Prove that the series $\sum_{k \in \mathbb{Z}}f(x+k) $ converges for all $x$ when $f$ is integrable and continuous

Question

Let $(\mathbb{R},\mathcal{L},m)$ be the real line equipped with the Lebesgue measure $m$. Assume that $f:\mathbb{R} \to [0,\infty]$ is continuous, non-negative, and that $$ \int_{\mathbb{R}} f \ dm < \infty $$ Then I would like to show that for any $x \in \mathbb{R}$,

$$ \sum_{k \in \mathbb{Z}}f(x+k) $$ is convergent. It follows by Fubinis theorem that $\sum_{k \in \mathbb{Z}}f(x+k)$ is convergent almost everywhere, which is shown here Prove that the series $\sum\limits_{n=-\infty}^{+\infty}f(x+n)$ converges absolutely for a.e. $x \in \mathbb{R}.$

I am wondering, with the added assumption of continuity, if we are able to extend the result such that it holds everywhere, not just almost everywhere. My effort so far has been the following:

Fix $x_0\in \mathbb{R}$. Assume for contradiction that

$$ \sum_{k \in \mathbb{Z}} f(x_0+k)=\lim_{n \to \infty}\sum_{k =-n}^{n} f(x_0+k) = \infty $$ Let $C=\int_{\mathbb{R}} f \ dm$. Let $N_1 \in \mathbb{N}$ be st. $$\sum_{k =-N_1}^{N_1} f(x_0+k) > C+1$$ My idea was then to use a Riemann sum that would be larger than $C$, but as this Riemann sum depends on the fineness of the partition, it seems like this is not a feasible strategy.

Answer 1

The claim is false even when $f$ is assumed to be smooth with bounded derivative of any order.

Counter-Example. Let $\varphi$ be a smooth function which is non-negative and supported on $[-\frac{1}{2}, \frac{1}{2}]$ and satisfies $\varphi(0) = 1$. Then define $f : \mathbb{R} \to \mathbb{R}$ by

$$ f(x) = \sum_{n=2}^{\infty} \frac{1}{n\log_2 n} \varphi((x-n)\log_2 n). $$

Then $f$ is smooth since the sum converges uniformly. Moreover, for each $m$,

\begin{align*} \left| f^{(m)}(x) \right| &= \left| \sum_{n=2}^{\infty} \frac{(\log_2 n)^{m-1}}{n} \varphi^{(m)}((x-n)\log_2 n) \right| \\ &\leq \left( \sup_{n\geq 2} \frac{(\log_2 n)^{m-1}}{n} \right) \|\varphi^{(m)}\| \end{align*}

where $\|\varphi^{(m)}\|$ denotes the supremum norm of $\varphi^{(m)}$ over $\mathbb{R}$, showing that $f^{(m)}$ is bounded over $\mathbb{R}$. Finally, we have

$$ \sum_{k\in\mathbb{Z}} f(k) = \sum_{n=2}^{\infty} \frac{1}{n\log_2 n} = \infty $$

as well as

$$ \int_{\mathbb{R}} f(x) \, \mathrm{d}x = \sum_{n=2}^{\infty} \frac{1}{n(\log_2 n)^2} \int_{\mathbb{R}} \varphi(y) \, \mathrm{d}y < \infty. $$

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On the other hand, the claim holds true if $f$ is has bounded variation.

Theorem. Let $f : \mathbb{R} \to \mathbb{R}$ be of bounded variation. Then the followings are equivalent:

1. $\int_{\mathbb{R}} f(x) \, \mathrm{d}x $ converges absolutely.
2. $\sum_{k\in\mathbb{Z}} f(k)$ converges absolutely.

The proof of this result hinges on the following two observations:

1. If $f : \mathbb{R} \to \mathbb{R}$ is of bounded variation on all of $\mathbb{R}$, then both $$ f(\infty) = \lim_{x\to\infty} f(x) \qquad\text{and}\qquad f(-\infty) = \lim_{x\to-\infty} f(x) $$ converge. This is easily proved by noting that $|f(a) - f(b)| \leq V_{a}^{b}(f)$, where $a < b$ and $V_{a}^{b}(f)$ is the variation of $f$ on $[a, b]$.

2. For any integrable function $f$ on $[a, b]$, we have $$ \left| \int_{[a, b]} f(x) \, \mathrm{d}x - \sum_{k \in [a, b]\cap\mathbb{Z}} f(k) \right| \leq \frac{1}{2} (V_{a}^{b}(f) + |f(a)| + |f(b)|). $$

Answer 2

The claim does not hold, which is shown by the following counter-example (provided by @Mariano Suárez-Álvarez)

Define the "triangle function" $\tilde{f}:[0,\infty) \to \mathbb{R}$ by $$ f(x) = \begin{cases} 1 & x\in \mathbb{Z} \\ 0 & n-1+\frac{1}{2^{n}}\leq x\leq n-\frac{1}{2^{n+1}} \\ \text{linear} & \text{otherwise} \end{cases} \ $$ and let $f:\mathbb{R}\to \mathbb{R}$ be the symmetric extension of $\tilde{f}$ to the whole real line. Then the integral of $f$ is given by the sum of the area of the triangles,

$$ \int_{\mathbb{R}}f \ dm=\frac{1}{2}+2\frac{1}{2}\cdot\sum_{n = 1}^{\infty}\frac{1}{2^n}< \infty$$

On the other hand, if $x=0$, then $\sum_{k \in \mathbb{Z}}f(x+k)=\sum_{k \in \mathbb{Z}}f(k)=\sum_{k \in \mathbb{Z}}1=\infty$