# Show that $\lim_{t \rightarrow \infty} tF(t) = 0.$

Let $f: [0, \infty) \rightarrow R$ be a continuous function and define $$F(t) = \int_{0}^{\frac{1}{t}} f(tx)\cos(t^2x) dx.$$

Show that $\lim_{t \rightarrow \infty} tF(t) = 0.$

This is part of a question from an old analysis qualifying at my university. I came across it while trying to study for my final exam (I'm an undergrad). I have honestly not seen a problem like this before, so I am not sure how to proceed. Can I please have a hint?

I tried integration by parts, but that was a disaster. Another technique I know to make integrals nice is the second mean value theorem for integrals, but that doesn't apply since neither $f(tx)$ nor $\cos(t^2x)$ are not necessarily non-negative for all $x$. Not sure what to try next, or if this question is even in the scope of what was covered in my class.

• Do you know the Riemann-Lebesgue Lemma? After the change of variables, RL finishes off the problem immediately. – zhw. Nov 29 '16 at 4:02

Make the change of variable $y=tx$ to see that $$tF(t) = \int_0^1 f(y) \cos(ty) dy.$$

Let $\epsilon >0$. Since $f$ is continuous on $[0,1]$, which is compact, we can use the Weierstrass approximation theorem to pick a polynomial $p$ such that $$\sup_{y \in [0,1]} |p(y) - f(y)| < \epsilon.$$ Then $$\left\vert tF(t) - \int_0^1 p(y) \cos(ty) dy \right \vert \le \int_0^1 \vert [f(y)-p(y) ]\cos(ty) \vert dy \le \int_0^1 \epsilon = \epsilon.$$

Now we integrate by parts: $$\int_0^1 p(y) \cos(ty) dy = \int_0^1 p(y) \frac{d}{dy} \frac{\sin(ty)}{t} dy = p(1) \frac{\sin(t)}{t} - \frac{1}{t} \int_0^1 p'(y) \sin(ty) dy.$$ Choose $T$ large enough so that $|p(1)|/T < \epsilon$ and so that $$\frac{1}{T} \int_0^1 \vert p'(y) \vert dy < \epsilon.$$ Then for $t > T$ we have that $$\left\vert \int_0^1 p(y) \cos(ty) dy \right\vert < 2 \epsilon.$$

Hence $t > T$ implies that $$\vert tF(t) \vert < 3 \epsilon,$$ and so $tF(t) \to 0$ as $t \to \infty$.

• Thank you. I don't quite follow how you changed the bounds on your integral, but the rest makes sense. – user389056 Nov 29 '16 at 3:36
• If $y=tx$ then $dy = t dx$ and so $dx = dy / t$, which explains where the $t$ goes when we multiply $t F(t)$. For the bounds, we have that if $x=0$ then $y = tx = t0 =0$ and if $x=1/t$ then $y = tx = t/t =1$, so in the "y coordinates" the bounds of integration are from $y=0$ to $y=1$. Make sense? – Glitch Nov 29 '16 at 3:47
• Yes, thank you! – user389056 Nov 29 '16 at 3:57
• Why not just use the Riemann Lebesgue lemma? – zhw. Nov 29 '16 at 4:02
• @zhw. This is effectively a proof of Riemann-Lebesgue. – Glitch Nov 29 '16 at 4:07

Some suggestion that might be useful. We can rewrite the integral as following: $$\lim_{t\to \infty}tF(t) = \lim_{t\to \infty}\int_{0}^{1}f(x)cos(tx)dx$$ You see, as $t \to \infty$, $cos(tx)$ oscillates very fast between -1 and +1. Because $f(x)$ is a continuous function, so the integral will be 0.