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For a continuous function, $f:[0,b] \to \Bbb{R}$ show that: $$ \lim_{\lambda\to\infty} \int^b_0 f(t) \frac{\sin(\lambda t)}{t}\,dt = \frac{\pi}{2}\,f(0) $$

I know it has something to do with the Riemann-Lebesgue lemma about Fourier series, but $\frac{f(t)}{t}$ is not an integrable function.

I tried to define $g(t) = \frac{2f(t)\sin(\frac{t}{2})}{t} $ which tends to $f(0)$ as $t\to0$ and $g(t)$ is integrable in $[0,b]$ so we get:

$$ \lim_{\lambda\to\infty} \int^b_0 f(t) \frac{\sin(\lambda t)}{t} \,dt= \lim_{\lambda\to\infty} \frac{1}{2}\int^b_0 g(t) \frac{\sin(\lambda t)}{\sin(\frac{t}{2})}\,dt $$ which kind of reminds $g(x)*\mathit{Dn}(0)$ ($\mathit{Dn}=$Dirichlet kernel), but it's not the same.

I guess it can be solved with the $\mathrm{Si}(x)$ function but I'm trying to solve it with Fourier series.

egreg
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MSm
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    You might try the change of variables $x=\lambda t$. – John Dawkins Jul 15 '16 at 13:22
  • @Clement C: I have deleted my answer as what I think to be a counterexample was not, and I need an additional hypothesis ($f$ has a derivative at $0$) to conclude. Sorry for that... – Kelenner Jul 15 '16 at 13:45
  • As David Ullrich has pointed out, this result is false. A simple condition under which it is true is that $\int_{0}^{b}\left|\frac{f(t)-f(0}{t}\right|dt < \infty$. For example, if $|f(t)-f(0)| \le Ct^{\alpha}$ for some $\alpha > 0$ and for all $t$ near $0$, then the result will hold. If the function is monotone near $0$, you'll also get what you want, or if the function is of bounded variation on an interval $[0,\delta]$ for some $\delta > 0$, you'll get what you want. It doesn't take much extra, but continuous is definitely out of the question. – Disintegrating By Parts Jul 26 '16 at 19:06

2 Answers2

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You can't prove this because it's false. (It's true if you assume just a tiny bit more smoothness than just continuity; see below.)

First, $$\lim_{\lambda\to\infty}\int_0^b\frac{|\sin(\lambda t)|}{t}\,dt =\lim_{\lambda\to\infty}\int_0^{\lambda b}\frac{|\sin(t)|}{t}\,dt=\infty.$$

But the $L^1$ norm of $\sin(\lambda t)/t$ is the same as its norm as a linear functional on $C([0,b])$. So the Uniform Boundededness Principle shows that there exists $f\in C([0,b])$ such that $$\sup_{n\in\Bbb N}\left|\int_0^bf(t)\frac{\sin(nt)}{t}\right|\,dt=\infty.$$

Below On the other hand it's easily seen to be true if you assume that $f$ satisfies $$|f(t)-f(0)|\le ct^\alpha$$for some $\alpha>0$.

First, it's true for $f=1$, since as has been pointed out $$\int_0^b\frac{\sin(\lambda t)}{t}\,dt=\int_0^{\lambda b}\frac{\sin(t)}{t}\,dt\to\frac\pi2.$$ So we may assume $f(0)=0$, and now $|f(t)|\le ct^\alpha$.

Let $\epsilon>0$. Choose $\delta>0$ so that $$\int_0^\delta\frac{ct^\alpha}{t}\,dt<\epsilon.$$So we have $$\int_0^\delta|f(t)|\frac{|\sin(\lambda t)|}{t}\,dt<\epsilon$$for every $\lambda$. And $$\left|\int_\delta^bf(t)\frac{\sin(\lambda t)}{t}\,dt\right|<\epsilon$$if $\lambda$ is large enough, by the Riemannn-Lebesgue lemma (note that $f(t)/t$ is integrable on $[\delta,b]$.)

