Cfrgtkky

Suppose $f:[a,b]tomathbb R$ be a function such that $|f(x)-f(y)|<epsilon_0$ for all $x,yin[a,b]$. Then $M-mleepsilon_0$, where $M=sup{f(x):xin[a,b]}$ and $m=inf{f(x):xin[a,b]}$.

I went this way: suppose $M-m>epsilon_0$, hence $M>epsilon_0+m$, hence there is an $xin[a,b]$ such that$$M>f(x)>epsilon_0+m,tag1$$and similarly $M-epsilon_0>m$ implies$$M-epsilon_0>f(y)>m.tag2$$Subtracting $(1)$ from $(2)$ says $-epsilon_0>f(y)-f(x)>-epsilon_0$, which is absurd.

But on a second thought, I realized that my argument says that $M-m>varepsilon$ is false for any $varepsilon>0$, not just for $epsilon_0$, implying $M-m=0$. What did I do wrong?

Your mistake is thinking you can just subtract inequalities. You can't do that. For example,

$$1>0$$

is true, and $$2>0$$ is also true, but subtracting these two inequalities, I get $$-1>0$$ which is not true.

The problem with subtracting inequalities is that when you subtract equations, you actually multiply one of them by $(-1)$ and then add them. With inequalities, you cannot do that because multiplication by a negative number reverses the inequality.

For an actual proof, a sketch of it would be this:

• Find some $x$ for which $f(x)$ is "near" $M$
  


• Find some $y$ for which $f(y)$ is "near" $m$
  


• Use the fact that $|f(x)-f(y)|<epsilon_0$ and the fact that $$|f(x)-f(y)| = |f(x)-M+M-f(y)+m-m|leq |f(x)-M| + |f(y)-m| + |M-m|$$ to reach a conclusion.

Naturally, the "near" in this sketch must, in the final proof, be a more rigorous statement. Good luck!

Last step is wrong. $aleq bleq c$ and $a'leq b'leq c'$ do not imply $a-a'leq b-b'leq c-c'$.

You may also proceed as follows:

• Note that $|x|$ is continuous.


• Choose sequences $(x_n), (y_n)$ with $lim_{n to infty} f(x_n) = M$ and $lim_{n to infty} f(y_n) = m$.

It follows:
$$|f(x_n) - f(y_n)| stackrel{n to infty}{longrightarrow} M-m$$
Now, as $|f(x_n) - f(y_n)| < epsilon_0 Rightarrow M-m leq epsilon_0$.