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Let $f : [0, 1] \to \mathbb{R}$ be a 'well behaved' function and define $c_n$ by the Riemann sums $$ c_n := \frac{1}{n} \sum_{k = 1}^n f(\frac{k - \theta}{n}) $$ for some given $\theta \in [0, 1]$. Then obviously $c_n$ will converge to the integral of $f$. Under which conditions on $f$ will this convergence be monotone? In particular, will the convergence be monotone if $f$ is twice differentiable with $f > 0$, $f' < 0$ and $f'' > 0$?

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One obvious class of functions whose Riemann sums have monotone convergence would be functions that do not cross the x-axis within the interval of integration. –  J. M. Aug 26 '10 at 13:32
    
@J.Mangaldan Maybe I'm misunderstanding something, but f having a definite sign is not enough to ensure monotonicity. As a counterexample, for theta = 1, take a function with f(0) = f(1/3) = f(2/3) = 1, and f(1/4) = f(1/2) = f(3/4) = 10. Then c_2 = 11/2, c_3 = 1, and c_4 = 31/4. –  Michael Ulm Aug 26 '10 at 14:02
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"Monotonously" means "in a boring fashion." "Monotonically" is probably the intended word ;-). –  whuber Aug 26 '10 at 15:22
    
@whuber: Thanks for the correction. –  Michael Ulm Aug 26 '10 at 18:16
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The answer to your final question is no for arbitrary $\theta$. Specifically, let $f(x) = 1/\sqrt(x)$ for $\epsilon < x <= 1$ and define $f$ however you like for $0 \le x < \epsilon$ << .01, say. (This can be done in a way that maintains the concavity and twice-differentiability of $f$.) With $\theta = 3/4$, the sequence of Riemann sums begins (with $n=1$):

2., 2.04667, 2.056, 2.0579, 2.05757, 2.05647, 2.05511, 2.05368, ...,

peaking at $n=4$. (The sequence is monotonic after that.)

The problem occurs because there is a section at the end of the graph of $f$, in this case near zero, that grows quickly enough to appreciably alter the very first term of the sum once $n$ grows large enough; this section is not sampled by the sum for smaller $n$. By modifying $f$ appropriately within disjoint intervals converging to 0, you should be able to construct an example whose sequence of Riemann sums wiggles arbitrarily many times. What this hand-waving argument does not settle is whether this process can be carried out ad infinitum: the concern is that the resulting $f$ might not be continuous at 0, which conceivably is part of your concept of "well-behaved" (although it is not necessary for the theory of Riemann integration, which permits some points of discontinuity).

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Thanks for the interesting counterexample. It's a pity that my assertion as stated is wrong - many interesting inequalities would have followed. –  Michael Ulm Aug 27 '10 at 16:56
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