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 with some other function and if
and control if this equals to 1 if m=n or 0 otherwise. Then we prove that is an orthogonal system right? but how do we choose /find the other function ??
To find , you just replace n by m in the formula that you obtained for But to compute , you should use Parseval's theorem, which tells you that That is much easier to calculate than the inner product of the Fourier transforms, and it will show you that the system is orthonormal. To show that it is not a complete system, you must find a nonzero -function that is orthogonal to each The way to do that is again by using Parseval's theorem, as indicated in my previous comment.
It is easier to find a counter example than you might think, since almost any element of that decays faster than cannot be represented. (the space spanned by this set of functions is a subset of a set of periodic function of period times a function - a bit of care and "set" can be replaced by "space" but I'm not minded to be careful today)
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.
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!
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.
That's what I originally wrote in post 7. I showed was incomplete. But that's ok, because if is incomplete, is incomplete because you can't create a complete basis using only an incomplete basis.
There is another way to show is incomplete using theorem from my previous post. Show (f, ) = 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 . 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.
But the are not all zeros for all less than some , or rather you seem to be assuming that rather than 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 , rather than . 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).
The easiest way to do this is to show that cannot be represented in the basis , and so cannot be represented in the basis
That's what I originally wrote in post 7. I showed was incomplete. But that's ok, because if is incomplete, is incomplete because you can't create a complete basis using only an incomplete basis.
There is another way to show is incomplete using theorem from my previous post. Show (f, ) = 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 . 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.
CB