Proving Monotonicity of f(x) in [0,1] with Limited Difference

[Zhujiao]

Homework Statement

Function f(x) defined in [0,1],and f(0)=f(1).If $$x_{1},x_{2}\in[0,1]$$ then $$|f(x_{1})-f(x_{2})|<|x_{1}-x_{2}|$$ (*)
Prove $$|f(x_{1}）-f(x_{2}）|<1/2$$

Homework Equations

I think maybe |a|-|b|$$\leq$$|a-b|$$\leq$$|a|+|b| will be helpful

The Attempt at a Solution

Well,I get some information from (*),I can prove that f(x)+x is monotone increasing and f（x)-x is monotone decreasing.Then I don't know what I should do. And I'm not sure whether I'm in the right way.

PS: why I can't see LaTex images clearly.They appear to be a black background and letters numbers or signs are not easy to recognize

 

[nate808]

.

Thank you for your post. I would be happy to help you with your problem. First, let's start with the given information. We know that f(x) is a function defined in the closed interval [0,1], and that f(0)=f(1). This means that the function is continuous and that the values at the endpoints are equal.

Next, we have the inequality |f(x1)-f(x2)|<|x1-x2|, which can also be written as |f(x1)-f(x2)|/|x1-x2|<1. This tells us that the absolute value of the slope of the secant line connecting any two points on the graph of f(x) is always less than 1.

Now, let's consider the points x1=0 and x2=1/2. Plugging these values into the inequality, we get |f(0)-f(1/2)|<|0-1/2|, which simplifies to |f(1/2)|<1/2. Since we know that f(0)=f(1), we can also write this as |f(0)-f(1)|<1/2.

From this, we can see that the maximum difference between any two points on the graph of f(x) is 1/2. This means that |f(x1)-f(x2)|<1/2 for any two points x1 and x2 in the interval [0,1]. Therefore, we have proved that |f(x1)-f(x2)|<1/2 for all x1 and x2 in [0,1].

As for your attempt at a solution, it seems like you are on the right track. However, instead of trying to prove that f(x)+x and f(x)-x are monotone increasing and decreasing, you can use the given inequality to directly prove the desired result.

I hope this helps. As for the issue with the LaTex images, it may be a problem with your internet connection or browser. You can try clearing your browser's cache and reloading the page. If the problem persists, you can try using a different browser or contacting the website administrator for assistance.

 

[Architeuthis Dux]

I would approach this problem by first defining the terms and variables involved. In this case, we are given a function f(x) defined in the interval [0,1] and we are also given the condition that f(0) = f(1). The statement (*) tells us that the difference between the values of f(x) at any two points in [0,1] is always less than the difference between those two points.

To prove that |f(x1) - f(x2)| < 1/2, we can start by using the triangle inequality, which states that for any two real numbers a and b, |a-b| ≤ |a| + |b|.

Using this inequality, we can rewrite the statement (*) as |f(x1) - f(x2)| ≤ |x1 - x2|. Since we are trying to prove that |f(x1) - f(x2)| is less than 1/2, we can set up the following inequality: |f(x1) - f(x2)| < 1/2.

Now, we can substitute the previous inequality into this one: |x1 - x2| < 1/2. This is true for any two points x1 and x2 in [0,1], so we can say that the difference between any two points in [0,1] is always less than 1/2.

Since f(x) is a continuous function, this means that the difference between any two values of f(x) in [0,1] is also less than 1/2. Therefore, we can conclude that |f(x1) - f(x2)| < 1/2 for all x1 and x2 in [0,1], which proves the statement.

As for the issue with the LaTeX images, it could be a problem with your browser or device. I would suggest trying to view the page on a different device or clearing your browser's cache and cookies. If the issue persists, you can try reaching out to the website's technical support team for assistance.