) \rvert \le \lvert x-y \rvert ^2$ for all $x$ and $y$ in $\mathbb{R}$ then $f$ is a constant function

Question

Assume that $\lvert f(x)-f(y) \rvert \le \lvert x-y \rvert ^2$ for all $x,y\in \mathbb{R}$. Prove that $f$ is the constant function.

My first thought was to use continuity of $f$, so that we get that $\forall x\in\mathbb{R},\forall \epsilon>0, \exists \delta>0, \forall y \in \mathbb{R}, \lvert x-y \rvert<\delta \implies \lvert f(x)-f(y) \rvert < \epsilon$.

So choosing $y$ so that the above property holds, we get that $\lvert f(x)-f(y) \rvert \le \lvert x-y \rvert ^2 <\delta^2$

Using the inequality I found (supposing it is correct), how would I continue to prove that $f(x)=f(y)$, and so their difference must be zero.

Asked many times on the site. Consider the points $x_k=x+\frac{k}n(y-x)$, the hypothesis applied to each $(x_{k-1},x_k)$ yields $|f(x_k)-f(x_{k-1})|\leqslant\frac1{n^2}|x-y|^2$. Summing these from $k=1$ to $k=n$ and applying the triangular inequality yields $|f(x)-f(y)|\leqslant n\cdot\frac1{n^2}|x-y|^2$ and you are done. — Did, Dec 13 '16 at 22:40
@fourierwho That one does not know a priori that it exists and that one does not need it to solve the exercise, perhaps? — Did, Dec 13 '16 at 22:41
The inequality explicitly shows that the derivative exists and is 0. — xyzzyz, Dec 13 '16 at 22:42
The proof using the derivative is fine, however it does require invoking the mean value theorem or similar to prove $f'(x) = 0 \implies f(x) = C$ to finish the proof. Thus it's a bit more of a complex proof than the primitive proof suggested above which just requires the triangle inequality. What is best is a matter of taste I guess. — Winther, Dec 13 '16 at 22:48
Related: How to show that every $\alpha$-Hölder function, with $\alpha>1$, is constant? ; Proving that a function satisfying $|f(x)-f(y)| \leq |x-y|^3$ is constant ; If $f$ satisfies $|f(x_1)-f(x_2)|\leqslant(x_1-x_2)^2$ on an interval — Winther, Dec 13 '16 at 22:55

score 3 · Accepted Answer · answered Dec 13 '16 at 22:44

3

Use the definition of the derivative.

$$0\le\lim_{x\to y} \left|{f(x) - f(y)\over x-y}\right|\le \lim_{x\to y}|x-y|=0$$

So the function is differentiable with derivative $0$ at all points.

answered Dec 13 '16 at 22:44

Adam Hughes

36,777

score 0 · Answer 2 · answered Dec 13 '16 at 22:47

0

Let $y=x+h$ with $h\ne 0$. Then $$0 \le \left| \frac{f(x+h)-f(x)}{h}\right|<|h|$$ and so $$-|h| \le\frac{f(x+h)-f(x)}{h} \le |h|.$$Take limit for $h$ to zero then $$f^\prime(x)=0.$$

answered Dec 13 '16 at 22:47

student forever

1,266

score 0 · Answer 3 · answered Dec 13 '16 at 22:48

0

An easy way to understand this is, when $x\neq y$, we can divide by $|x-y|$ to get that: $$\frac{|f(x)-f(y)|}{|x-y|}\leq |x-y|$$ Now, one characterization of the derivative of $f$ is: $$f'(x) = \lim_{y\to x}\frac{f(x)-f(y)}{x-y}$$ As out inequality is valid for all $x,y$, we can pick a sequence $y_k$ with $y_k\to x$, $y_k\neq x$ and look at: $$\lim_{k\to\infty} \left|\frac{f(x)-f(y_k)}{x-y_k}\right|\leq \lim_{k\to\infty} |x-y_k|$$ The left-hand side of this is just the derivative at $x$, so we have that: $$|f'(x)|\leq 0$$ From this, it's clear that $f'(x) = 0$, and so $f(x)$ is constant.

answered Dec 13 '16 at 22:48

Mark Schultz-Wu

6,986

take a function takes the values $\pm3$ (rational/irrational) and then absolute value will have limit but not the function. Saying left hand side is $|f^\prime(x)|$ is assuming that $f^\prime(x)$ exists. – student forever Dec 13 '16 at 22:51
I don't believe that function satisfies the initial constraint, as if you have $i, r$ (irrational and rational) within epsilon of eachother, then the difference of your function would be $3-(-3)=6$, which for suitably small epsilon would contradict our assumptions on $f$. – Mark Schultz-Wu Dec 13 '16 at 23:52
I am saying it for general. Here it is true, since it exists. But you used it without knowing/showing it exists. – student forever Dec 13 '16 at 23:54
Indeed this is no proof that $f'(x)$ exists, only that if it exists then it is $0$. – Did Dec 14 '16 at 10:40

If $\lvert f(x)-f(y) \rvert \le \lvert x-y \rvert ^2$ for all $x$ and $y$ in $\mathbb{R}$ then $f$ is a constant function

3 Answers3