Thread: Proving completeness in $L_2(E)$...

The journal I am reading through contains the following theorem...

Theorem: The set $\{w_n(y,\tau)\}$ is linearly independent and complete in $L_2(E).$

(If you are interested the work is on the heat equation and $w_n$ represents the fundamental solution given at a denumerable number of points and $E$ is the lateral surface of a cylinder under consideration).

I am fine with the proof of linear independence, my question is how do you prove a sequence is complete in functional analysis in general?

In the proof he lets an $f(y,\tau)$ be an arbitrary function in $E$ and
$$\int_{0}^{T}d\tau\int_{S}f(y,\tau)w_n(y,\tau) \,dS = 0, \, n=1,2,3,...$$

He then states that if $f(y,\tau) = 0$ then the sequence $\{w_n(y,\tau)\}$ is complete.

Sorry if this is a stupid question above, just like details of why we do the above in the proof.

Originally Posted by KroneckerDelta

The journal I am reading through contains the following theorem...

Theorem: The set $\{w_n(y,\tau)\}$ is linearly independent and complete in $L_2(E).$

(If you are interested the work is on the heat equation and $w_n$ represents the fundamental solution given at a denumerable number of points and $E$ is the lateral surface of a cylinder under consideration).

I am fine with the proof of linear independence, my question is how do you prove a sequence is complete in functional analysis in general?

In the proof he lets an $f(y,\tau)$ be an arbitrary function in $E$ and
$$\int_{0}^{T}d\tau\int_{S}f(y,\tau)w_n(y,\tau) \,dS = 0, \, n=1,2,3,...$$ Shouldn't that dS be dy?

He then states that if $f(y,\tau) = 0$ then the sequence $\{w_n(y,\tau)\}$ is complete.

Sorry if this is a stupid question above, just like details of why we do the above in the proof.

The space $\displaystyle L_2(E)$ is a Hilbert space. In a Hilbert space, the closed linear subspace spanned by a set W is equal to the second orthogonal complement $\displaystyle W^{\perp\perp}$ of W. Clearly if $\displaystyle W^\perp = \{0\}$ then $\displaystyle W^{\perp\perp}$ will be the whole space. So to prove that a set W is complete, you need to show that $\displaystyle W^\perp = \{0\}$.

To do that, you take an arbitrary element $\displaystyle f\in L_2(E)$ and show that if the inner product $\displaystyle \langle f,w\rangle=0$ for all w in W then f must be 0.

The equation $\displaystyle \int_{0}^{T}d\tau\int_{S}f(y,\tau)w_n(y,\tau) \,dy = 0$ is exactly the condition that that inner product should be 0.

Thanks for the reply Opalg, really helpful and makes total sense now, I should most definately brush up on my linear algebra .