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I didn't get any reaction on that question on Math.stackexchange. So I hope to get a hint from the experts.

I have to prove the following inequality. Let $\nu$ be a function on $[0,1]$ with one discontinuity point $x^{*}$. Furthermore, we set $\Delta\nu_{x}:=\nu_{x}-\underset{t\uparrow x}{\text{lim }}\nu_{t}$. We do the following assumptions. For $w_{x}:=\nu_{x}-\Delta\nu_{x^{*}}\mathbb{I}_{\{x\geq x^{*}\}}$ it holds that $\underset{x,y}{\text{sup}}\left|w_{x}-w_{y}\right|\leq L\delta^{\alpha}$ with $\alpha\in\left(0,1\right)$ and the sup is taken over all $x,y$ s.t. $\left|x-y\right|<\delta$. Finally we assume $\Delta\nu_{x^{*}}>b$. Then I have to prove: $$ \frac{1}{z}\left|\int_{x^{*}-z}^{x^{*}}\nu_{s}ds-\int_{x^{*}}^{x^{*}+z}\nu_{s}ds\right|\geq b-2Lz^{\alpha}. $$ I think it should be a straightforward application of the reverse triangle inequality, but I am not really able to work it out rigorously.

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  • $\begingroup$ You have a strange definition of $w_x$: it should really be $\nu_x-\Delta\nu_{x^*}I_{\{x\ge x^*\}}$ where $I_E$ is the indicator function of the set $E$. That's probably what is confusing you. Voting to close. $\endgroup$
    – fedja
    Commented Jul 24, 2017 at 14:46
  • $\begingroup$ This is not my own creation. If you have editor rights, feel free to correct my post. Nevertheless, could you give me a hint, how to tackle this task? Edit: I have corrected the definition of $w$. $\endgroup$
    – stochic
    Commented Jul 24, 2017 at 14:55
  • $\begingroup$ The inequality is trivial, but I suspect you confused yourself (and others, over at MSE) with the not very reader friendly notation (for starters, why do you write $f_x$ for what the rest of the world calls $f(x)$). Your function $\nu=w+b\theta(x-x^*)$ has a Holder continuous part and a jump at $x^*$, and now just look at the contributions these two parts make to your integrals. $\endgroup$
    – Christian Remling
    Commented Jul 24, 2017 at 17:08
  • $\begingroup$ @ChristianRemling I am a statistician, and that's why I am used to apply that notation. I am very sorry for that. $\endgroup$
    – stochic
    Commented Jul 24, 2017 at 17:15
  • $\begingroup$ @ChristianRemling Would you like to write down an answer on MSE? As I think that this question isn't really research level. Here is the link to the MSE question: math.stackexchange.com/questions/2338000/… Then I think one could close/delete this question in order to keep MO "clean". $\endgroup$
    – stochic
    Commented Jul 24, 2017 at 17:27

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