Why is this function not ##L^1(\mathbb{R} \times \mathbb{R})##?

[laurabon]

Hi everyone in the following expression
$f(t)=\frac{1}{2 \pi} \int\left(\int f(u) e^{-i \omega u} d u\right) e^{i \omega t} d \omega$

the book says I can't swap integrals bacause the function

$f(u) e^{i \omega(t-u)}$ is not $L^1(\mathbb{R} \times \mathbb{R})$

why ? complex exponential is not always bounded?

 

[FactChecker]

What properties does $f(u)$ have?

 

[laurabon]

What properties does $f(u)$ have?

it is Fuorier inverse theorem with $f$ in $L^1(\mathbb{R})$

 

[PeroK]

Hi everyone in the following expression
$f(t)=\frac{1}{2 \pi} \int\left(\int f(u) e^{-i \omega u} d u\right) e^{i \omega t} d \omega$

the book says I can't swap integrals bacause the function

$f(u) e^{i \omega(t-u)}$ is not $L^1(\mathbb{R} \times \mathbb{R})$

why ? complex exponential is not always bounded?

Note that $\sin$ and $\cos$ are not integrable over $\mathbb R$.

 

[laurabon]

Note that $\sin$ and $\cos$ are not integrable over $\mathbb R$.

I came up with this . Is correct? $\begin{aligned} & \left|f(u) e^{i \omega(t-u)}\right|=|f(u)| \\ & \iint|f(u)|=\int_{-\infty}^{\infty} c = \infty \end{aligned}$

how should I use the fact that sin and cosine are not integrable ?I mean to verify that is in L1 I always have to use the fact that $|e^{i \omega(t-u)}|=1$ I think

 

[PeroK]

I came up with this . Is correct? $\begin{aligned} & \left|f(u) e^{i \omega(t-u)}\right|=|f(u)| \\ & \iint|f(u)|=\int_{-\infty}^{\infty} c = \infty \end{aligned}$

how should I use the fact that sin and cosine are not integrable ?I mean to verify that is in L1 I always have to use the fact that $|e^{i \omega(t-u)}|=1$ I think

I don't think that works. First, let's take $t = 0$, where the integral reduces to:
$$f(0) = \int_{-\infty}^{+\infty} \bigg ( \int_{-\infty}^{+\infty}f(u)e^{-iwu}\ du \bigg ) \ dw$$If we try to swap the integrals and look at:$$\int_{-\infty}^{+\infty} f(u) \bigg ( \int_{-\infty}^{+\infty}e^{-iwu}\ dw \bigg ) \ du$$We can see immediately that the inner integral does not converge and does not represent a well-defined function of $u$. Because neither $\cos(wu)$ nor $\sin(wu)$ is integrable with respect to $w$ on all of $\mathbb R$.

 

[WWGD]

Hi everyone in the following expression
$f(t)=\frac{1}{2 \pi} \int\left(\int f(u) e^{-i \omega u} d u\right) e^{i \omega t} d \omega$

the book says I can't swap integrals bacause the function

$f(u) e^{i \omega(t-u)}$ is not $L^1(\mathbb{R} \times \mathbb{R})$

why ? complex exponential is not always bounded?

Notice f(x)==1 is also bounded. To be able to be in $L^1(X)$ , where $X$ itself is unbounded, your f has to decrease fast-enough.