I am trying to find suitable conditions (integrability, growth...) on a function $f:\mathbb{R}\to \mathbb{R}$ such that: \begin{equation} \sum_{k\in\mathbb{Z}}f(kh)h= \mathcal{O}(1),\qquad h\to 0^+. \end{equation}
In other words I am trying to find conditions on $f$ such that the above sum is bounded for $h$ small enough. Alternatively, can I impose conditions of $f$ that make $\sum_{k\in\mathbb{Z}}f(kh)h \to_{h\to0} \int_R f (x)dx$? Many thanks.
EDIT: Following on the comment, note that this would trivially work for a continuous function on a closed and bounded interval. Here's an example where things may go wrong: Let $\mathcal{S}(f, [0,1], \mathcal{P}, \mathcal{P}^\prime)$ be the Riemann sum $\sum_{k = 1}^n f(\xi_k)(x_k - x_{k-1})$, where $\mathcal{P}= \{x_0, \dots, x_n\}$ is a partition of $[0,1]$ and $\mathcal{P}^\prime= \{\xi_1, \dots, \xi_n; x_{n-1}\leq \xi_k \leq x_n\}$. Let $f(x) = 1/\sqrt{x}$ on $(0,1]$. Then $\lim_{\| \mathcal{P} \| \to 0} \mathcal{S}(f, [0,1], \mathcal{P}, \mathcal{P}^\prime)$ does not exist. In fact, consider the sums: $\mathcal{S}(f, [0,1], \mathcal{P}_n, \mathcal{P}_n^\prime)$ where the points of $\mathcal{P}_n$ are chosen such that $x_1 = 1/n$ and the points of $\mathcal{P}_n^\prime$ are chosen such that $\xi_1 = 1/n^4$. Then \begin{equation} \mathcal{S}(f, [0,1], \mathcal{P}_n, \mathcal{P}_n^\prime) = \frac{1}{\sqrt{1/n^4}} \, \frac{1}{n} +\sum_{k = 2}^n f(\xi_k)(x_k - x_{k-1}) \geq n \end{equation}• and hence the sum can be arbitrarily large.