# Baby Rudin problem

1. Jan 24, 2013

### Government\$

1. The problem statement, all variables and given/known data
Let f be defined for all real x and suppose that,
$\left|f(x) - f(y)\right|\leq(x-y)^2$

for all real x and y. Prove that f is constant.

3. The attempt at a solution
First of all, is following allowed. Since f is constant then $\left|f(x) - f(y)\right|=0$, and form here to build my proof?

Here is what i dont understand. If i suppose that $\left|f(x) - f(y)\right|\leq(x-y)^2$ then does that imply that $f(x)$ is always greater then $f(y)$?
Since $(x-y)^2>0$ when $x≠y$ then that means that $\left|f(x) - f(y)\right|> 0$ because if it were to be $\left|f(x) - f(y)\right|= 0$ we would violate $x≠y$ . Onliy way that $\left|f(x) - f(y)\right|> 0$ is that $f(x)$ is always greater then $f(y)$

But this leads me to contradiction since $f$ should be constant i.e. $\left|f(x) - f(y)\right|=0$.

On the other hand $(x-y)^2=0$ iff $x=y$ then that means that $\left|f(x) - f(y)\right|= 0$.

So perhaps my question boils down to is can i take that $x=y$ or am i only allowed to that $x≠y$???

Also i am trying to think of all things that will prove that some function is constant. First thing that comes to mind is that first derivative is 0. Can that be a starting point?

2. Jan 24, 2013

### Staff: Mentor

No, this isn't allowed. You can't assume that f is constant - that's what you need to show. Starting with the given inequality, you need to arrive at the conclusion that f is constant.
No it doesn't imply that. |f(x) - f(y)| is always >= 0, regardless of which function is larger.

3. Jan 24, 2013

### jbunniii

It sounds promising, but you aren't given that the function is differentiable. Can you show it is differentiable based on the given inequality?

One observation is that the function is uniformly continuous, because if $|x - y| \leq 1$ we have
$$|f(x) - f(y)| \leq (x-y)^2 \leq |x - y|$$
In fact, this shows that $f$ is Lipschitz, a stronger condition than uniform continuity, but weaker than differentiability. Perhaps with a bit more work you can show differentiability?

4. Jan 25, 2013

### ArcanaNoir

Try dividing both sides by $\mid x-y\mid$ and then find a way to look at it as $f'(y)$.

5. Jan 25, 2013

### lurflurf

This is a dumb question (like so many in baby Rudin, what a terible book). Clearly f is differentiable, but I prefer to note

$$|\mathrm{f}(x)-\mathrm{f}(y)|=\left| \sum^n_{k=1}(\mathrm{f}(z_k)-\mathrm{f}(z_{k-1}))\right|\le \sum^n_{k=1}|\mathrm{f}(z_k)-\mathrm{f}(z_{k-1})|\le \sum^n_{k=1}(z_k-z_{k-1})^2=(x-y)^2/n\le \epsilon$$
where
$$z_k=y+(x-y)(k/n)$$