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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.

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    Yes, I'd deleted the comment because I realized it was more subtle than I first thought...2012-08-10
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    If someone has encountered a similar problem any reference would be greatly appreciated.2012-08-10
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    Basically, you need the convergence of the Riemann sums to be uniform with respect to the improper bound of the integral, that is, the limiting operations of the norm of the partition going to zero and the improper bound going to its limit should commute.2012-08-10
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    @joriki: I dont really understand, could you expand on that a bit more? Obtaining a bound on the sum by giving conditions on $f$ would be enough for me. Thanks for your comment2012-08-10

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