Fourier transform and orthogonal system

**rayman** · December 12th 2011, 09:47 AM

Let $\phi_{n}(x)=\begin{cases} 1 ,\ x\in[n-\frac{1}{2},n+\frac{1}{2}]\\ 0, \ elsewhere \end{cases}$

a) Compute the Fourier transform of the function
$(\hat{\phi_{n}})(\xi)=\int_{n-\frac{1}{2}}^{n+\frac{1}{2}}e^{-i\xi x}dx=-\frac{1}{i\xi}\Bigl[e^{-i\xi x}\Bigr]_{n-\frac{1}{2}}^{n+\frac{1}{2}}=-\frac{1}{i\xi}\Bigl[e^{-i\xi n}e^{-i\xi\frac{1}{2}}-e^{-i\xi n}e^{i\xi\frac{1}{2}}\Bigr]=\frac{1}{\xi i}\Bigl[-e^{-i\xi n}e^{-i\xi\frac{1}{2}}+e^{-i\xi n}e^{i\xi\frac{1}{2}}\Bigr]=\frac{2e^{-i\xi n}}{\xi}\Bigl[\frac{e^{i\xi \frac{1}{2}}-e^{-i\xi \frac{1}{2}}}{2i}\Bigr]=\frac{2e^{-i\xi n}}{\xi}\sin(\frac{\xi}{2})$

is this part correct?

b) prove that $(\hat{\phi_{n}})_{n=1}^{\infty}$ is an orthogonal system in $L^2(\mathbb{R})$ . Prove that it is not complete.

Could someone show me how to do the part b?
thank you very much

**CaptainBlack** · December 12th 2011, 10:58 PM

Originally Posted by rayman

Let $\phi_{n}(x)=\begin{cases} 1 ,\ x\in[n-\frac{1}{2},n+\frac{1}{2}]\\ 0, \ elsewhere \end{cases}$

a) Compute the Fourier transform of the function
$(\hat{\phi_{n}})(\xi)=\int_{n-\frac{1}{2}}^{n+\frac{1}{2}}e^{-i\xi x}dx=-\frac{1}{i\xi}\Bigl[e^{-i\xi x}\Bigr]_{n-\frac{1}{2}}^{n+\frac{1}{2}}=-\frac{1}{i\xi}\Bigl[e^{-i\xi n}e^{-i\xi\frac{1}{2}}-e^{-i\xi n}e^{i\xi\frac{1}{2}}\Bigr]=\frac{1}{\xi i}\Bigl[-e^{-i\xi n}e^{-i\xi\frac{1}{2}}+e^{-i\xi n}e^{i\xi\frac{1}{2}}\Bigr]=\frac{2e^{-i\xi n}}{\xi}\Bigl[\frac{e^{i\xi \frac{1}{2}}-e^{-i\xi \frac{1}{2}}}{2i}\Bigr]=\frac{2e^{-i\xi n}}{\xi}\sin(\frac{\xi}{2})$

is this part correct?

b) prove that $(\hat{\phi_{n}})_{n=1}^{\infty}$ is an orthogonal system in $L^2(\mathbb{R})$ . Prove that it is not complete.

Could someone show me how to do the part b?
thank you very much

(a) looks OK

(b) Start by showing that $\langle \widehat{\phi}_n, \widehat{\phi}_m\rangle=0$ when $n\ne m$ .

To show that it is not compelete I think you need to find a function in $L^2(\mathbb{R})$ that cannot be represented as a series in the $\widehat{\phi}_n$ convergent in norm.

CB

**Opalg** · December 12th 2011, 11:47 PM

For (b), I would start by taking f to be a function in $L^2(\mathbb{R})$ that is orthogonal to all the functions $\phi_n$ . For example, $f(x) = \begin{cases} x & x\in[-\frac{1}{2},\frac{1}{2}],\\ 0 & \text{elsewhere.} \end{cases}$

Parseval's theorem then tells you that $\hat{f}$ is orthogonal to all the functions $\hat{\phi}_n$ .

**rayman** · December 13th 2011, 12:04 AM

I am not sure If I understand what I am supposed to do in part b)
I thougt than to prove the system to be orthogonal we need to compute the inner product of our transformed function $\hat{(\phi_{n})}$ with some other function and if
$\Bigl<\hat{(\phi_{n})},\hat{(\phi_{m})}\Bigr>$ and control if this equals to 1 if m=n or 0 otherwise. Then we prove that $\hat{(\phi_{n})}$ is an orthogonal system right? but how do we choose /find the other function $\hat{(\phi_{m})}$ ??

