# Prove Baby Rudin Problem: Constant f If |f(x)-f(y)|≤(x-y)^2

• Government$This is the mean value theorem, just a different way of writing it.In summary, the conversation discusses the attempt to prove that a function, f, is constant based on the given inequality. The individual asks if assuming that |f(x) - f(y)| ≤ (x-y)^2 implies that f(x) is always greater than f(y), and if it is allowed to take x = y. They also suggest using the first derivative being 0 as a starting point, but it is noted that the function is not given to be differentiable. Another approach is mentioned, which involves dividing both sides by |x-y| and looking at it as f'(y Government$

## 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?

Government$said: ## 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. Government$ said:
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.
Government$said: 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? Government$ said:
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?

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

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

## 1. What is the Baby Rudin Problem?

The Baby Rudin Problem is a mathematical problem that involves proving a certain property for a function called a "constant f". This problem is commonly encountered in undergraduate level mathematics courses and is named after mathematician Walter Rudin.

## 2. Can you explain the notation in the problem statement?

The notation |f(x)-f(y)|≤(x-y)^2 means that the absolute value of the difference between the function values at two points, f(x) and f(y), is less than or equal to the squared difference between the two points, (x-y)^2. In other words, the function values cannot vary too much as the points get closer together.

## 3. How do you approach solving this problem?

The first step in solving the Baby Rudin Problem is to carefully read and understand the problem statement. Then, it is important to consider the properties of the function f and try to manipulate the given inequality to prove the desired property. It may also be helpful to use mathematical techniques such as the Mean Value Theorem or the Triangle Inequality.

## 4. Is this problem difficult to solve?

The difficulty of solving the Baby Rudin Problem can vary depending on an individual's mathematical background and experience. It may be challenging for those who are new to mathematical proof writing, but with practice and perseverance, it can be solved.

## 5. What are the real-world applications of this problem?

The Baby Rudin Problem may not have direct real-world applications, but it is an important concept in understanding the behavior of functions and their properties. It also helps develop critical thinking and problem-solving skills, which are useful in various fields such as science, engineering, and economics.

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