Let $\Pi=\left\{ t_{k}\right\} _{k\in\mathbb{Z}}$ a sequence of real numbers such that $-\infty<t_{k}<t_{k+1}<+\infty$ for every $k\in\mathbb{Z}$, $\lim_{k\rightarrow\pm\infty}t_{k}=\pm\infty$ and such that there are two positive constants $\Delta,\,\delta $ with $\delta\leq t_{k+1}-t_{k}\leq\Delta$. We also define $\Delta_{k}:=t_{k+1}-t_{k}$, for every $k\in\mathbb{Z}$. We call a function $\chi:\,\mathbb{R}\rightarrow\mathbb{R}$ a kernel if it belongs to $L^{1}\left(\mathbb{R}\right)$, is bounded in a neighbourhood of the origin, and satisfies the following conditions :

$(\chi_{1})$ for every $u\in\mathbb{R}$ $$\sum_{k\in\mathbb{Z}}\chi\left(u-t_{k}\right)=1;$$

$(\chi_{2})$ $$m_{1,\,\Pi}\left(\chi\right)=\sup_{u\in\mathbb{R}}\sum_{k\in\mathbb{Z}}\left|\chi\left(u-t_{k}\right)\right|\left|u-t_{k}\right|<+\infty.$$ Let us define the following operator: $$\left(K_{w}^{\chi}f\right)\left(x\right):=\sum_{k\in\mathbb{Z}}\chi\left(wx-t_{k}\right)\left[\frac{w}{\Delta}_{k}\int_{t_{k}/w}^{t_{k+1}/w}f\left(u\right)du\right],\,x\in\mathbb{R}$$where $f:\,\mathbb{R}\rightarrow\mathbb{R}$ is locally integrable function such that the above series is a convergent for every $x\in\mathbb{R}$ and $\chi$ and $w \geq \overline{w}>0$. Assume that exists a kernel belonging to $C^{1}(\mathbb{R})$ such that $\left\Vert K_{w}^{\chi}f-f\right\Vert _{\infty}=O\left(w^{-1}\right)$ as $w \rightarrow +\infty$, where the implicit constant depends only on $f$ and $\chi$ and $\left\Vert \cdot\right\Vert _{\infty}$ is the classical sup norm.

I would like to prove that exists some $s \neq 0$ such that $$\left\Vert K_{w}^{\chi_{s}}f-f\right\Vert _{\infty}=O\left(w^{-1}\right)$$ as $w \rightarrow +\infty$, where $$\chi_{s}(\cdot)=\chi(\cdot+s)$$ (which is obviously a kernel).

I'm quite sure that it is true but I'm not able to prove it. I tried to use $$\left\Vert K_{w}^{\chi_{s}}f-f\right\Vert _{\infty}\leq\left\Vert K_{w}^{\chi}f-f\right\Vert _{\infty}+\left\Vert K_{w}^{\chi_{s}}f-K_{w}^{\chi}f\right\Vert _{\infty}$$ but it not seems very helpful. Thank you.


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