Closed subspace of finite dimension
-
- Community Team
- Posts: 426
- Joined: Mon Nov 09, 2015 1:52 pm
Closed subspace of finite dimension
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.
such that \(\displaystyle{X\subseteq C^1(\left[0,1\right])}\) .
Prove that the subspace \(\displaystyle{X}\) has finite dimension.
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 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 .
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 .
-
- Community Team
- Posts: 426
- Joined: Mon Nov 09, 2015 1:52 pm
Create an account or sign in to join the discussion
You need to be a member in order to post a reply
Create an account
Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute
Sign in
Who is online
Users browsing this forum: No registered users and 2 guests