Let $f \in L^1(\mathbb{R})$. Find $$ \lim_{n \rightarrow \infty} \int_{-\infty}^\infty f(x)\sin(nx) dx \,. $$

LDCT is a no go, as well as MCT and FL, which are really the only integration techniques we've developed thus far in the course. Integration by parts isn't applicable either since we just have $f \in L^1$. I've tried splitting both $f$ and $\sin(nx)$ into unsigned parts but that didn't seem to help. I don't think using the density of $C_c(\mathbb{R}) \subset L^1$ will help either. I've tried approximating $f$ by simple functions, but that didn't seem to do the trick either..

To give you a gauge in regards to where we are at, we just finished basic properties of $L^p$ spaces and will begin Hilbert Space Theory as in Big Rudin.

Any hints are greatly appreciated.

Edit: Let $h \in C^1_c(\mathbb{R})$ be such that $\sup \limits_{x \in X}|f(x)-h(x)| < \epsilon.$ Let $\operatorname{supp}(h) = [a,b]$. Then,

\begin{eqnarray*} \int_{\mathbb{R}} h(x)\sin(nx) \, dx &=& \int_{[a,b]} h(x)\sin(nx) \, dx \\ &=& \left[h(x) \frac{\cos(nx)}{n} \right] + \int_{[a,b]} \frac{1}{n} f'(x) \cos(nx) \, dx \\ \end{eqnarray*}

note that since $f'$ is continuous on $[a,b]$, it is bounded by some $M$. Thus, $\frac{1}{n}f'(x)\cos(nx) \leq \frac{M}{n}$, which converges to $0$ uniformly. Taking limits yields that $$\lim_{n \rightarrow \infty} \int_{-\infty}^\infty f(x)\sin(nx) dx = 0.$$

Alternatively, if we use a simple function approximation, we end up with something of the form $$\int_\mathbb{R} \sum_{i=1}^{k}c_i \chi_{E_i}(x)\sin(nx)dx = \sum_{i=1}^k c_i \int_{E_i} \sin(nx)dx$$ which goes to $0$ as $n \to \infty$ (The $E_i$'s can be chosen to be disjoint), but the below question still remains.

How does this follow from the fact that we've only used a dense subset?

  • 3
    $\begingroup$ Are you able to prove the result for a dense subset of functions ? You may for example use the density of $\mathcal{C}^1_c(\mathbb{R})$ and use integration by parts. $\endgroup$ – Joel Cohen Oct 19 '15 at 18:42
  • $\begingroup$ @JoelCohen I understand that this would make the first term vanish, but why does $f$ need be differentiable? $\endgroup$ – Anthony Peter Oct 19 '15 at 18:48
  • $\begingroup$ You need the differentiability to perform the integration by parts in the first place (differentiate $f$ and integrate the sine, yielding a $\frac{1}{n}$ factor). $\endgroup$ – Joel Cohen Oct 19 '15 at 18:51
  • 1
    $\begingroup$ Well, denoting by $I_n(f)$ the integrals you are interested in, for every $g$ simple, one gets $$|I_n(f)|\leqslant |I_n(g)|+\int_\mathbb R|f(x)-g(x)|\,|\sin(nx)|\,dx\leqslant|I_n(g)|+\|f-g\|_1,$$ hence $$\limsup_{n\to\infty}|I_n(f)|\leqslant\limsup_{n\to\infty}|I_n(g)|+\|f-g\|_1=0+\|f-g\|_1,$$ and since the RHS can be made as small as one wants, the proof is over. $\endgroup$ – Did Oct 19 '15 at 19:41
  • 1
    $\begingroup$ This is playing with words: either one reproves this (and this is short and easy to do, as I showed) or one admits it, saying it is well known. $\endgroup$ – Did Oct 19 '15 at 20:29

Your Answer

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

Browse other questions tagged or ask your own question.