$L_d = { phi in L_{[-a,a]}^2 vert phi(t) = -phi(-t) }$, prove that is a closed subspace of $L^2_{[-a,a]}$
$L_d = { phi in L_{[-a,a]}^2$ $vert$ $phi(t) = -phi(-t)$ $forall t in [-a,a] }$, prove that is a closed subspace of $L^2_{[-a,a]}$.
Now, the subspace part is always the same and I managed.
The part on closeness is where I stumble. This is what I did:
I want to prove it using closeness by sequences, so:
Let ${phi_n} subset L_d$ be a sequence, such that: $phi_n to phi$ in $L^2$. I want to prove that $phi in L_d$.
I know that $phi_n to phi$ in $L^2$ implies that there exists an extracted subsequence $phi_{n_k}$ such that $phi_{n_k} to phi'$ almost everywhere. I also know that $phi' = phi$ almost everywhere for uniqueness of the limit. But still, I do not know how to say that $phi' = phi in L^d$. I feel like I am just missing this last step but I am not sure on how to proceed.
functional-analysis
asked Nov 27 '18 at 19:26
qcc101
458113
add a comment |
$L_d = { phi in L_{[-a,a]}^2$ $vert$ $phi(t) = -phi(-t)$ $forall t in [-a,a] }$, prove that is a closed subspace of $L^2_{[-a,a]}$.
Now, the subspace part is always the same and I managed.
The part on closeness is where I stumble. This is what I did:
I want to prove it using closeness by sequences, so:
Let ${phi_n} subset L_d$ be a sequence, such that: $phi_n to phi$ in $L^2$. I want to prove that $phi in L_d$.
I know that $phi_n to phi$ in $L^2$ implies that there exists an extracted subsequence $phi_{n_k}$ such that $phi_{n_k} to phi'$ almost everywhere. I also know that $phi' = phi$ almost everywhere for uniqueness of the limit. But still, I do not know how to say that $phi' = phi in L^d$. I feel like I am just missing this last step but I am not sure on how to proceed.
functional-analysis
asked Nov 27 '18 at 19:26
qcc101
458113
add a comment |
$L_d = { phi in L_{[-a,a]}^2$ $vert$ $phi(t) = -phi(-t)$ $forall t in [-a,a] }$, prove that is a closed subspace of $L^2_{[-a,a]}$.
Now, the subspace part is always the same and I managed.
The part on closeness is where I stumble. This is what I did:
I want to prove it using closeness by sequences, so:
Let ${phi_n} subset L_d$ be a sequence, such that: $phi_n to phi$ in $L^2$. I want to prove that $phi in L_d$.
I know that $phi_n to phi$ in $L^2$ implies that there exists an extracted subsequence $phi_{n_k}$ such that $phi_{n_k} to phi'$ almost everywhere. I also know that $phi' = phi$ almost everywhere for uniqueness of the limit. But still, I do not know how to say that $phi' = phi in L^d$. I feel like I am just missing this last step but I am not sure on how to proceed.
functional-analysis
asked Nov 27 '18 at 19:26
qcc101
458113
$L_d = { phi in L_{[-a,a]}^2$ $vert$ $phi(t) = -phi(-t)$ $forall t in [-a,a] }$, prove that is a closed subspace of $L^2_{[-a,a]}$.
Now, the subspace part is always the same and I managed.
The part on closeness is where I stumble. This is what I did:
I want to prove it using closeness by sequences, so:
Let ${phi_n} subset L_d$ be a sequence, such that: $phi_n to phi$ in $L^2$. I want to prove that $phi in L_d$.
I know that $phi_n to phi$ in $L^2$ implies that there exists an extracted subsequence $phi_{n_k}$ such that $phi_{n_k} to phi'$ almost everywhere. I also know that $phi' = phi$ almost everywhere for uniqueness of the limit. But still, I do not know how to say that $phi' = phi in L^d$. I feel like I am just missing this last step but I am not sure on how to proceed.
functional-analysis
functional-analysis
asked Nov 27 '18 at 19:26
qcc101
458113
asked Nov 27 '18 at 19:26
qcc101
458113
asked Nov 27 '18 at 19:26
qcc101
458113
asked Nov 27 '18 at 19:26
qcc101
458113
asked Nov 27 '18 at 19:26
qcc101
458113
458113
add a comment |
add a comment |
1 Answer
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Let $ phi^-(t) = phi(-t). $
We want to show that $phi+phi^- = 0$.
