Let \(\displaystyle{X}\) be a closed subspace of \(\displaystyle{\left(C(\left[0,1\right],\cdot_{\infty}\right)}\)
such that \(\displaystyle{X\subseteq C^1(\left[0,1\right])}\) .
Prove that the subspace \(\displaystyle{X}\) has finite dimension.
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Closed subspace of finite dimension

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Re: Closed subspace of finite dimension
It is enough to prove that the $B_{X} $ ( unit ball) is compact . From arzela ascoli we know that if we have a family $ F $ of uniformly bounded and equicontinuous functions ( in $C[0,1]$) then every sequence $ \left \{ f_{n} \right \}^{\infty}_{n=1} \subseteq F $ has a convergent subsequence .
Given the fact that we are in a subset of continuously differentiable functions it would convenient if the derivatives of all functions in $ B_{X} $ had uniformly bounded derivatives .
So i consider the operator
$ T \quad : \quad X \rightarrow C[0,1] $
$Tf=f'$
$ X $ and $ C[0,1] $ are banach spaces ( $ X $ is closed subset of a banach space)
It is a known lemma that if $ f_{n} \in C^{1}[a,b] $ converges uniformly to $ f $ and $ f'_{n}$ converges uniformly in $ [a,b]$ then $ f'_{n} \rightarrow f'$ in $[a,b] $
If $ f_{n} \rightarrow f $ and $ T f_{n} \rightarrow g $
Using the lemma above $ g=f'$ hence $Graph(T) $ is closed .
So using closed graph theorem $ T $ is bounded hence $ \exists M>0 \quad : \quad Tf \leq Mf $ for all $ f \in X $
Hence $ f' \leq M f $
So $ \forall f \in B_{X} $ it is $ f' \leq M $
Hence given $ \epsilon >0 $ we choose $ \delta = \frac{\epsilon}{2M} $
$xy < \delta \Rightarrow f(x)f(y) \leq f' \delta < \epsilon $ $ \forall f \in B_{X} $ so $ B_{X} $ is equicontinuous and uniformly bounded so by arzelaascoli $ B_{X} $ is compact so X has finite dimension .
Given the fact that we are in a subset of continuously differentiable functions it would convenient if the derivatives of all functions in $ B_{X} $ had uniformly bounded derivatives .
So i consider the operator
$ T \quad : \quad X \rightarrow C[0,1] $
$Tf=f'$
$ X $ and $ C[0,1] $ are banach spaces ( $ X $ is closed subset of a banach space)
It is a known lemma that if $ f_{n} \in C^{1}[a,b] $ converges uniformly to $ f $ and $ f'_{n}$ converges uniformly in $ [a,b]$ then $ f'_{n} \rightarrow f'$ in $[a,b] $
If $ f_{n} \rightarrow f $ and $ T f_{n} \rightarrow g $
Using the lemma above $ g=f'$ hence $Graph(T) $ is closed .
So using closed graph theorem $ T $ is bounded hence $ \exists M>0 \quad : \quad Tf \leq Mf $ for all $ f \in X $
Hence $ f' \leq M f $
So $ \forall f \in B_{X} $ it is $ f' \leq M $
Hence given $ \epsilon >0 $ we choose $ \delta = \frac{\epsilon}{2M} $
$xy < \delta \Rightarrow f(x)f(y) \leq f' \delta < \epsilon $ $ \forall f \in B_{X} $ so $ B_{X} $ is equicontinuous and uniformly bounded so by arzelaascoli $ B_{X} $ is compact so X has finite dimension .

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