Functions in reals such that inequality holds

[maximus101]

for which f: $$R$$ $$\rightarrow$$ $$R$$ such that
$$\forall$$ $$x,y$$$$\in$$ $$R$$

does

| $$f(y)$$ - $$f(x)$$ | $$\mid$$ $$\leq$$ $$(y-x)^2$$

hold

 

[Buri]

Which inequality?

 

[maximus101]

sorry just fixed it

 

[maximus101]

so I did the following:

$$\dfrac{|f(y) - f(x)|}{|y-x|}\leqslant |y-x|$$

was obtained by $$(y-x)^2$$=$$|y-x||y-x|$$ and dividing both sides by $$|y-x|$$.


then trying to use the fact that

$$\displaystyle\lim_{y\to x}\frac{f(y)-f(x)}{y-x}$$ is the derivative at x

but not sure what's next how to combine the two?

 

[Buri]

Is the function differentiable? If it isn't you can't use the limit thing. But if it is, apply the limit to both sides. You should then get something very familiar and be able to deduce what kind of function f must be.

 

[Buri]

Okay, if f isn't differentiable, here's a hint. Divide the interval from [x,y] into n equal sub intervals. See if you can get anything from that.

 

[Yuqing]

Wouldn't this statement
$$\dfrac{|f(y) - f(x)|}{|y-x|}\leqslant |y-x|$$
automatically imply the function is differentiable?

 

[JG89]

Wouldn't this statement
$$\dfrac{|f(y) - f(x)|}{|y-x|}\leqslant |y-x|$$
automatically imply the function is differentiable?

Yes. Using the squeeze theorem you can see that the derivative is actually 0 everywhere, implying the only functions satisfying the inequality are constant functions.

 

[TylerH]

Am I missing something? I don't see any derivatives anywhere, just the difference quotient.

If you indented for it to be the difference quotient, then a counter example would be $$r \mapsto r^3$$. For all |x-y| >= 1, $$\left| x^3 - y^3 \right| > \left( x - y \right)^2$$

 

[JG89]

Assume there exists a function f satisfying $$0 \le |f(y) - f(x)| \le (y-x)^2 = |y-x|^2$$ for any y, x in the domain of f. This implies that $$0 \ le \frac{|f(y) - f(x)|}{|y-x|} \le |y-x|$$.

Let y approach x. By the squeeze theorem, we see that the limit exists, is equal to 0, and is in fact the derivative of f evaluated at x. So f'(x) = 0 for all x in its domain, which means that f must be a constant function.