Dimension for a closed subspace of $C[0,1]$.

Let $X \subset C^1[0,1]$ be a closed subspace of $C[0,1]$ (with sup norm).

Prove that $X$ has to be finite-dimensional.

Solution 1:

Since $V$ is a closed subspace of $C[0,1]$, then it is a Banach space. Let $A : V \to \mathrm{C}[0,1]$ be the operator $Au=u'$. In order to show that $A$ is bounded, it suffices, using Closed Graph Theorem, to show that:

If $\{u_n\}_{n\in\mathbb N}\subset V$ and $$ u_n\to u \,\,\,\&\,\,\, u_n'\to v, \,\,\, \text{both uniformly on $[0,1]$}, $$ then $u\in \mathrm{C}^1[0,1]$ and $u'=v$.

Since $V$ is closed subspace, then $u\in V$, hence $u$ is continuously differentiable and $$ u_n(x) = u_n(0)+\int_0^x u_n'(t)\,dt \,\to\, u(0)+\int_0^x v(t)\,dt=u(x), $$ as $n\to\infty$, due to the uniform convergence of the sequence $\{u_n^\prime\}_{n\in\mathbb N}$.

Meanwhile, $u(x) = u(0)+\int_0^x u'(t)\,dt$, and thus $v=u'$.

The boundedness of $A$ implies that there exists a $c>0$, such that $$ \|u'\|\le c\|u\|, \tag{$\star$} $$ for all $u\in V$, where $\|\cdot\|$ is the norm of $\mathrm{C}[0,1]$. Inequality ($\star$) implies that the closed unit ball of $V$: $$ B_1 \,=\, \{u\in V: \|u\|\le 1\}, $$ is equicontinuous, i.e., for every $u\in B_1$ and $x,y\in[0,1]$, $$ |u(x)-u(y)| \,=\, \Big|\int_x^y u'(t)\,dt\,\Big| \,\le\, c|x-y|, $$ and by virtue of Lemma Arzelà–Ascoli, $B_1$ is compact. Hence $V$ is locally compact. But is a Banach space (more generally, a topological vector space) is locally compact, then it has finite dimension.