# If $f$ is monotone increasing on an interval and has a jump discontinuity at $x_0$, show that the jump is bounded above by $f(x_1)-f(x_2)$

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!

$$x_1$$ and $$x_2$$ are supposed to be given points. You cannot let $$x_2 \to x_1$$. If $$x_0 then $$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)$$.

• Thanks! That is very helpful – Nolando Nov 19 '18 at 5:26

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.

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