3
$\begingroup$

Consider the series $$f_n=\sum_{k=0}^{n-1}\frac{\sin^2(x+\frac{k}{n})}{n}.$$ Since $\lim_{n\rightarrow\infty}f_n=\int_x^{x+1}\sin^2(y)dy$, we know that $f_n$ converges.
And my question is: will the sequence converge uniformly?
I tried to integrate term by term, and the resulting series does converge to the integral of the limiting function. So the series might converge uniformly, but I do not know how to check this directly. It looks too bizarre to check directly, so I am wondering if there is any theorem that might come in handy.

In general, is there a criterion telling us when a Riemann sum converges uniformly?

If something above is wrong, tell me. Thanks in advance.

$\endgroup$
2
  • 4
    $\begingroup$ "Since this $f_n$ is the Riemann sum of the function $\sin^2(x)$ in the interval $[x,x+1]$, we know that..." Actually we know that you are mixing much too much functions and their values! You might want to explain what you mean by the quoted sentence. $\endgroup$
    – Did
    Jul 26, 2013 at 16:00
  • $\begingroup$ @Did I meant to say that $\lim_{n\rightarrow\infty}f_n=\int_x^{x+1}\sin^2(y)dy$ so that the sequence $f_n$ does converge. $\endgroup$
    – awllower
    Jul 27, 2013 at 1:51

2 Answers 2

4
$\begingroup$

Here, the key is going to be the $t\mapsto\sin^2(t)$ is uniformly continuous.

Suppose we're given $\epsilon>0$. Then there exists $\delta>0$ so that $\lvert \sin^2(t)-\sin^2(a)\rvert<\epsilon$ whenever $0<\lvert t-a\rvert<\delta$.

Choose $N\in\mathbb{N}$ large enough that $0<\frac{1}{N}<\delta$. Use uniform continuity to show that for $n\geq N$, for each $0\leq k\leq n-1$, $$ \left\lvert \frac{\sin^2\left(x+\frac{k}{n}\right)}{n}-\int_{x+\frac{k}{n}}^{x+\frac{k+1}{n}}\sin^2(t)\,dt\right\rvert<\frac{\epsilon}{n}, $$ no matter what $x$ you choose. It follows by the triangle inequality that for any $x$, for $n\geq N$ you have $$ \left\lvert\sum_{k=0}^{n-1}\frac{\sin^2(x+\frac{k}{n})}{n}-\int_{x}^{x+1}\sin^2(t)\,dt\right\rvert<\epsilon. $$ In general, when you're dealing with Riemann sums, this will be a good place to start; at the least it will help you think about where possible problems could occur.

$\endgroup$
3
  • $\begingroup$ Hence, the Riemann sums of a uniformly continuous functions converges uniformly to the integral, right? Thanks for the answer. :) $\endgroup$
    – awllower
    Jul 27, 2013 at 1:54
  • 1
    $\begingroup$ Yes, that is the case - in fact, you could set up a 2-variable function which would converge uniformly to the integral on $[x,y]$, and use a similar proof that it converged uniformly. $\endgroup$ Jul 27, 2013 at 12:47
  • $\begingroup$ And: my pleasure! $\endgroup$ Jul 27, 2013 at 12:48
3
$\begingroup$

A general argument is the following theorem of Dini : An increasing sequence of continuous functions on a compact space such that its limit is still a continuous function, converges not only pointwise but uniformly.

$\endgroup$
1
  • $\begingroup$ Interesting to hear that theorem! That is another direction than the above. Thanks for the answer. $\endgroup$
    – awllower
    Oct 21, 2013 at 16:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.