# Linear span of weighted powers

I am reading Functional Analysis by Peter Lax, and I do not understand the passage where it says that $w(t)e^{i\zeta t}$ belongs to $C$, where:

• $\zeta$ is a complex variable, and
• $C$ is the set of continuous functions on $\mathbb R$ that vanish at $\infty$, as is stated above (2).

Thank you anyone for help.

• Could you provide a clearer picture? Thank you. / Può fornire un quadro più chiaro? Grazie. Jun 26, 2016 at 7:18

Define $x(t) = w(t) e^{i \zeta t}$. We want to show that $x \in C$, i.e, $\lim_{|t| \rightarrow \infty} x(t) = 0$ according to (2), or, equivalently, $|x(t)| \rightarrow 0$ as $|t| \rightarrow \infty$.

Note that $e^{i \zeta t} = e^{i (Re \zeta + i Im \zeta) t} = e^{-(Im \zeta) t} e^{i (Re \zeta) t}$, and note also that the text requires $|Im \zeta| < c$, so that $c + Im \zeta > 0$ and $c - Im \zeta > 0$.

According to (1), we have that $w$ satisfies $0 < w(t) < ae^{-c |t|}$ for some $a > 0$ and $c > 0$.

We analyze the two cases when $|t| \rightarrow \infty$:

When $t \rightarrow \infty$, we have $$|x(t)| = |w(t) e^{i \zeta t}| = |w(t)| |e^{i \zeta t}| = |w(t)| |e^{-(Im \zeta) t}| |e^{i (Re \zeta) t}| = w(t) e^{-(Im \zeta) t} < ae^{-c |t|} e^{-(Im \zeta) t} = ae^{-c t} e^{-(Im \zeta) t} = a e^{-(c + Im \zeta) t} \rightarrow 0$$ since $t > 0$ and $-(c + Im \zeta) < 0$.

Similarly, when $t \rightarrow -\infty$, we have $$|x(t)| = |w(t) e^{i \zeta t}| = |w(t)| |e^{i \zeta t}| = |w(t)| |e^{-(Im \zeta) t}| |e^{i (Re \zeta) t}| = w(t) e^{-(Im \zeta) t} < ae^{-c |t|} e^{-(Im \zeta) t} = ae^{c t} e^{-(Im \zeta) t} = a e^{(c - Im \zeta) t} \rightarrow 0$$ since $t < 0$ and $c - Im \zeta > 0$.

• Thank you Svinto for answering my question. The only fact that I still do not understand is that x(t) is not a real valued function while every function in C is from R to R Jun 26, 2016 at 12:11
• Where does it say that $C: \mathbb{R} \rightarrow \mathbb{R}$? Usually the set of continuous functions on $\mathbb{R}$ vanishing at infinity, commonly denoted as $C_0(\mathbb{R})$, has $\mathbb{C}$ as range. Jun 26, 2016 at 15:44
• Thank you very much Svinto. I just did not know it. Jun 26, 2016 at 16:17