$\lim_{\lambda \to \infty} \int^b_0 f(t) \frac{\sin(\lambda t)}{t} $

Question

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.

Answer 1

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 $$\sup_{n\in\Bbb N}\left|\int_0^bf(t)\frac{\sin(nt)}{t}\right|\,dt=\infty.$$

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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 $[\delta,b]$.)

Answer 2

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.