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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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We can write $v(x)=w(x)+\Delta\theta(x-x^*)$, with $\Delta=\Delta v_x$ and $\theta(t)=1$ for $t>0$ and $=0$ for $t<0$ (I've slightly changed notations).

Now $\int_{x^*-z}^{x^*}\theta(x-x^*)\, dx=0$, $\int_{x^*}^{x^*+z}\theta(x-x^*)\, dx= z$, so the $\Delta\theta$ part of $v$ makes a contribution of $-\Delta z$ to the difference of the integrals, and this is $>bz$ in absolute value, by our assumption on $\Delta$.

On the other hand, $$ \left| \int_{x^*-z}^{x^*}w(x)\, dx - \int_{x^*}^{x^*+z}w(x)\, dx\right| = \left| \int_{x^*-z}^{x^*}(w(x)-w(x^*))\, dx - \int_{x^*}^{x^*+z}(w(x)-w(x^*))\, dx\right| \le \int_{x^*-z}^{x^*}L|x-x^*|^{\alpha}\, dx + \int_{x^*}^{x^*+z}L|x-x^*|^{\alpha}\, dx \le 2Lz^{1+\alpha} . $$ Putting these together, we'll obtain the desired inequality.

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