# A $\| \cdot \|_2$-closed subspace of $C[0,1]$ is always Banach.

I was doing a problem and I realize that if I prove that given a Y $\|\cdot \|_2$-closed vectorial subspace of $\mathcal{C}[0,1]$ is Banach, I'm done.

Well, what I've been trying is establishing a relation between the $\|\cdot \|_2$ and $\|\cdot \|_\infty$ (that makes complete the $\mathcal{C}[0,1]$ space). I defined the $\|\cdot \|_* = \|\cdot \|_\infty + \|\cdot \|_2$. Then, it is not difficult to show that $(Y, \|\cdot \|_*)$ is Banach.

Let me show you how. Let $(f_n)_{n\in \mathbb{N}}$ a $\|\cdot \|_*$ -Cauchy sequence. Since then, it is also a $\|\cdot \|_\infty$-Cauchy and there exists $g \in \mathcal{C}[0,1]$ that $f_n \to g$. Using the fact that $Y$ is a $\|\cdot \|_2$-closed subspace, follows that $g \in Y$.

After that, my main idea was trying to use the open mapping theorem, but it is here where I'm stuck.

• $\|\cdot\|_2$-closed subspace of $C[0, 1]$ is Banach for which norm? – user66081 Feb 12 '15 at 0:18
• It is closed respect to the norm 2 – user195451 Feb 12 '15 at 0:32

This is not true. Let $Y=C([0,1])$ itself. Define $$f_n(x):=\begin{cases} 0&\text{if}\ 0\le x\le\tfrac{n-2}{2n},\\ nx-\tfrac{n-2}2&\text{if}\ \tfrac{n-2}{2n}<x\le\frac12,\\ 1&\text{if}\ \tfrac12<x\le1. \end{cases}$$ Define $f=\mathbf1_{[\frac12,1]}$. Clearly $f_n\to f$ pointwise, and $|f_n-f|^2\le2\in L^1(0,1)$, so by the dominated convergence theorem $f_n\to f$ in $L^2$. Since $f_n\in Y$ for each $n$ this implies $(f_n)$ is an $L^2$-Cauchy sequence in $Y$. If $f_n\stackrel{L^2}\to g\in Y$ then $g=f$ a.e. which can easily be seen to be impossible. Hence $Y$ is not Banach.
However, it is true that $(Y,\|\cdot\|_*)$ is Banach. If $(f_n)$ is $\|\cdot\|_*$-Cauchy in $Y$ then it is also $\|\cdot\|_\infty$-Cauchy, so there exists $f\in Y$ such that $f_n\stackrel{\|\cdot\|_\infty}\to f$. This implies $$\int_0^1|f_n-f|^2\ \mathrm{d}x\leq\|f_n-f\|^2_\infty\int_0^11\ \mathrm{d}x\to0$$ as $n\to\infty$. Hence $\|f_n-f\|_*=\|f_n-f\|_\infty+\|f_n-f\|_2\to0$, so $(Y,\|\cdot\|_*)$ is complete.
• so this $Y$ is not an $\|\cdot\|_2$-closed subspace of $C([0, 1])$ – user66081 Feb 12 '15 at 8:52
• @user66081: it seems that something is confusing you. $(Y,\lVert\cdot\rVert_2) = (C[0,1],\lVert\cdot\rVert_2)$. And $Y$ is certainly $\lVert\cdot\rVert_2$-closed in $C[0,1]$ because the complement is the empty set which is open regardless of whatever metric we're working with. – kahen Feb 12 '15 at 9:35
• @kahen: thanks; yes. So Jason's argument is in essence: $C^0 \subsetneq L^2$ is dense. – user66081 Feb 12 '15 at 10:15