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We are given $\int_{\mathbb{R}}\vert f\vert <\infty$. Want to show that for almost every $x$, $\sum_{n=1}^\infty\vert n\int_n^{n+1/n}f(x+y)dy\vert<\infty$.

I have two ideas. The first is to let $f_n=\int_n^{n+1/n}\vert f\vert$. I can show by DCT that $f_n\to0$ so for a given $\varepsilon$ there is $N$ so that $n\ge N \rightarrow f_n < \varepsilon$. So $\sum_{n=1}^\infty\vert n\int_n^{n+1/n}f(x+y)dy\vert =\sum_{n=1}^N\vert nf(x+y)\vert+\sum_{n=N+1}^\infty n\varepsilon$

however, the last term tends to infinity so either I'm using the wrong $f_n$ or DCT doesn't apply.

Another idea is to apply Cauchy Criterion. Choose $n>m>N$ then $$\vert n\int_n ^{n+1}f-m\int_m ^{m+1}f\vert \\\le \vert n\int_n ^{n+1/n} f\vert + \vert m\int_m ^{m+1/m} f\vert$$ and if I want to show these two terms go to zero, I bump into the same problem above ($n\cdot\varepsilon$) unless I find a better upper bound.

How can I find upper bound for $\int_n ^{n+1/n}\vert f\vert$ that allows $n\int_n ^{n+1}\vert f\vert\to0?$

cap
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  • Is there any other information given about $f$ or $x$? – robjohn Mar 03 '15 at 01:16
  • All I know about $f$ is that it's in $L^1(\mathbb{R})$. The sum should converge for almost all $x$, will edit. – cap Mar 03 '15 at 01:21
  • see related post http://math.stackexchange.com/questions/1770888/prove-series-converge-for-almost-every-x/1823517#1823517 –  Jun 14 '16 at 01:57

1 Answers1

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Hint: Without loss of generality, assume $f(x)\ge 0$.

Define $$ \psi(y)=\left\{\begin{array}{} n&\text{if }n\le y\lt n+1/n&\text{for some }n\in\mathbb{Z}\\ 0&&\text{otherwise}\\ \end{array}\right. $$ Then $$ \begin{align} \sum_{n=1}^\infty n\int_n^{n+1/n}f(x+y)\,\mathrm{d}y =\int_\mathbb{R}\psi(y)f(x+y)\,\mathrm{d}y \end{align} $$ Consider $$ \begin{align} \int_k^{k+1}\int_\mathbb{R}\psi(y)f(x+y)\,\mathrm{d}y\,\mathrm{d}x &=\int_k^{k+1}\int_\mathbb{R}\psi(y-x)f(y)\,\mathrm{d}y\,\mathrm{d}x\\ &=\int_\mathbb{R}\left(\int_k^{k+1}\psi(y-x)\,\mathrm{d}x\right)f(y)\,\mathrm{d}y\\ &=\int_\mathbb{R}\underbrace{\left(\int_{y-k-1}^{y-k}\psi(x)\,\mathrm{d}x\right)}_{\le2}f(y)\,\mathrm{d}y \end{align} $$

robjohn
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  • how do you get the upper bound of $2$ for that integral? I would give a crude upper bound the ceiling of $(y-k-1)$ based on your definition of $\psi.$ – cap Mar 03 '15 at 02:34
  • @cap Remember that $phi(x)$ takes two values on $(y-k-1, y-k)$. The interval where it takes on $y-k-1$ has length $1/(y-k-1)$. So actually you should be able to compute the integral exactly. – abnry Mar 03 '15 at 02:46
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    @cap: any interval $[y-k-1,y-k]$ of length $1$ must be contained in an an interval $[m,m+2]$ where $m\in\mathbb{Z}$. Note that $$\int_m^{m+2}\psi(x),\mathrm{d}x=\left{\begin{array}{}2&\text{if }m\ge1\1&\text{if }m=0\0&\text{if }m\lt0\end{array}\right.$$ therefore, $$\int_{y-k-1}^{y-k}\psi(x),\mathrm{d}x\le2$$ – robjohn Mar 03 '15 at 02:55