Prove that $ \sup_{s,t\in J}|f\left(t\right)-f\left(s\right)|=\sup_{x\in J}f\left(x\right)-\inf_{x\in J}f\left(x\right) $

Question

Assume $ f $ is bounded function in an interval $ J $, and prove that:

$ \sup_{s,t\in J}|f\left(t\right)-f\left(s\right)|=\sup_{x\in J}f\left(x\right)-\inf_{x\in J}f\left(x\right) $

Here's what I tried:

let $ \varepsilon>0 $. We can find $ s,t\in J $ such that

$ \sup_{x\in J}f\left(x\right)<f\left(s\right)+\frac{\varepsilon}{2} $ and $ f\left(t\right)-\frac{\varepsilon}{2}<\inf_{x\in J}f\left(x\right) $

Thus,

$ 0\leq\sup_{x\in J}f\left(x\right)-\inf_{x\in J}f\left(x\right)<f\left(s\right)-f\left(t\right)+\varepsilon\leq|f\left(s\right)-f\left(t\right)|+\varepsilon\leq\sup_{t,s\in J}|f\left(s\right)-f\left(t\right)|+\varepsilon $

And therefore

$ \left(\sup_{x\in J}f\left(x\right)-\inf_{x\in J}f\left(x\right)\right)-\sup_{s,t\in J}|f\left(t\right)-f\left(s\right)|<\varepsilon $

Now, I will argue that $ \sup_{s,t\in J}|f\left(s\right)-f\left(t\right)|\leq\sup_{x\in J}f\left(x\right)-\inf_{x\in J}f\left(x\right) $.

Let $ t,s\in J $

Then, $ f\left(s\right)-f\left(t\right)\leq\sup_{s\in J}f\left(s\right)-\inf_{t\in J}f\left(t\right)=\sup_{x\in J}f\left(x\right)-\inf_{x\in J}f\left(x\right) $

And also

$ f\left(t\right)-f\left(s\right)\leq\sup_{t\in J}f\left(t\right)-\inf_{s\in J}f\left(s\right)=\sup_{x\in J}f\left(x\right)-\inf_{x\in J}f\left(x\right) $

Therefore,

$ |f\left(s\right)-f\left(t\right)|\leq\sup_{x\in J}f\left(x\right)-\inf_{x\in J}f\left(x\right) $

for any, $ t,s \in J $. So $ \sup_{x\in J}f\left(x\right)-\inf_{x\in J}f\left(x\right) $ is an upper bound, and therefore $ \sup_{t,s\in J}|f\left(s\right)-f\left(t\right)|\leq\sup_{x\in J}f\left(x\right)-\inf_{x\in J}f\left(x\right) $

And finally we have:

$ 0\leq\left(\sup_{x\in J}f\left(x\right)-\inf_{x\in J}f\left(x\right)\right)-\sup_{t,s\in J}|f\left(s\right)-f\left(t\right)|<\varepsilon $

for any $ \varepsilon >0 $.

So we can conclude that

$ \left(\sup_{x\in J}f\left(x\right)-\inf_{x\in J}f\left(x\right)\right)=\sup_{t,s\in J}|f\left(s\right)-f\left(t\right)| $

I feel like I made it complicated and one can find much simpler proof.

So, I'll be glad to hear if this proof works at all, and if anyone can suggest simpler solution that would be helpful. Thanks in advance

Answer 1

First of all notice that you can drop the absolute value. Then you have

$$\sup _{s,t} (f(t)-f(s)) = \sup _t\sup_s (f(t)-f(s)) = \sup_t (f(t)-\inf_s f(s)) = \sup_t f(t)-\inf_s f(s)$$ and that is it. I just used the fact that if something does not depend on $s$, than can be brought outside of the sup.