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Post Closed as "Not suitable for this site" by Christian Remling, Chris Godsil, user6976, Stefan Waldmann, Stefan Kohl♦
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stochic
stochic
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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}$$w_{x}:=\nu_{x}-\Delta\nu_{x^{*}}\mathbb{I}_{\{x\geq x^{*}\}}$ it holds that $\underset{x,y}{\text{sup}}\left|w_{x}-w_{y}\right|\leq L\delta^{\alpha}$ with $\alpha\in\left(0,1\right)$ and the sup is taken over all $x,y$ s.t. $\left|x-y\right|<\delta$. Finally we assume $\Delta\nu_{x^{*}}>b$. Then I have to prove: $$ \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.

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 $w_{x}:=\nu_{x}-\Delta\nu_{x}$ 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.

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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stochic
stochic
  • 115
  • 5

Inequality concerning integral w.r.t. Hölder functions

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 $w_{x}:=\nu_{x}-\Delta\nu_{x}$ 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.