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.