**Opalg** · December 13th 2011, 03:31 AM

Originally Posted by rayman

I am not sure If I understand what I am supposed to do in part b)
I thougt than to prove the system to be orthogonal we need to compute the inner product of our transformed function $\hat{(\phi_{n})}$ with some other function and if
$\Bigl<\hat{(\phi_{n})},\hat{(\phi_{m})}\Bigr>$ and control if this equals to 1 if m=n or 0 otherwise. Then we prove that $\hat{(\phi_{n})}$ is an orthogonal system right? but how do we choose /find the other function $\hat{(\phi_{m})}$ ??

To find $\hat{\phi}_m$ , you just replace n by m in the formula that you obtained for $\hat{\phi}_n.$ But to compute $\bigl\langle\hat{\phi}_n,\hat{\phi}_m\bigr\rangle$ , you should use Parseval's theorem, which tells you that $\bigl\langle\hat{\phi}_n,\hat{\phi}_m\bigr\rangle = \bigl\langle\phi_n,\phi_m\bigr\rangle.$ That is much easier to calculate than the inner product of the Fourier transforms, and it will show you that the system $\hat{(\phi_{n})}$ is orthonormal. To show that it is not a complete system, you must find a nonzero $L^2$ -function that is orthogonal to each $\hat{\phi}_n.$ The way to do that is again by using Parseval's theorem, as indicated in my previous comment.

**CaptainBlack** · December 13th 2011, 06:01 AM

Originally Posted by CaptainBlack

(a) looks OK

(b) Start by showing that $\langle \widehat{\phi}_n, \widehat{\phi}_m\rangle=0$ when $n\ne m$ .

To show that it is not compelete I think you need to find a function in $L^2(\mathbb{R})$ that cannot be represented as a series in the $\widehat{\phi}_n$ convergent in norm.

CB

It is easier to find a counter example than you might think, since almost any element of $L^2(\mathbb{R})$ that decays faster than $1/\xi$ cannot be represented. (the space spanned by this set of functions is a subset of a set of periodic function of period $2\pi$ times a $\rm {sinc}$ function - a bit of care and "set" can be replaced by "space" but I'm not minded to be careful today)

**Hartlw** · December 13th 2011, 11:38 AM

b) Incomplete because you can't represent an arbitrary function with basis functions that are 0 for all x < x0.

Like trying to represent an arbitrary vector in 3d with basis vectors that have 0 as the first component.

**CaptainBlack** · December 13th 2011, 11:21 PM

Originally Posted by Hartlw

b) Incomplete because you can't represent an arbitrary function with basis functions that are 0 for all x < x0.

Like trying to represent an arbitrary vector in 3d with basis vectors that have 0 as the first component.

I thought I had already posted this but ...

The basis functions are not all zero beyoud some point, they are a sinc function times the elements of the standard fourier series basis $\exp(-{\rm{i}} n \xi), \ \ n= ... -1,0,1, ...$ .
CB

**Hartlw** · December 14th 2011, 09:03 AM

Originally Posted by CaptainBlack

I thought I had already posted this but ...

The basis functions are not all zero beyoud some point, they are a sinc function times the elements of the standard fourier series basis $\exp(-{\rm{i}} n \xi), \ \ n= ... -1,0,1, ...$ .
CB

You can't create a complete basis using only an incomplete basis.

Edit: While I was responding on my computer the screen went blank and locked. After I shut down and restarted I could access the forum home pg but nothing else. Went to another computer (this one, with later version of Windows) and managed to respond, but when I shut down after having done nothing else I got notice of down-loading fifteen updates. That's usually a sign of a detected attack. When I rebooted there was a long list of registry updates. Scary. Concidence? Censorship?

Edit Update: Now can access MHF from my computer. Looks like whatever happened was unrelated to Forum, but detected by security software and corrected. Whew!

**Hartlw** · December 14th 2011, 11:47 AM

This is awkward but I simply can't get TEX to do this. So

xn = phi sub n
x'n = hat phi sub n

You can't create a complete basis x'n using only an incomplete basis xn. xn was previously shown to be incomplete.

There is also a direct test of x'n which might work out:

"An orthonormal sequence in H is complete iff the condition (x,x'n) = 0 for all n implies x = 0."

