1. ## Finding functions

So far, I have this:

Proving $|f(x)-f(y)|$ is continuous:

Need to show: $\forall \ \epsilon>0 \exists \delta>0 s.t \ |y-x| < \delta \ \Rightarrow |f(x)-f(y)|< \epsilon$.

$|y-x|< \delta \Rightarrow |y-x|^2< \delta^2$

Therefore: $|f(x)-f(y)| \leq (y-x)^2< \delta^2$

Let $\delta^2=\epsilon$

Hence $|f(x)-f(y)| \leq (y-x)^2< \delta^2=\epsilon$ so continuity is proved.

However, i'm sure this helps but I can't see why. I've also tried expanding the right and left hand sides but that doesn't help either.

Does anyone have any hints?

2. Hello,

You're right, the functions $f$ satisfying $|f(x)-f(y)|\leqslant (x-y)^2$ for all $x,y\in\mathbf{R}$ are continuous on $\mathbf{R}$. In fact they are even differentiable on $\mathbf{R}$. Can you show it ?

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

$\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \leq \frac{h^2}{h}=h \rightarrow 0$.

Setting $h=y-x$.

Therefore:

$|f(x)-f(y)| \leq (x-y)^2 \ \forall x,y \in \mathbb{R}$

Hence $|f(x)-f(y)| \leq 0$

$0 \leq f(x)-f(y) \leq 0 \Rightarrow \ f(x)=f(y)$

Therefore $f(x)$ and $f(y)$ are the same number.

Therefore function f is a constant polynomial.

4. Originally Posted by Showcase_22
$|f(x)-f(y)| \leq (x-y)^2$

$\lim_{h \rightarrow 0} \frac{f(x+h)}{h} \leq \frac{h^2}{h}=h \rightarrow 0$.
You forgot one term: $\lim_{h \rightarrow 0} \frac{f(x+h){\color{blue}-f(x)}}{h} \leqslant \frac{h^2}{h}=h \rightarrow 0$.
Anyway this does not show that $f$ is differentiable at $x$ since it is possible for $\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} =-\infty$ and $\lim_{h \rightarrow 0} \frac{f(x+h){-f(x)}}{h} \leqslant 0$ to hold simultaneously.

Hint: use the same approach to show that $|f'(x)|\leqslant 0$.

5. Sorry about the first part, it was a typo.

Originally Posted by flyingsquirrel
You forgot one term: [tex]
Anyway this does not show that $f$ is differentiable at $x$ since it is possible for $\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} =-\infty$ and $\lim_{h \rightarrow 0} \frac{f(x+h){-f(x)}}{h} \leqslant 0$ to hold simultaneously.
Since we're taking the modulus of $f(x)-f(y)$, wouldn't this imply $|-\infty| \leq 0$ giving $+\infty \leq 0$ which is a contradiction?

6. Originally Posted by Showcase_22
Since we're taking the modulus of $f(x)-f(y)$, wouldn't this imply $|-\infty| \leq 0$ giving $+\infty \leq 0$ which is a contradiction?
If you are thinking about saying that $\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\leqslant 0$ implies $\left| \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\right|\leqslant 0$, no, it would not lead to a contradiction because $a\leqslant b$ does not imply $|a|\leqslant |b|$. (the reason is that $t\mapsto |t|$ is not an increasing function)