Proof that in the Banach space $C^0[0,1]$ the set of functions in $C^1[0,1]$ is a meagre set

Question

How can I prove that in the Banach space $C^0[0,1]$ the set of functions in $C^1[0,1]$ is a meagre set?

How do I go about proving this statement? I can't find any direct union of nowhere dense set I could write it as, nor an indirect trick.

Answer 1

hint: You may have a look at the sets: $E_n = \{ \phi \in C^1([0,1]): \sup |\phi'|\leq n \}$, $n\geq 1$

Answer 2

Define $E_n$ to be the set of functions $f\in C[0,1]$ such that $\|f\|_\infty \le n,$ and such that $|f(y)-f(x)|\le n|y-x|$ for all $x,y\in [0,1].$ Then each $E_n$ is closed in $C[0,1].$ But each $f\in E_n$ can be uniformly approximated by functions not in $E_n,$ say by looking at $f(x)+\epsilon\sqrt x$ for small $\epsilon >0.$ Thus each $E_n$ is nowhere dense in $C[0,1].$ Hence $\cup E_n$ is meager in $C[0,1].$ Since $\cup E_n$ contains $C^1[0,1],$ the latter set is also meager.