Let $\varphi_{f,\delta}$ be a function defined by $\varphi_{f,\delta}(u)=\sup\left\{\left|f(y)-f(x)\right|:x,y\in\left[u-\delta,u+\delta\right]\right\}$ for a bounded and symmetric function $f:\left[-1,1\right]\rightarrow\mathbb{R}$ with compact support. Furthermore, the following holds. $\int_{\mathbb{R}}^{}\varphi_{f,\delta}(u)du=\mathcal{O}(\delta)$ as $\delta\rightarrow 0$. Question: How can I calculate upper bounds for following quantities using the functions $\varphi_{f,\delta}$.

(a) $\int_{-c}^{c}f(x+p)-f(y+p)dp$ and

(b) $\int_{-c}^{c}(f(x+p)-f(y+p))^{k}dp$ with $k\in\mathbb{N}$.

My own effort:

(a) It obviously holds that $f(x+p)-f(y+p)\leq\left|f(x+p)-f(y+p)\right|\leq \sup\left\{\left|f(z)-f(z')\right|:z,z'\in\left[x+p,y+p\right]\right\}$. In order to apply the functions $\varphi_{f,\delta}$ for a suitable $\delta$, we have to symmetrize the interval $\left[x+p,y+p\right]$. Therefore, we could replace the interval by $\left[\xi_{1},\xi_{2}\right]$ with $\xi_{1}=\frac{x+y+2p}{2}-(\frac{x+y+2p}{2}-(x+p))$ and $\xi_{2}=\frac{x+y+2p}{2}+(\frac{x+y+2p}{2}-(x+p))$. Where is my mistake?

(b) This should follow from (a) via Cauchy Schwarz or Jensen's inequality.

  • $\begingroup$ Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments. $\endgroup$ – José Carlos Santos Sep 11 '18 at 11:15
  • $\begingroup$ @JoséCarlosSantos Thx for your comment. I have written down what I have tried so far. $\endgroup$ – Knivel Sep 11 '18 at 11:41

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