# Baby Rudin problem

Government\$

## Homework Statement

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.

## 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 don't 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?

Mentor

## Homework Statement

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.

## The Attempt at a Solution

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

Here is what i don't 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 $f(y)$?
No it doesn't imply that. |f(x) - f(y)| is always >= 0, regardless of which function is larger.
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?

Homework Helper
Gold Member
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?
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?

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

Homework Helper
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
$$z_k=y+(x-y)(k/n)$$