David C. Ullrich
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    I really don't get this. Why is it relevant that $\text{sinc}(x)\not\in L^1$? $$\lim_{\lambda\to +\infty}\int_{0}^{b}1\cdot\frac{\sin(\lambda t)}{t},dt=\frac{\pi}{2}$$ is true, just like $$\lim_{\lambda \to +\infty}\int_{0}^{\lambda}\frac{\sin t}{t},dt = \frac{\pi}{2}.$$ – Jack D'Aurizio Jul 15 '16 at 15:05
  • Since polynomials are dense in $C^0[0,b]$ with respect to the uniform norm, the claim is true. – Jack D'Aurizio Jul 15 '16 at 15:06
  • The reason $\sinc\notin L^1$ is relevant is that it shows that$\int_0^b|\sin(\lambda t)|/t\to\infty$. The reason that shows the result is false is Uniform Boundedness (aka Banach-Steinhaus). The fact that those $L^1$ norms are unbounded also show directly why uniform approximation doesn't help: If $|f-f_n|<\epsilon$ it follows that blah blah is no larger than $\epsilon\int_0^b|\sin(\lambda t)|/t$, but that tends to infinity. – David C. Ullrich Jul 15 '16 at 15:14
  • I still cannot see any absolute value in OP's question... – Jack D'Aurizio Jul 15 '16 at 15:15
  • @JackD'Aurizio Do you know the Banach-Steinhaus theorem? – David C. Ullrich Jul 15 '16 at 15:16
  • I know the statement but I still cannot see why it disproves OP's claim. Can you show a $f\in C^0(0,b)$ such that the given equality does not hold? It looks to me that your choice of the $L_1$ norm is rather arbitrary. – Jack D'Aurizio Jul 15 '16 at 15:19
  • @JackD'Aurizio The situation is exactly the same as with the standard result that there exists a continuous periodic function with a divrgent Fourier series: Writing down an explicit example is hard, but using Banach-Steinhaus to show an example exists is easy. I didn't "choose" the $L^1$ norm. The $L^1$ norm of $|\sin(\lambda t)/t$ is the norm of $\sin(\lambda t)/t$ as a linear functional on $C([0,b])$. (By the Riesz Representation Theorem.) – David C. Ullrich Jul 15 '16 at 15:25
  • All right, got it, thanks for the explanation. – Jack D'Aurizio Jul 15 '16 at 15:27
  • Sorry if that's daft, but by any chance would it be possible to expand the answer above to give more detail? This seems like a quite interesting fact (or, at least, non-trivial to me), and I'm not familiar with the argument developed in the comments... – Clement C. Jul 15 '16 at 15:30
  • @ClementC. Not that it matters, but the argument wasn't really develooped in the comments - things I said about the wrong argument are irrelevant,, and everything I said about the right argument in the comments was a repetition of something I said in the answer. Anyway, what part don't you get? As far as I can see there are only two things here: (i) The Uniform Boundedness Principle, (ii) the $L^1$ norm is the same as the norm in $(C([0,b]))^*$. – David C. Ullrich Jul 15 '16 at 15:36
  • That is the daft part coming: I am not familiar enough with the UBP to fill in the details. But in light of your comment, it sounds like it's on me :) – Clement C. Jul 15 '16 at 15:38
  • @ClementC. I'll think about how to add more detail... – David C. Ullrich Jul 15 '16 at 15:44
  • @DavidC.Ullrich Thank you! – Clement C. Jul 15 '16 at 15:48
  • @ClementC. I started a Q&A on this and people objected it was well known: https://en.wikipedia.org/wiki/Uniform_boundedness_principle#Example:_pointwise_convergence_of_Fourier_series – David C. Ullrich Jul 15 '16 at 16:25
  • Just saw that... oh well, thanks for trying, at least (and for the link). – Clement C. Jul 15 '16 at 16:28
  • @ClementC. The voting's mixed - I'm leaving it up until things get nasty... – David C. Ullrich Jul 15 '16 at 16:41
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Sketch of a more computational proof that the result fails: Let's take $b=1$ for simplicity. Verify that

$$\int_0^1 \frac{\sin^2(\lambda x)}{x}\, dx \to \infty$$

as $\lambda \to \infty.$ This and the Riemann Lebesgue lemma allow us to inductively choose $\lambda_1 < \lambda_2 < \dots $ so that

$$\tag 1 \int_0^1 \frac{\sin^2(\lambda_n x)}{x}\, dx > n^3, \,\,\text { and }$$

$$\tag 2 \left | \int_0^1 \sin(\lambda_m x)\frac{\sin(\lambda_n x)}{x}\, dx\right | \le 1\,\,\text { for } n \ne m.$$

Property $(2)$ is an "almost orthogonality" that's good enough to construct an example. Define

$$f(x) = \sum_{n=1}^{\infty} \frac{\sin (\lambda_n x)}{n^2}.$$

The series converges uniformly on $[0,1],$ so $f$ is continuous there by the Weierstrass M test. Fix $\lambda_N$ for the moment. We then have

$$\int_0^1 f(x) \frac{\sin(\lambda_N x)}{x} \, dx = \sum_{n=1}^{\infty}\frac{1}{n^2}\int_0^1 \sin(\lambda_n x) \frac{\sin(\lambda_N x)}{x} \, dx.$$

Call the $n$th integral on the right above $I_n.$ The sum above is then

$$\sum_{n=1}^{\infty}\frac{1}{n^2}I_n \ge \frac{1}{N^2}I_N - \sum_{n\ne N}\frac{1}{n^2}|I_n| \ge \frac{N^3}{N^2} - \pi^2/6,$$

where $(1),(2)$ above have been used. This $\to \infty$ as $N\to \infty,$ completing the proof.

zhw.
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