0

Good day,

Currently I am working with the book "Introduction to Fourier Series" by R. Lasser. There we have the following theorem (on page 227):

Let $f \in L^1(\mathbb{R})$. Set $$F(x):=\int_{-\infty}^x f(y) dy=\int_{\mathbb{R}}f(y) \chi_{]-\infty,x]} dy$$ for $x \in \mathbb{R}$. If $F \in L^1(\mathbb{R})$, then $$\hat{F}(a)=\frac{1}{ia} \hat f (a)$$ for all $a \in \mathbb{R}\backslash \{0\}$.

Proof: We know that $F'(x)=f(x)$ holds for almost all $x \in \mathbb{R}$. Integration by parts yields $$\hat F (a) = \lim_{A \to \infty} F(x) \frac{1}{-ia} e^{-iax} \bigg|_{-A}^A+\int_{\mathbb{R}} f(x) \frac{1}{ia} e^{-iax} dx$$ where we have also used Lebesgue's dominated convergence theorem. Because of its integrability $F$ vanishes in infinity, that is $$\lim_{|A|\rightarrow \infty} F(A)=0.$$

I don't get why $F$ vanishes at infinity. It is stated because of its integrability. But I have found an example of an integrable function that doesn't vanish at infinity, see: Does an absolutely integrable function tend to $0$ as its argument tends to infinity?

It might be because of the integral representation of $F$.

Does someone have an idea? I would really appreciate a proof of this statement.

Thanks a lot, Marvin

Community
  • 1
Cahn
  • 4,621

1 Answers1

2

Since $f\in L^1(\mathbb{R})$, it follows that $F(x) = \int\limits_{-\infty}^{x}{f(y)\text{ d}y}$ has a finite limit as $x\rightarrow\infty$. This limit must be zero in order for $F$ to be integrable on $\mathbb{R}$.

Joey Zou
  • 8,466
  • Ahhh, thank you, why didn't I see that, it was right before me. Thanks for your help, I appreciate it :) – Cahn Jul 28 '16 at 20:53