1
$\begingroup$

If $f$ is monotone increasing on an interval and has a jump discontinuity at $x_0$ in the interior of the domain show that the jump is bounded above by $f(x_1) - f(x_2)$ for any two points $x_1$, $x_2$ of the domain surrounding $x_0$, $x_1 < x_0 < x_2$.

So I've I tried solving this here is what I have:

Let $f$ be monotone increasing on an interval $A$ that has a jump discontinuity at $x_0$ on the interior of the domain where $x_0 \in A$. Let there be any $x_1, x_2 \in A$ where $x_1< x_0 < x_2$. Then by definition of monotone increasing $f(x_1) \leq f (x_2)$.

From here I want to say that $f(x_1) \leq f(x_0) \leq f(x_2) \rightarrow f(x_1) - f(x_1) \leq f(x_0) \leq f(x_2) - f(x_1)$ and that $f(x_0) \leq f(x_2) -f(x_2)$. But I'm not sure if I'm going in the right direction since I can't say for sure that $f(x_0)$ is really less than $f(x_2)-f(x_1)$. Or would I need to do something like $f(x_0)-f(x_1)\leq f(x_2)-f(x_1)$ and go from there. Any help would be appreciated as I'm somewhat unsure on this problem!

$\endgroup$
1
$\begingroup$

$x_1$ and $x_2$ are supposed to be given points. You cannot let $x_2 \to x_1$. If $x_0<t<x_2$ then $f(t) \leq f(x_2)$. Let $t \to x_0$ to get $f(x_0+) \leq f(x_2)$. [ $f(x_0+)$ is the right hand limit of $f$ at $x_0$]. Similarly prove that $f(x_0-) \geq f(x_1)$. From these two se get $f(x_0+)-f(x_0-) \leq f(x_2)-f(x_1)$.

$\endgroup$
  • $\begingroup$ Thanks! That is very helpful $\endgroup$ – Nolando Nov 19 '18 at 5:26
0
$\begingroup$

First of all you have to define " the jump at $x_0$ "

It means $$\lim _{x\to x_0+}f(x) -\lim _{x\to x_0-}f(x) $$

The above limits exist due to the monotonicity of $f(x)$

The rest is not too complicated to prove and you can do it.

$\endgroup$
  • $\begingroup$ Thanks you! That definitely helps $\endgroup$ – Nolando Nov 19 '18 at 5:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.