Show that C^α([0, 1]) is of first category in C([0, 1]).

Question

Recall that for any $0 < α < 1,$ the space $C^\alpha ([0, 1])$ is the set of continuous functions on $[0, 1]$ with norm of f = sup |f| + $ sup \ x\ne y$ |f(x) − f(y)|/|x − y|^α< ∞, equipped with the norm of f. a. Show that the unit ball of C^α([0, 1]) has compact closure in C([0, 1]). b. Show that C^α([0, 1]) is of first category in C([0, 1]).

I know by applying Arzela Ascoli theorem, P art a) can be done easily. but I got stuck in part b) how can we show this. I am sure we need Baire Category theorem.

Answer 1

Yes, you are right about Baire. Sketch: Let $E_n = \{f\in C^\alpha : \|f\|_\alpha \le n\}.$ Show each $E_n$ is closed in $C.$ We'll be done if we show the interior of each $E_n$ is empty. So let $f\in E_n.$ It suffices to show $f$ is the limit in $C$ of functions not in $C^\alpha.$ Consider $f +x^{\alpha/2}/m, m=1,2,\dots.$