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PostPosted: Mon Feb 01, 2016 3:00 pm 
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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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PostPosted: Sun Apr 23, 2017 3:05 am 

Joined: Tue Nov 24, 2015 7:47 pm
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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 M||f|| $ 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} $


$|x-y| < \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 arzela-ascoli $ B_{X} $ is compact so X has finite dimension .


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PostPosted: Mon Apr 24, 2017 4:23 pm 
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