Prove that a certain closed subspace of $C[0,1]$ has finite dimension












1














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







share|cite|improve this question
























  • 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


















1














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







share|cite|improve this question
























  • 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
















1












1








1







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







share|cite|improve this question















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






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share|cite|improve this question













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edited Nov 18 at 16:47









Taroccoesbrocco

5,05761839




5,05761839










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




















  • 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












1 Answer
1






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3














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.






share|cite|improve this answer























  • Thanks a lot, it really worked!
    – Anton Zagrivin
    Nov 18 at 17:26











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3














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.






share|cite|improve this answer























  • Thanks a lot, it really worked!
    – Anton Zagrivin
    Nov 18 at 17:26
















3














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.






share|cite|improve this answer























  • Thanks a lot, it really worked!
    – Anton Zagrivin
    Nov 18 at 17:26














3












3








3






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.






share|cite|improve this answer














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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 18 at 16:54

























answered Nov 18 at 16:13









user25959

1,573816




1,573816












  • 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




Thanks a lot, it really worked!
– Anton Zagrivin
Nov 18 at 17:26


















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