Is $C[0,1]$ equipped with $\lVert \cdot \rVert_1$ a countable union of nowhere dense sets?

Question

Let's consider the space of all continuous function $C[0,1]$ on the intervall $[0,1]$. But instead of using the usual supremum norm we use the $L^1$-Norm $: \lVert f \rVert_1=\int_0^1 \lvert f(x) \rvert dx$ for $f \in C[0,1]$

It is well known that this space is not complete with respect to the $L^1$-Norm.

The Baire Category Theorem states that if a metric space is Banach then it cannot be a countable union of nowhere dense sets.

What can we say about the converse in this case? Is $(C[0,1],\lVert \cdot \rVert_1) $ a countable union of nowhere dense sets?

Answer 1

Yes,

$$C([0,1]) = \bigcup_{n = 1}^\infty \underbrace{\{ f \in C([0,1]) : \lVert f\rVert_\infty \leqslant n\}}_{A_n},$$

and $A_n$ is closed for each $n$ - if $\lVert g\rVert_\infty > n$, then there is a $\delta > 0$ and a non-degenerate interval $[a,b] \subset [0,1]$ such that $\lvert g(x)\rvert \geqslant n+\delta$ for all $x\in [a,b]$, and hence $\lVert g-f\rVert_1 \geqslant \delta\cdot (b-a)$ for all $f\in A_n$, and $A_n$ has empty interior - that follows for example by the open mapping theorem, if it had non-empty interior the two norms would be equivalent. But one can also argue elementarily that each $A_n$ has empty interior.