Prove that a certain closed subspace of $C[0,1]$ has finite dimension
I need to show that if $L subset C[0, 1]$ is a closed subspace in the $C[0,1]$ norm (i.e. $||f||_{L}=sup|f(x)|$), and all elements of $L$ are continuously differentiable, then L has finite dimension.
I can show:
$ exists n in mathbb{N}$ such that $forall f in L, lVert f rVert=1
> Rightarrow lVert f'rVert le n$
But I need to conclude from this that the mapping $psi :L rightarrow mathbb{R}^{4n+1}$ is injective, where $psi: f mapsto(f(0), f(1/4n), f(2/4n), ... , f(1))$
And I have no idea how to prove the injectivity of $psi$ and I even think that it may not be injective in general. Could you give me a hint how to prove injectivity?
functional-analysis
edited Nov 18 at 16:47
Taroccoesbrocco
5,05761839
asked Nov 18 at 15:20
Anton Zagrivin
1598
add a comment |
I need to show that if $L subset C[0, 1]$ is a closed subspace in the $C[0,1]$ norm (i.e. $||f||_{L}=sup|f(x)|$), and all elements of $L$ are continuously differentiable, then L has finite dimension.
I can show:
$ exists n in mathbb{N}$ such that $forall f in L, lVert f rVert=1
> Rightarrow lVert f'rVert le n$
But I need to conclude from this that the mapping $psi :L rightarrow mathbb{R}^{4n+1}$ is injective, where $psi: f mapsto(f(0), f(1/4n), f(2/4n), ... , f(1))$
And I have no idea how to prove the injectivity of $psi$ and I even think that it may not be injective in general. Could you give me a hint how to prove injectivity?
functional-analysis
edited Nov 18 at 16:47
Taroccoesbrocco
5,05761839
asked Nov 18 at 15:20
Anton Zagrivin
1598
I know that I can proove that $L$ is finite dimensional easier, but I need to prove exactly that $dimL le 4n+1$ where $n$ is from my post
– Anton Zagrivin
Nov 18 at 15:21
what happen if $L=C^1[a,b]$ ?
– Tsemo Aristide
Nov 18 at 15:34
Oh, I meant $L subset C[0, 1]$-closed subspace and all functions from $L$ are continuously differentiable
– Anton Zagrivin
Nov 18 at 15:40
add a comment |
I need to show that if $L subset C[0, 1]$ is a closed subspace in the $C[0,1]$ norm (i.e. $||f||_{L}=sup|f(x)|$), and all elements of $L$ are continuously differentiable, then L has finite dimension.
I can show:
$ exists n in mathbb{N}$ such that $forall f in L, lVert f rVert=1
> Rightarrow lVert f'rVert le n$
But I need to conclude from this that the mapping $psi :L rightarrow mathbb{R}^{4n+1}$ is injective, where $psi: f mapsto(f(0), f(1/4n), f(2/4n), ... , f(1))$
And I have no idea how to prove the injectivity of $psi$ and I even think that it may not be injective in general. Could you give me a hint how to prove injectivity?
functional-analysis
edited Nov 18 at 16:47
Taroccoesbrocco
5,05761839
asked Nov 18 at 15:20
Anton Zagrivin
1598
I need to show that if $L subset C[0, 1]$ is a closed subspace in the $C[0,1]$ norm (i.e. $||f||_{L}=sup|f(x)|$), and all elements of $L$ are continuously differentiable, then L has finite dimension.
I can show:
$ exists n in mathbb{N}$ such that $forall f in L, lVert f rVert=1
> Rightarrow lVert f'rVert le n$
But I need to conclude from this that the mapping $psi :L rightarrow mathbb{R}^{4n+1}$ is injective, where $psi: f mapsto(f(0), f(1/4n), f(2/4n), ... , f(1))$
And I have no idea how to prove the injectivity of $psi$ and I even think that it may not be injective in general. Could you give me a hint how to prove injectivity?
functional-analysis
functional-analysis
edited Nov 18 at 16:47
Taroccoesbrocco
5,05761839
asked Nov 18 at 15:20
Anton Zagrivin
1598
edited Nov 18 at 16:47
Taroccoesbrocco
5,05761839
asked Nov 18 at 15:20
Anton Zagrivin
1598
edited Nov 18 at 16:47
Taroccoesbrocco
5,05761839
edited Nov 18 at 16:47
Taroccoesbrocco
5,05761839
edited Nov 18 at 16:47
Taroccoesbrocco
5,05761839
5,05761839
asked Nov 18 at 15:20
Anton Zagrivin
1598
asked Nov 18 at 15:20
Anton Zagrivin
1598
asked Nov 18 at 15:20
Anton Zagrivin
1598
1598
I know that I can proove that $L$ is finite dimensional easier, but I need to prove exactly that $dimL le 4n+1$ where $n$ is from my post
– Anton Zagrivin
Nov 18 at 15:21
what happen if $L=C^1[a,b]$ ?
– Tsemo Aristide
Nov 18 at 15:34
Oh, I meant $L subset C[0, 1]$-closed subspace and all functions from $L$ are continuously differentiable
– Anton Zagrivin
Nov 18 at 15:40
add a comment |
I know that I can proove that $L$ is finite dimensional easier, but I need to prove exactly that $dimL le 4n+1$ where $n$ is from my post
– Anton Zagrivin
Nov 18 at 15:21
what happen if $L=C^1[a,b]$ ?
– Tsemo Aristide
Nov 18 at 15:34
Oh, I meant $L subset C[0, 1]$-closed subspace and all functions from $L$ are continuously differentiable
– Anton Zagrivin
Nov 18 at 15:40
I know that I can proove that $L$ is finite dimensional easier, but I need to prove exactly that $dimL le 4n+1$ where $n$ is from my post
– Anton Zagrivin
Nov 18 at 15:21
I know that I can proove that $L$ is finite dimensional easier, but I need to prove exactly that $dimL le 4n+1$ where $n$ is from my post
– Anton Zagrivin
Nov 18 at 15:21
what happen if $L=C^1[a,b]$ ?