You could write out the explicit expression for (x,x'n) and see if there is an obvious non-zero x which satrisfies the condition.

**CaptainBlack** · December 14th 2011, 07:32 PM

Originally Posted by Hartlw

This is awkward but I simply can't get TEX to do this. So

xn = phi sub n
x'n = hat phi sub n

You can't create a complete basis x'n using only an incomplete basis xn. xn was previously shown to be incomplete.

Where? Not in this thread, you are appealing to a result that we do not know that you know, though you could use the properties of the FT to justify this on the fly.

CB

**Hartlw** · December 14th 2011, 09:06 PM

Originally Posted by Hartlw

b) Incomplete because you can't represent an arbitrary function with basis functions that are 0 for all x < x0.

Like trying to represent an arbitrary vector in 3d with basis vectors that have 0 as the first component.

That's what I originally wrote in post 7. I showed $\phi_n$ was incomplete. But that's ok, because if $\phi_n$ is incomplete, $\widehat{\phi}_n$ is incomplete because you can't create a complete basis using only an incomplete basis.

There is another way to show $\phi_n$ is incomplete using theorem from my previous post. Show (f, $\phi_n$ ) = 0 for a non-zero f. take f(x) = 0 for x>0 and f(x) =1 for a<= x <=0, any a.

NOTE: finally figured out how to write $\phi_n$ . I copied the code and tried it in a practice post and wrapped it in [tex] tags and it simply did not work. Then tried [TEX] tags and it worked. Now I can communicate a little more intelligibly.

BY the way, when I tried to reply on my other computer, Forum locked up on me again. I noticed a current thread about hacking. Maybe something is around.

**CaptainBlack** · December 14th 2011, 10:46 PM

Originally Posted by Hartlw

b) Incomplete because you can't represent an arbitrary function with basis functions that are 0 for all x < x0.

Like trying to represent an arbitrary vector in 3d with basis vectors that have 0 as the first component.

But the $\phi_n$ are not all zeros for all $x$ less than some $x_0$ , or rather you seem to be assuming that $n \in \mathbb{N}$ rather than $n \in \mathbb{Z}$ and there seems to be no reason to assume so (It was rather sloppy of the OP not to specify what the index set was/is).

Also, it is sloppy not to specify which basis you are talking about, I originally assumed you were talking about $\{\widehat{\phi}_n, n\in \mathbb{Z}\}$ , rather than $\{{\phi}_n, n\in \mathbb{Z}\}$ . It is also sloppy to rely on results in other threads without explicitly referencing them and giving a link to the thread (which I pressume was your intention from some of what you have written in other posts in this thread).

That's what I originally wrote in post 7. I showed $\phi_n$ was incomplete. But that's ok, because if $\phi_n$ is incomplete, $\widehat{\phi}_n$ is incomplete because you can't create a complete basis using only an incomplete basis.

There is another way to show $\phi_n$ is incomplete using theorem from my previous post. Show (f, $\phi_n$ ) = 0 for a non-zero f. take f(x) = 0 for x>0 and f(x) =1 for a<= x <=0, any a.

NOTE: finally figured out how to write $\phi_n$ . I copied the code and tried it in a practice post and wrapped it in [tex] tags and it simply did not work. Then tried [TEX] tags and it worked. Now I can communicate a little more intelligibly.

BY the way, when I tried to reply on my other computer, Forum locked up on me again. I noticed a current thread about hacking. Maybe something is around.

The easiest way to do this is to show that $\chi_{{[-1/4,1/4]}}$ cannot be represented in the basis $\{\phi_n, n\in \mathbb{Z}\}$ , and so $\mathfrak{F} \chi_{[-1/4,1/4]}$ cannot be represented in the basis $\{\widehat{\phi}_n, n\in \mathbb{Z}\}$

CB

**Opalg** · December 15th 2011, 12:07 AM

Originally Posted by CaptainBlack

Originally Posted by Hartlw

This is awkward but I simply can't get TEX to do this. So

xn = phi sub n
x'n = hat phi sub n

You can't create a complete basis x'n using only an incomplete basis xn. xn was previously shown to be incomplete.

Where? Not in this thread

Has my comment #3 become invisible then?

**CaptainBlack** · December 15th 2011, 01:46 AM

Originally Posted by Opalg

Has my comment #3 become invisible then?

Your $f$ is $\phi_0$ under my assumption about the unspecified index set for the $\phi$ s (but the basic idea can be made to work)

CB

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