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## Closed subspace of finite dimension

Functional Analysis
Papapetros Vaggelis
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### 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.
dr.tasos
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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 $\exists M>0 \quad : \quad ||Tf|| \leq M||f||$ 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 .
Papapetros Vaggelis
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