– Tsemo Aristide
Nov 18 at 15:34
what happen if $L=C^1[a,b]$ ?
– Tsemo Aristide
Nov 18 at 15:34
Oh, I meant $L subset C[0, 1]$-closed subspace and all functions from $L$ are continuously differentiable
– Anton Zagrivin
Nov 18 at 15:40
Oh, I meant $L subset C[0, 1]$-closed subspace and all functions from $L$ are continuously differentiable
– Anton Zagrivin
Nov 18 at 15:40
add a comment |
1 Answer
1
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Let $fin L$ satisfy $psi(f) = (0,ldots,0)$, i.e., $f(k/4n)=0$ for $0leq k leq 4n$.
Suppose for contradiction that $fneq 0$. By compactness we may find $x_0$ such that $|f(x_0)| = |f|$ for some $x_0in (0,1)$. Since $L$ is a subspace we may rescale so that $|f(x_0)|=1$.
Now you can use the mean value theorem applied to the subinterval containing $x_0$, to find $x_k$ such that $|f'(x_k)| > frac{1}{1/4n}$ and hence $|f'| > 4n$, a contradiction to the fact about $L$ which you were able to prove.
answered Nov 18 at 16:13
user25959
1,573816
Thanks a lot, it really worked!
– Anton Zagrivin
Nov 18 at 17:26
add a comment |
Your Answer
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
Let $fin L$ satisfy $psi(f) = (0,ldots,0)$, i.e., $f(k/4n)=0$ for $0leq k leq 4n$.
Suppose for contradiction that $fneq 0$. By compactness we may find $x_0$ such that $|f(x_0)| = |f|$ for some $x_0in (0,1)$. Since $L$ is a subspace we may rescale so that $|f(x_0)|=1$.
Now you can use the mean value theorem applied to the subinterval containing $x_0$, to find $x_k$ such that $|f'(x_k)| > frac{1}{1/4n}$ and hence $|f'| > 4n$, a contradiction to the fact about $L$ which you were able to prove.
answered Nov 18 at 16:13
user25959
1,573816
Thanks a lot, it really worked!
– Anton Zagrivin
Nov 18 at 17:26
add a comment |
Let $fin L$ satisfy $psi(f) = (0,ldots,0)$, i.e., $f(k/4n)=0$ for $0leq k leq 4n$.
Suppose for contradiction that $fneq 0$. By compactness we may find $x_0$ such that $|f(x_0)| = |f|$ for some $x_0in (0,1)$. Since $L$ is a subspace we may rescale so that $|f(x_0)|=1$.
Now you can use the mean value theorem applied to the subinterval containing $x_0$, to find $x_k$ such that $|f'(x_k)| > frac{1}{1/4n}$ and hence $|f'| > 4n$, a contradiction to the fact about $L$ which you were able to prove.
answered Nov 18 at 16:13
user25959
1,573816
Thanks a lot, it really worked!
– Anton Zagrivin
Nov 18 at 17:26
add a comment |
Let $fin L$ satisfy $psi(f) = (0,ldots,0)$, i.e., $f(k/4n)=0$ for $0leq k leq 4n$.
Suppose for contradiction that $fneq 0$. By compactness we may find $x_0$ such that $|f(x_0)| = |f|$ for some $x_0in (0,1)$. Since $L$ is a subspace we may rescale so that $|f(x_0)|=1$.
Now you can use the mean value theorem applied to the subinterval containing $x_0$, to find $x_k$ such that $|f'(x_k)| > frac{1}{1/4n}$ and hence $|f'| > 4n$, a contradiction to the fact about $L$ which you were able to prove.
answered Nov 18 at 16:13
user25959
1,573816
Let $fin L$ satisfy $psi(f) = (0,ldots,0)$, i.e., $f(k/4n)=0$ for $0leq k leq 4n$.
Suppose for contradiction that $fneq 0$. By compactness we may find $x_0$ such that $|f(x_0)| = |f|$ for some $x_0in (0,1)$. Since $L$ is a subspace we may rescale so that $|f(x_0)|=1$.
Now you can use the mean value theorem applied to the subinterval containing $x_0$, to find $x_k$ such that $|f'(x_k)| > frac{1}{1/4n}$ and hence $|f'| > 4n$, a contradiction to the fact about $L$ which you were able to prove.
answered Nov 18 at 16:13
user25959
1,573816
edited Nov 18 at 16:54
answered Nov 18 at 16:13
user25959
1,573816
answered Nov 18 at 16:13
user25959
1,573816
answered Nov 18 at 16:13
user25959
1,573816
1,573816
Thanks a lot, it really worked!
– Anton Zagrivin
Nov 18 at 17:26
add a comment |
Thanks a lot, it really worked!
– Anton Zagrivin
Nov 18 at 17:26
Thanks a lot, it really worked!
– Anton Zagrivin
Nov 18 at 17:26
Thanks a lot, it really worked!
– Anton Zagrivin
Nov 18 at 17:26
add a comment |
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I know that I can proove that $L$ is finite dimensional easier, but I need to prove exactly that $dimL le 4n+1$ where $n$ is from my post
– Anton Zagrivin
Nov 18 at 15:21
what happen if $L=C^1[a,b]$ ?
– Tsemo Aristide
Nov 18 at 15:34
Oh, I meant $L subset C[0, 1]$-closed subspace and all functions from $L$ are continuously differentiable
– Anton Zagrivin
Nov 18 at 15:40