Fourier transform and orthogonal system

**Opalg** · December 15th 2011, 02:02 AM

Originally Posted by CaptainBlack

Your $f$ is $\phi_0$ under my assumption about the unspecified index set for the $\phi$ s

I don't understand that. The OP's $\phi_0$ is given by $\phi_0(x) = \begin{cases} 1 & x\in[-\frac{1}{2},\frac{1}{2}],\\ 0 & \text{elsewhere.} \end{cases}$

My f is given by $f(x) = \begin{cases} x & x\in[-\frac{1}{2},\frac{1}{2}],\\ 0 & \text{elsewhere.} \end{cases}$

They are orthogonal to each other.

**CaptainBlack** · December 15th 2011, 04:42 AM

Originally Posted by Opalg

I don't understand that. The OP's $\phi_0$ is given by $f(x) = \begin{cases} 1 & x\in[-\frac{1}{2},\frac{1}{2}],\\ 0 & \text{elsewhere.} \end{cases}$

My f is given by $f(x) = \begin{cases} x & x\in[-\frac{1}{2},\frac{1}{2}],\\ 0 & \text{elsewhere.} \end{cases}$

They are orthogonal to each other.

I'm lost, $f(x)$ is in $\{\phi_n,\ n\in \mathbb{Z}\}$ but arn't you looking for a $L_2$ function not expressible in terms of these functions?

CB

**Opalg** · December 15th 2011, 05:43 AM

Originally Posted by CaptainBlack

I'm lost, $f(x)$ is in $\{\phi_n,\ n\in \mathbb{Z}\}$ but arn't you looking for a $L_2$ function not expressible in terms of these functions?

Sorry to keep harping on about this. There was a typo (now corrected) in my previous comment, which made nonsense of it. What I should have said is

$\phi_0(x) = \begin{cases} 1 & x\in[-\frac{1}{2},\frac{1}{2}],\\ 0 & \text{elsewhere,} \end{cases}\qquad f(x) = \begin{cases} x & x\in[-\frac{1}{2},\frac{1}{2}],\\ 0 & \text{elsewhere.} \end{cases}$

The difference is that $\phi_0(x)$ has a 1 in its definition where $f(x)$ has an x.

**CaptainBlack** · December 15th 2011, 06:11 AM

Originally Posted by Opalg

Sorry to keep harping on about this. There was a typo (now corrected) in my previous comment, which made nonsense of it. What I should have said is

$\phi_0(x) = \begin{cases} 1 & x\in[-\frac{1}{2},\frac{1}{2}],\\ 0 & \text{elsewhere,} \end{cases}\qquad f(x) = \begin{cases} x & x\in[-\frac{1}{2},\frac{1}{2}],\\ 0 & \text{elsewhere.} \end{cases}$

The difference is that $\phi_0(x)$ has a 1 in its definition where $f(x)$ has an x.

OK that makes sense, I still prefer $f(x)=\chi_{[-1/4,1/4]}(x)$ , which is obviously not in the space spanned by the $\phi_n$ (though not orthogonal to $\phi_0$ ) and does not require any great though to take the FT of.

CB

**Hartlw** · December 15th 2011, 08:54 AM

Originally Posted by Opalg

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 $\hat{\phi}_n$ .

Parseval's Theorem: An orthonormal sequence {xn} in H is complete iff:

llxll^2 = (sum n=1 to inf) l(x,xn)l^2.. for every x in H.

Using your premise above, how does Parseval's theorem tell you $\widehat{\phi}_n$ is incomplete? ie, how do you show " $\hat{f}$ is orthogonal to all the functions $\hat{\phi}_n$ "

**Opalg** · December 15th 2011, 12:15 PM

Originally Posted by Hartlw

Parseval's Theorem: An orthonormal sequence {xn} in H is complete iff:

llxll^2 = (sum n=1 to inf) l(x,xn)l^2.. for every x in H.

Using your premise above, how does Parseval's theorem tell you $\widehat{\phi}_n$ is incomplete? ie, how do you show " $\hat{f}$ is orthogonal to all the functions $\hat{\phi}_n$ "

There is more than one result that carries Parseval's name. The result that you state as Parseval's Theorem is what I call Parseval's Identity. For me, Parseval's Theorem is the result which says that if f, g are functions in $L^2(\mathbb{R})$ with Fourier transforms $\widehat{f},\,\widehat{g}$ then $\langle f,g\rangle = \langle \widehat{f},\widehat{g}\rangle.$

It follows that if you have a (nonzero) function $f\inL^2(\mathbb{R})$ such that $\langle f,\phi_n\rangle = 0$ for all n, then $\langle \widehat{f},\widehat{\phi}_n\rangle = 0$ for all n.

**Hartlw** · December 16th 2011, 05:38 AM

Originally Posted by Opalg

There is more than one result that carries Parseval's name. The result that you state as Parseval's Theorem is what I call Parseval's Identity. For me, Parseval's Theorem is the result which says that if f, g are functions in $L^2(\mathbb{R})$ with Fourier transforms $\widehat{f},\,\widehat{g}$ then $\langle f,g\rangle = \langle \widehat{f},\widehat{g}\rangle.$

It follows that if you have a (nonzero) function $f\inL^2(\mathbb{R})$ such that $\langle f,\phi_n\rangle = 0$ for all n, then $\langle \widehat{f},\widehat{\phi}_n\rangle = 0$ for all n.

That wraps it up. Thanks, very nice.

I note belatedly that f is given by you in post #3, which also invokes Parseval's theorem.

The theorem that if ( $\widehat{f}$ , $\widehat{\phi}_n$ ) = 0 for all n and some $\widehat{f}$ not zero implies $\widehat{\phi}_n$ is incomplete, might help the novice.

Finally, I found many versions of Parsevals theorem, with various (formula, theorem, identity) interchanged names, including for Fourier Series and one that states the Fourier transform is bijective, but none with your version. Could you please give a source for your version, preferably internet?

**Opalg** · December 16th 2011, 07:39 AM

Originally Posted by Hartlw

That wraps it up. Thanks, very nice.

I note belatedly that f is given by you in post #3, which also invokes Parseval's theorem.

The theorem that if ( $\widehat{f}$ , $\widehat{\phi}_n$ ) = 0 for all n and some $\widehat{f}$ not zero implies $\widehat{\phi}_n$ is incomplete, might help the novice.

Finally, I found many versions of Parsevals theorem, with various (formula, theorem, identity) interchanged names, including for Fourier Series and one that states the Fourier transform is bijective, but none with your version. Could you please give a source for your version, preferably internet?

Parseval's theorem - Wikipedia, the free encyclopedia (see especially the section headed Equivalence of the norm and inner product forms.

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