# Show that an integral is well defined

• April 20th 2010, 07:43 AM
Ase
Show that an integral is well defined
Hi

I am trying to show that for f belonging to L^2(-pi;pi) the integral that defines the complex Fourier Coefficients is well defined. In other words what I need to show is that

int_from -pi to pi(|f(x)*exp(-i*k*x)|dx) < infinity (limited)

I was thinking that since f belongs to L^2(-pi;pi) then the integral of this will be finite. Further more since the inteval is limited (-pi;pi) and k belonging to the set of integers the complex exponential function would also be finite.

Am I right?
• April 20th 2010, 07:51 AM
Laurent
Quote:

Originally Posted by Ase
Hi

I am trying to show that for f belonging to L^2(-pi;pi) the integral that defines the complex Fourier Coefficients is well defined. In other words what I need to show is that

int_from -pi to pi(|f(x)*exp(-i*k*x)|dx) < infinity (limited)

I was thinking that since f belongs to L^2(-pi;pi) then the integral of this will be finite. Further more since the inteval is limited (-pi;pi) and k belonging to the set of integers the complex exponential function would also be finite.

Am I right?

You didn't prove anything. Basically, since $|e^{-ikx}|=1$, you have to prove that if $\int_{-\pi}^\pi |f(x)|^2 dx<\infty$, then $\int_{-\pi}^\pi |f(x)| dx<\infty$. The advice is to use Cauchy-Schwarz inequality (or the elementary inequality $a\leq\frac{a^2+1}{2}$).
• April 20th 2010, 08:10 AM
Ase
Hi Laurent

Thanks a lot. I will get right at it.
• April 20th 2010, 12:52 PM
surjective
well-defined integral
For $f \in L^{2}(-\pi,\pi)$ I would like to show that:

$\int_{-\pi}^{\pi}|f(x)e^{-ikx}|< \infty$

I have done the following:

$\int_{-\pi}^{\pi}|f(x)e^{-ikx}|dx= \int_{-\pi}^{\pi}|f(x)(\cos(kx)-i\sin(kx)|dx\leq \int_{-\pi}^{\pi}|f(x)dx|$

How do I continue (I tried to use Cauchy-Sjcwarz inequality as suggested but it did not lead my anywhere)

Thanks a bunch.
• April 20th 2010, 12:58 PM
Laurent
Quote:

Originally Posted by surjective
For $f \in L^{2}(-\pi,\pi)$ I would like to show that:

$\int_{-\pi}^{\pi}|f(x)e^{-ikx}|< \infty$

I have done the following:

$\int_{-\pi}^{\pi}|f(x)e^{-ikx}|dx= \int_{-\pi}^{\pi}|f(x)(\cos(kx)-i\sin(kx)|dx\leq \int_{-\pi}^{\pi}|f(x)dx|$

How do I continue (I tried to use Cauchy-Sjcwarz inequality as suggested but it did not lead my anywhere)

Thanks a bunch.

To apply Cauchy-Schwarz, write $\int f(x) dx = \int f(x)g(x) dx$ where $g(x)=1$.

Alternatively, use the simple inequality I gave in the previous post: $\int f\leq \frac{1}{2}\Big((\int f^2)+(\int 1)\Big)=\frac{1}{2}\Big((\int f^2) + 2\pi\Big)<\infty$.
• April 20th 2010, 01:24 PM
surjective
well-defined intergral
Here is what I have done:

$
\int_{-\pi}^{\pi}|f(x)|dx=\int_{-\pi}^{\pi}|f(x)g(x)|dx \leq \int_{-\pi}^{\pi}|f(x)||g(x)|dx \leq \int_{-\pi}^{\pi}|f(x)||f(x)|dx < \infty
$

Correct?
• April 20th 2010, 01:29 PM
Laurent
Quote:

Originally Posted by surjective
Here is what I have done:

$
\int_{-\pi}^{\pi}|f(x)|dx=\int_{-\pi}^{\pi}|f(x)g(x)|dx \leq \int_{-\pi}^{\pi}|f(x)||g(x)|dx \leq \int_{-\pi}^{\pi}|f(x)||f(x)|dx < \infty
$

Correct?

You should take a look at what Cauchy-Schwarz inequality for integrals is; because I can't see how you're applying it here.
• April 20th 2010, 02:12 PM
surjective
well-defined integral revisited
Hello,

Let me try one more time:

$\int_{-\pi}^{\pi}|f(x)|dx= \int_{-\pi}^{\pi}|f(x)g(x)|dx \leq \left(\int_{-\pi}^{\pi}|f(x)|^{2}dx \right)^{\frac{1}{2}}\left(\int_{-\pi}^{\pi}|g(x)|^{2}dx \right)^{\frac{1}{2}} \leq$

$\left(\int_{-\pi}^{\pi}|f(x)|^{2}dx \right)^{\frac{1}{2}}\left(\int_{-\pi}^{\pi}|f(x)|^{2}dx \right)^{\frac{1}{2}} < \infty$

What do you think?
• April 21st 2010, 12:15 AM
Laurent
Quote:

Originally Posted by surjective
Hello,

Let me try one more time:

$\int_{-\pi}^{\pi}|f(x)|dx= \int_{-\pi}^{\pi}|f(x)g(x)|dx \leq \left(\int_{-\pi}^{\pi}|f(x)|^{2}dx \right)^{\frac{1}{2}}\left(\int_{-\pi}^{\pi}|g(x)|^{2}dx \right)^{\frac{1}{2}} \leq$

$\left(\int_{-\pi}^{\pi}|f(x)|^{2}dx \right)^{\frac{1}{2}}\left(\int_{-\pi}^{\pi}1 dx \right)^{\frac{1}{2}} =\sqrt{2\pi}\left(\int_{-\pi}^{\pi}|f(x)|^{2}dx \right)^{\frac{1}{2}}< \infty$

What do you think?

I did the corrections above.