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Let {$g_{n}$}be a bounded sequence of functions on $[0,1]$ which is uniformly Lipschitz. That is, there is a constant $M$ (independent of $n$) such that for all $n$, $|g_{n}(x)-g_n(y)|\leq M|x-y|$ for all $x,y\in [0,1]$ and $|g_{n}(x)|\leq M$ for all $x\in [0,1]$. Then I have the following two questions:

(a) prove for all any $0\leq a\leq b\leq 1$, $$\lim_{n\rightarrow \infty } \int_{a}^{b}g_{n}(x)\sin (2n\pi x)\,dx=0. $$ (b) prove that for any $f\in L^{1}[0,1]$, $$\lim_{n\rightarrow \infty } \int_{0}^{1}f(x)g_{n}(x)\sin (2n\pi x)\,dx=0.$$

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This is the Riemann Lebesgue lemma. Smooth functions that vanish at $\{a,b\}$ are dense in $L^1[a,b]$. Combine this with integration by parts to conclude. – copper.hat Dec 30 '12 at 22:34
Thanks for reminding me of that, yes, that make sense, but we don't prove it before, so I wonder if there is some other elegant proof without using that? – user53800 Dec 30 '12 at 22:40
Yes, I was wondering about that. Nothing jumps out at me, except perhaps you could estimate the integral between the zero crossings of $\sin$, and use the Lipschitz constant to show the limit is zero? This would prove b) for simple $f$ and the conclusion follows from the DCT? – copper.hat Dec 30 '12 at 22:49
I think that if you can give me some details that will help me more, thanks a lot. – user53800 Dec 31 '12 at 0:41
up vote 3 down vote accepted

There is a discrete analog of "integrate by parts to kill the periodic term". Namely, "translate by half-period and cancel". Like this: $$\int_a^b g_n(x)\sin (2n \pi x)\,dx = - \int_a^b g_n(x)\sin (2n \pi (x+1/(2n))\,dx \\ =- \int_{a+1/(2n)}^{b+1/(2n)} g_n(y-1/(2n))\sin (2n \pi y)\,dy $$ The right hand side is nearly the same as $-\int_{a}^{b} g_n(y)\sin (2n \pi y)\,dy$: the discrepancy of intervals of integration contributes $O(1/n)$, and the difference of integrands is also $O(1/n)$, due to the Lipschitz condition. Conclusion: $\int_a^b g_n(x)\sin (2n \pi x)\,dx = O(1/n)$.

I'll leave it for you to adapt this to (b). You'll need the usual "estimate the difference of products" trick, plus the fact that translation is continuous in $L^1$: $\|f(\cdot)-f(\cdot+1/n)\|_{L^1}\to 0$ as $n\to \infty$.

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Thanks very much, I have already got all the proof, it's a elegant proof. Your hint and explanation is cool! Can you tell me how do you come up with such an idea to translate by half-period and cancel? :) – user53800 Dec 31 '12 at 13:29

Here's an alternative proof of (a):

Since $g_n$ is Lipschitz, it is absolutely continuous, hence differentiable a.e. with $g_n' \in L^1[0,1]$, and $g_n(x) = \int_0^x g_n'(t) dt$. Let $\phi(x) = g_n(x) \frac{\cos(2 \pi nx)}{2 \pi n}$. We have $\phi'(x) = -g_n(x)\sin(2 \pi nx)+g_n'(x) \frac{\cos(2 \pi nx)}{2 \pi n}$. Then $\phi(b)-\phi(a) = \int_a^b \phi'(t) dt$, which gives $\frac{1}{n}(g_n(b) \frac{\cos(2 \pi n b)}{2 \pi}-g_n(a) \frac{\cos(2 \pi n a)}{2 \pi}) = - \int_a^b g_n(x)\sin(2 \pi nx) dx + \int_a^b g_n'(x) \frac{\cos(2 \pi nx)}{2 \pi n} dx$. Rearranging, and using the fact that $|g_n(x)| \leq M$, gives $|\int_a^b g_n(x)\sin(2 \pi nx) dx| \leq \frac{1}{2n\pi}(2M+\|g_n'\|_1)$, from which the desired limit (a) follows.

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