1

I'm going through old exercises of a real analysis course to prepare for an exam and there I found the following problem concerning the Riemann-Lebesgue lemma:

Let $f \in \operatorname{C}^k([a,b], \mathbb{R})$, such that $f$ and all existing derivatives of $f$ vanish at the endpoints $a,b$. That is $f^{(j)}(a) = f^{(j)}(b) = 0$ for all $0 \leq j \leq k$. Show that:

$$\lim_{\omega \to \infty} \int_{a}^{b} f(t)\sin(\omega t)dt = \mathcal{O}(1/|\omega|^k)$$

Addendum: What happens if $f$ is smooth?

Now, I know that there is a similar question on here; however, thus far I was unable to connect the dots. Here's what I've come up with so far:

We need to find a $C > 0$ and a $t_C > 0$, such that for all $t \geq t_C$

$$\left| \lim_{\omega \to \infty} \int_{a}^{b} f(t)\sin(\omega t)dt \right| \leq C \frac{1}{|\omega|^k}$$

Since $f$ is $C^k$ on a bounded interval, we know that all $f^{j}$ are integrable for $j \leq k$. And using integration by parts we can make the following estimate

$$\left| \int_{a}^{b} f(t)\sin(\omega t)dt \right| \leq \frac{1}{|w|} \int_{a}^{b} |f'(t)\cos(\omega t)|dt$$

The integral on right side is finite and thus the whole expression tends to $0$ as $\omega \to \infty$. My idea is to somehow iterate this estimation and the proceed by induction over $k$ to prove the lemma. Maybe it's possible to choose $C$ as whatever integral appears on the right side of this estimation, because this integral will be positive and finite. However, I need some help on how to formalize all of this or some more explanations if my approach doesn't make sense at all. Any help is appreciated!

Community
  • 1
jazzinsilhouette
  • 1,218
  • 1
    What do you have after $k$ times parts? – Nosrati Aug 16 '17 at 18:35
  • 1
    Notice that $\int_{a}^{b} f(t)e^{i\omega t}dt = (-i\omega)^{-1}\int_a^b f'(t)e^{i\omega t}dt$ by integration by parts. Applying it several times you obtain $\int_{a}^{b} f(t)e^{i\omega t}dt = (-i\omega)^{-k}\int_a^b f^{(k)}(t)e^{i\omega t}dt$.

    Then, use the identity above to prove the fact you want.

     – Hugo Aug 16 '17 at 18:35
  • @MyGlasses: I get something like $\int_{a}^{b} f(t)sin(\omega t)dt = \frac{1}{|\omega|^k} \int_{a}^{b}$. Where the exact integrand depends on parity of $k$. I guess this ambiguity can be circumvented by using the exponential function as Hugocito suggests. Thanks, to you both! The only thing left that's still confusing me is the choice of $C$. Can I just take the integral as my $C$? – jazzinsilhouette Aug 16 '17 at 18:53
  • Thinking about the addendum I noticed that there's another thing I don't get. If we take $\omega \to \infty$ then $\lim_{\omega \to \infty} \int_{a}^{b} f(t)\sin(\omega t)dt = 0$. So what is the big deal here? Is it just that the expression will tend to zero faster, if $f$ is more often differentiable? – jazzinsilhouette Aug 16 '17 at 19:09
  • 1
    First $ \int_{a}^{b} f(t)\sin(\omega t)dt ={\bf Im} \int_{a}^{b} f(t)e^{i\omega t}dt \leq \frac{1}{|\omega|^k}\int_{a}^{b} |f^{(k)}(t)| dt|=C\frac{1}{|\omega|^k} $ and then $\lim_{\omega \to \infty} \int_{a}^{b} f(t)\sin(\omega t)dt = 0$ and Riemann-Lesbegue shows the coefficients vanish faster! – Nosrati Aug 16 '17 at 19:44

0 Answers0