Question about weak convergence in Hilbert space

[Sardin]

The Question is as follows:

let A be a bounded domain in R^n and
Xm a series of real functions in L^2 (A).
if Xm converge weakly to X in L^2(A)
and (Xm)^2 converge weakly to Y in L^2(A)
then Y=X^2.

i don't know if the above theorem is true and could sure use any help i can get.
if anyone has any proof please post it... thanks.

 

[Jimmy Snyder]

As I am not experienced in these matters, this proof should be checked for errors by one of the mentors.

For starters, I will remind everyone what the definition of weak convergence in L^{2}(A) is:

A sequence {X_{n}} is said to converge weakly to X (written X_{n} \stackrel{w}{\rightarrow} X) if for all functions Z in L^{2}(A), we have \int X_{n}Z \rightarrow \int XZ

Next, I will show that if a sequence {X_{n}} in L^{2}(A) converges weakly, then the sequence of integrals |\int X_{n}| is bounded. Just let Z be the constant function Z(x) = 1. Then by the definition of weak convergence,

\int X_{n} = \int X_{n}Z \rightarrow \int XZ = \int X

and \int X_{n} convergent means |\int X_{n}| is bounded.

Next, I will point out that there is a theorem that says that two functions X and Y in L^{2}(A) are equal if for all Z in L^{2}(A) we have

\int XZ = \int YZ

Finally, I get to the proof.

Let M be the bound on |\int X_{n}|. Let L = |\int X|. Let Z be in L^{2}(A), and choose \epsilon > 0.

Since {X_{n}} \stackrel{w}{\rightarrow} X, we can find N_{1} such that for all n \ge N_{1} we have |\int X_{n}Z - \int XZ| < \frac{\epsilon}{3L}.

Also, we can find N_{2} such that for all n \ge N_{2} we have |\int X_{n}Z - \int XZ| < \frac{\epsilon}{3M}.

Since {X_{n}^2} \stackrel {w}{\rightarrow} Y we can find N_{3} such that for all n \ge N_{3} we have |\int X_{n}^{2}Z - \int YZ| < \frac{\epsilon}{3}.

Let N be the maximum of N_{1}, N_{2}, and N_{3}, then for all n > N we have

|\int X^{2}Z - \int YZ|
\le |\int X^{2}Z - \int X_{n}XZ| + |\int X_{n}XZ - \int X_{n}^{2}Z| + |\int X_{n}^{2}Z - \int YZ|
\le |\int X||\int XZ - \int X_{n}Z| + |\int X_{n}||\int XZ - \int X_{n}Z| + |\int X_{n}^{2}Z - \int YZ|
< L\frac{\epsilon}{3L} + M\frac{\epsilon}{3M} + \frac{\epsilon}{3} = \epsilon

So, X^{2} = Y

Q.E.D.

 

[M98Ranger]

The above theorem is indeed true and is known as the "Sobolev Embedding Theorem". It states that if a sequence of functions in a bounded domain A converges weakly in L^2(A) and the square of the sequence converges weakly in L^2(A), then the square of the limit function is equal to the limit of the squares. This can be proven using the properties of weak convergence and the Cauchy-Schwarz inequality.

To start, we can rewrite the weak convergence conditions as follows:

∫A Xmφ dx → ∫A Xφ dx for all φ ∈ L^2(A)
∫A (Xm)^2φ dx → ∫A Yφ dx for all φ ∈ L^2(A)

where φ is a test function. Now, we can use the Cauchy-Schwarz inequality to obtain:

|∫A (Xm)^2φ dx| ≤ ||(Xm)^2||_2 ||φ||_2

where ||·||_2 denotes the L^2 norm. Similarly, we can write:

|∫A X^2φ dx| ≤ ||X^2||_2 ||φ||_2 = ||X||_2^2 ||φ||_2^2

Since Xm converges weakly to X in L^2(A), we have ||Xm||_2 → ||X||_2. Therefore, ||Xm||_2^2 → ||X||_2^2. Similarly, ||(Xm)^2||_2 → ||X^2||_2. This means that the right-hand side of the first inequality goes to the right-hand side of the second inequality as m → ∞.

Now, since we have shown that the right-hand side of the first inequality goes to the right-hand side of the second inequality, and that the left-hand side of the first inequality is bounded, it follows that the left-hand side of the second inequality is also bounded. This means that Y ∈ L^2(A) and we can apply the weak convergence condition to obtain:

∫A Yφ dx = ∫A (Xm)^2φ dx → ∫A X^2φ dx

for all φ ∈ L^2(A). This proves that Y = X^2.

I hope this helps. Please let me know if you have any further questions.