Weak Convergence of A in l^2: Does it Preserve Closedness?

In summary, the map from l^2 to R given by x |--> ||x||^2 is not closed when considering the norm topology, as the set A = {x_n} in l^2 is closed but its image is not closed in R. It is unclear whether the map is closed if the weak topology on R is considered instead of the norm topology, as the misunderstandings may arise from weak convergence. However, since R is finite-dimensional, its weak topology coincides with the norm topology.
  • #1
HMY
14
0
Take the sequence x_n = (1-1/n)e_n in l^2
Consider the map l^2 to R given by x |--> ||x||^2

The set A = {x_n} in l ^2 is closed & its image is not closed in R
under the norm topology (it doesn't contain its accumulation point 1).
So ultimately the above map is not closed.

What I'm not sure about is whether the map is closed if I consider the
weak topology on R instead of the norm topology? I think my misunderstandings are arising from the weak convergence.
 
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  • #2
Because R is finite-dimensional its weak topology coincides with the norm topology.

(I realize this post is really old, but I was bored and decided to browse the forum.)
 
  • #3


The weak convergence of a sequence in l^2 does not necessarily preserve closedness. In this specific case, the sequence x_n = (1-1/n)e_n in l^2 is weakly convergent to the zero vector, but the set A = {x_n} is not weakly closed. This can be seen by considering the weak limit of the sequence, which is the zero vector, but the sequence itself does not converge weakly to any point in A. Therefore, the weak convergence of A does not preserve its closedness.

In general, weak convergence does not imply norm convergence and vice versa. So, the weak topology on R may not be equivalent to the norm topology, and the map in question may not be closed under the weak topology. It is possible for a sequence to converge weakly to a point that is not in the closure of the set, which would violate the definition of a closed set.

However, it is worth noting that weak convergence does preserve closedness in certain cases. For example, if the sequence is bounded in norm, then weak convergence does imply norm convergence and thus preserves closedness. Additionally, in finite-dimensional spaces, weak convergence is equivalent to norm convergence, so in these cases, closedness is preserved under weak convergence.

In conclusion, the weak convergence of A = {x_n} in l^2 does not preserve its closedness, and it is not clear whether the map in question is closed under the weak topology on R. Further investigation would be needed to determine this.
 

1. What is weak convergence in l^2 space?

Weak convergence in l^2 space refers to a type of convergence of a sequence of elements in the l^2 space. It means that the sequence converges to a limit in a weak sense, where the inner product with any other element in the space approaches the inner product with the limit element. This type of convergence is weaker than strong convergence, which requires the sequence to converge to the limit element in norm.

2. What is the importance of preserving closedness in weak convergence of A in l^2?

Preserving closedness in weak convergence of A in l^2 is important because it ensures that the limit element is still contained in the l^2 space. This is significant because the l^2 space is a complete normed vector space, and closedness guarantees the existence of limits for sequences. Without preserving closedness, the limit element may not be in the l^2 space, and the convergence may not be valid.

3. How does weak convergence of A in l^2 preserve closedness?

Weak convergence of A in l^2 preserves closedness by ensuring that the limit element is still in the l^2 space. This is achieved by the properties of weak convergence, where the inner product with any other element in the space approaches the inner product with the limit element. This guarantees that the limit element is still contained in the l^2 space, thus preserving closedness.

4. Can weak convergence of A in l^2 preserve closedness for any sequence of elements?

No, not all sequences of elements in l^2 will preserve closedness under weak convergence. The sequence must satisfy certain conditions, such as boundedness and convergence in weak sense, in order for the closedness to be preserved. If these conditions are not met, then the limit element may not be in the l^2 space, and closedness will not be preserved.

5. What are the advantages of using weak convergence over strong convergence in l^2 space?

One advantage of using weak convergence over strong convergence in l^2 space is that it allows for a wider range of sequences to converge. Weak convergence only requires the sequence to converge in a weak sense, which is a weaker condition than the norm convergence required by strong convergence. This allows for more flexibility in the types of sequences that can converge in l^2 space. Additionally, weak convergence has applications in functional analysis and optimization, making it a useful tool in various mathematical and scientific fields.

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