We have
$$ Vert phi+phi^- Vert leq Vert phi-phi_n Vert + Vert phi_n+phi_n^- Vert + Vert phi^--phi_n^- Vert. $$
The middle term is 0 since $phi_n in L_d$, and the other two terms converge to 0 by assumption.
answered Nov 27 '18 at 19:31
MisterRiemann
5,7791624
I got your proof, but can you elaborate on the last step of mine?
– qcc101
Nov 27 '18 at 19:44
@qcc101 I'm not sure to be honest, but I don't think that the pointwise argument will work since you're assuming convergence in norm. See this post.
– MisterRiemann
Nov 27 '18 at 20:16
add a comment |
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1 Answer
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1 Answer
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Let $ phi^-(t) = phi(-t). $
We want to show that $phi+phi^- = 0$.
We have
$$ Vert phi+phi^- Vert leq Vert phi-phi_n Vert + Vert phi_n+phi_n^- Vert + Vert phi^--phi_n^- Vert. $$
The middle term is 0 since $phi_n in L_d$, and the other two terms converge to 0 by assumption.
answered Nov 27 '18 at 19:31
MisterRiemann
5,7791624
I got your proof, but can you elaborate on the last step of mine?
– qcc101
Nov 27 '18 at 19:44
@qcc101 I'm not sure to be honest, but I don't think that the pointwise argument will work since you're assuming convergence in norm. See this post.
– MisterRiemann
Nov 27 '18 at 20:16
add a comment |
Let $ phi^-(t) = phi(-t). $
We want to show that $phi+phi^- = 0$.
We have
$$ Vert phi+phi^- Vert leq Vert phi-phi_n Vert + Vert phi_n+phi_n^- Vert + Vert phi^--phi_n^- Vert. $$
The middle term is 0 since $phi_n in L_d$, and the other two terms converge to 0 by assumption.
answered Nov 27 '18 at 19:31
MisterRiemann
5,7791624
I got your proof, but can you elaborate on the last step of mine?
– qcc101
Nov 27 '18 at 19:44
@qcc101 I'm not sure to be honest, but I don't think that the pointwise argument will work since you're assuming convergence in norm. See this post.
– MisterRiemann
Nov 27 '18 at 20:16
add a comment |
Let $ phi^-(t) = phi(-t). $
We want to show that $phi+phi^- = 0$.
We have
$$ Vert phi+phi^- Vert leq Vert phi-phi_n Vert + Vert phi_n+phi_n^- Vert + Vert phi^--phi_n^- Vert. $$
The middle term is 0 since $phi_n in L_d$, and the other two terms converge to 0 by assumption.
answered Nov 27 '18 at 19:31
MisterRiemann
5,7791624
Let $ phi^-(t) = phi(-t). $
We want to show that $phi+phi^- = 0$.
We have
$$ Vert phi+phi^- Vert leq Vert phi-phi_n Vert + Vert phi_n+phi_n^- Vert + Vert phi^--phi_n^- Vert. $$
The middle term is 0 since $phi_n in L_d$, and the other two terms converge to 0 by assumption.
answered Nov 27 '18 at 19:31
MisterRiemann
5,7791624
edited Nov 27 '18 at 20:12
answered Nov 27 '18 at 19:31
MisterRiemann
5,7791624
answered Nov 27 '18 at 19:31
MisterRiemann
5,7791624
answered Nov 27 '18 at 19:31
MisterRiemann
5,7791624
5,7791624
I got your proof, but can you elaborate on the last step of mine?
– qcc101
Nov 27 '18 at 19:44
@qcc101 I'm not sure to be honest, but I don't think that the pointwise argument will work since you're assuming convergence in norm. See this post.
– MisterRiemann
Nov 27 '18 at 20:16
add a comment |
I got your proof, but can you elaborate on the last step of mine?
– qcc101
Nov 27 '18 at 19:44
@qcc101 I'm not sure to be honest, but I don't think that the pointwise argument will work since you're assuming convergence in norm. See this post.
– MisterRiemann
Nov 27 '18 at 20:16
I got your proof, but can you elaborate on the last step of mine?
– qcc101
Nov 27 '18 at 19:44
I got your proof, but can you elaborate on the last step of mine?
– qcc101
Nov 27 '18 at 19:44
@qcc101 I'm not sure to be honest, but I don't think that the pointwise argument will work since you're assuming convergence in norm. See this post.
– MisterRiemann
Nov 27 '18 at 20:16
@qcc101 I'm not sure to be honest, but I don't think that the pointwise argument will work since you're assuming convergence in norm. See this post.
– MisterRiemann
Nov 27 '18 at 20:16
add a comment |
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