PDE: Proving that a set is an orthogonal bases for L2

[RJLiberator]

Homework Statement

Show that the set {sin(nx)} from n=1 to n=∞ is orthogonal bases for L^2(0, π).

Homework Equations

The Attempt at a Solution

Proof: Let f(x)= sin(nx), consider scalar product in L^2(0, π)

$(ƒ_n , ƒ_m) = \int_{0}^π ƒ_n (x) ƒ_m (x) \, dx = \int_{0}^π sin(nx)sin(mx) \, dx = \frac{1}{2} \int_{0}^π cos(n-m)x - cos(n+m)x \, dx = \frac{1}{2} ( \int_{0}^π cos(n-m)x - \int_{0}^π cos(n+m)x \, dx = \frac{1}{2} ( {\frac{\sin(n-m)x}{(n-m)x}} - {\frac{\sin(n+m)x}{(n+m)x}} ) = 0$

Of course the last result is from 0 to pi for both parts, but didn't know how to latex it :p.

My question is, is this everything to show that the set is orthogonal bases for L2 (0, π)?

I think it is, but definitions in the crazy world of PDE are confusing to me. I do not believe I need to show completeness per the question.

Thank you.

 

[andrewkirk]

My question is, is this everything to show that the set is orthogonal bases for L2 (0, π)?

That's part of the answer, as it proves the elements of the set are pairwise orthogonal.
But now you need to prove it's a basis, which involves proving:

1. That the set spans $\mathscr{L}^2(0,\pi)$, ie that any element of $\mathscr{L}^2(0,\pi)$ can be written as the limit of a sequence of linear combinations of elements of the set.
2. That the set is linearly independent, ie that there is no nontrivial finite linear combination of elements that is equal to zero.

 

[RJLiberator]

@andrewkirk I see what you are saying.

I only proved that the system was orthogonal, but did not show that it was a bases.

I'm having a little trouble going further.

I guess I can set d = sin(nx)/||sin(nx)||, where {d} from 1 to infinity is a complete orthogonal set in L^2.

I can expand this in terms of {sin(nx)} for any f in L^2[0,pi]

f = sum from 1 to infinity <f,sin(nx)>sin(nx)/||sin(nx)||^2

So I can show that it is complete orthogonal set in L^2

 

[andrewkirk]

If you've shown it's complete then all that's left to prove is linear independence, and that follows easily from the orthogonality. If we have a finite set of nonzero functions in the set, for which there is a linear combination that is zero, how can we derive a contradiction from that using the fact that they're all orthogonal?

 

[RJLiberator]

Could we take two elements and take the inner product and by orthogonality we show that element*element = 0. thus we see linear independence?

 

[andrewkirk]

That's the broad idea. But it's not necessarily just two elements.

All that's needed for the set to not be linearly independent is for there to exist some integer $N>0$ and set of complex numbers $c_1,...,c_N$, such that $c_N\neq 0$ and
$$\sum_{k=1}^N c_k\sin kx=0$$

Following your suggestion of taking an inner product, what could you take an inner product of with both sides of this equation, to obtain (via a few more steps) a contradiction?

 

[RJLiberator]

Ugh, I understand exactly what you are trying to do, but my mind is blown.

The right side will be 0 as anything product 0 = 0.
The left side, perhaps the set of complex numbers (c1, c2,...,ck)

 

[andrewkirk]

We want to take the inner product of the left hand side with a vector/function that will give us a nonzero left side. That will give us a contradiction because the right side is zero.

The inner product of the left side with a function f is the sum of the inner products of each term on the LHS with f.
What should we choose f to be so that we know that at least one of those N inner products is nonzero?
Bear in mind that there is only one of the $c_k$ that we know to be nonzero. Which one?

 

[Samy_A]

@andrewkirk I see what you are saying.

I only proved that the system was orthogonal, but did not show that it was a bases.

I'm having a little trouble going further.

I guess I can set d = sin(nx)/||sin(nx)||, where {d} from 1 to infinity is a complete orthogonal set in L^2.

I can expand this in terms of {sin(nx)} for any f in L^2[0,pi]

f = sum from 1 to infinity <f,sin(nx)>sin(nx)/||sin(nx)||^2

So I can show that it is complete orthogonal set in L^2

Maybe I'm missing something, but how did you show the set is complete?

 

[RJLiberator]

We know that C_N is non zero.
So we can multiply both sides by C_N*sinkx ?

As the inner product of C_N and C_N is 1.

 

[RJLiberator]

@Samy_A

From my notes, I can take say g(x) = sin(nx)/||sin(nx)|| then {g(x)} from N = 1 to infinity is an orthonormal set in L^2.

f = sum from n =1 to n=infinity of <f, sin(nx)>sin(nx) / ||sin(nx)||^2
||f||^2 = sum from n = 1 to n = infinity of |<f,sin(nx)>|^2/||sin(nx)||^2

{sin(nx)} from n = 1 to n = infinity is thus a complete orthogonal set in L^2 (0,pi).

 

[Samy_A]

@Samy_A

From my notes, I can take say g(x) = sin(nx)/||sin(nx)|| then {g(x)} from N = 1 to infinity is an orthonormal set in L^2.

f = sum from n =1 to n=infinity of <f, sin(nx)>sin(nx) / ||sin(nx)||^2
||f||^2 = sum from n = 1 to n = infinity of |<f,sin(nx)>|^2/||sin(nx)||^2

{sin(nx)} from n = 1 to n = infinity is thus a complete orthogonal set in L^2 (0,pi).

Ok, if it is given (or proven) in your course that the set is complete, then that's it.

 

[andrewkirk]

We know that C_N is non zero.
So we can multiply both sides by C_N*sinkx ?

If by 'multiply' you mean 'take the inner product of' then yes. Can you do that and get to the solution?

 

[RJLiberator]

Yes, my apologies @andrewkirk , I mean take the inner product of it.

I'm at work now, so I'm trying to work this out in my head, but...
If we take the inner product of both sides, we would see that all the terms cancel out to be left with
Cn*Cnsin^2(kx)

where Cn*Cn is Cn multiplied by the conjugation of cn (inner product)

(I will try this out in more detail upon returning home in a number of hours)