# Thread: Show a function f is constant, given...

1. ## Show a function f is constant, given...

Q: f from the the reals to the reals satisfies |f(x)-f(y)|<=|x-y|^2 (*), for all x,y in the reals. Show f is constant.

Suppose, without loss of generality, that y>x. Suppose also that f is strictly increasing, i.e. f(y)>f(x).

Then LHS of (*) is f(y)-f(x), and the RHS is (y-x)^2, with y-x>0,

Thus, (f(y)-f(x))/(y-x) <= y - x. Now, the limit of the LHS of inequality, as y tends to x, is f'(x), which is strictly postive, whereas the lim of the RHS as y tends to x is 0, so we have a contradiction, as the right hand side is always greater than the left hand side.

Similarly, we obtain a contradiction if we assume f is strictly decreasing, and thus I conclude f is constant.

Is this argument correct?

Thanks.

2. Originally Posted by Rudipoo
Q: f from the the reals to the reals satisfies |f(x)-f(y)|<=|x-y|^2 (*), for all x,y in the reals. Show f is constant.
Let $c\in \mathbb{R}$, then we know $|f(x) - f(c)| \leq (x-c)^2$.
For $x\not = c$ we have $\left| \frac{f(x) - f(c)}{x-c} \right| \leq |x-c|$.
Thus, the limit of the quotient exists and $\lim_{x\to c} \frac{f(x)-f(c)}{x-c} = 0$.
Thus, $f'(c) = 0$ for all $c\in \mathbb{R}$.

The function $f$ has vanishing derivative everywhere which means we must conclude that $f$ is a konstant function.

3. Originally Posted by ThePerfectHacker
Let $c\in \mathbb{R}$, then we know $|f(x) - f(c)| \leq (x-c)^2$.
For $x\not = c$ we have $\left| \frac{f(x) - f(c)}{x-c} \right| \leq |x-c|$.
Thus, the limit of the quotient exists and $\lim_{x\to c} \frac{f(x)-f(c)}{x-c} = 0$.
Thus, $f'(c) = 0$ for all $c\in \mathbb{R}$.

The function $f$ has vanishing derivative everywhere which means we must conclude that $f$ is a konstant function.
Or using the notation in the problem, divide both sides by $|x-y|$. We get $\left|\frac{f(x)-f(y)}{x-y} \right| \leq |x-y|$. Then $-|x-y| \leq \frac{f(x)-f(y)}{x-y} \leq |x-y|$. So $\lim\limits_{x \to y} -|x-y| \leq \lim\limits_{x \to y} \frac{f(x)-f(y)}{x-y} \leq \lim\limits_{x \to y} |x-y|$.

So by Squeeze theorem $f'(x) = 0$ for all $x \in \mathbb{R}$.

4. Originally Posted by manjohn12
Or using the notation in the problem, divide both sides by $|x-y|$. We get $\left|\frac{f(x)-f(y)}{x-y} \right| \leq |x-y|$. Then $-|x-y| \leq \frac{f(x)-f(y)}{x-y} \leq |x-y|$. So $\lim\limits_{x \to y} -|x-y| \leq \lim\limits_{x \to y} \frac{f(x)-f(y)}{x-y} \leq \lim\limits_{x \to y} |x-y|$.

So by Squeeze theorem $f'(x) = 0$ for all $x \in \mathbb{R}$.
Using squeeze theorem is one way to do it. Another way is to notice that the limit of an absolute value is zero if and only if the the limit of the function is zero. This follows straightforward by the definition.