Let $\alpha\in(0,1)$ and define the $\alpha$-Hölder norm of a function $f : [0,1]\to\mathbb{R}$ by
$$
\lVert f\rVert_\alpha:=|f(0)| + \sup_{0\le s<t\le 1}\frac{|f(t)-f(s)|}{|t-s|^\alpha}.
$$
Let $f_n : [0,1]\to\mathbb{R}$ be a sequence of $\alpha$-Hölder functions satisfying $\sup_n\lVert f_n\rVert_\alpha<\infty$, and suppose $f_n\to f$ uniformly. It follows immediately that $\lVert f\rVert_\alpha<\infty$. Does it also follow that $\lim_{n\to\infty}\lVert f_n-f\rVert_\alpha =0$? My intuition says no, but I am struggling to find a counterexample.
If in fact the convergence in Hölder norm follows, I am curious to hear whether it also follows when the $f_n$ take values in an arbitrary Banach space.
Best Answer
Let $f_n(x)={\sin( {\pi\over 2} n x)\over n^{\alpha}}.$ Then $f_n$ is $\alpha$-Hölder on the real line and the $\alpha$-Hölder norms there are all equal (by substitution $x=ns,\ y=nt$, see also below). In particular $\|f_n\|_\alpha $ are bounded. Furthermore $$\|f_n\|_{\alpha} \ge \sup_{0\le s<t\le {1\over n}}\frac{f_n(t)-f_n(s)}{(t-s)^\alpha}\\ \underset{x=ns,y=nt}{\ = \ }\ \sup_{0\le x<y\le 1}\frac{f_1(y)-f_1(x)}{(y-x)^\alpha} \\ \ge {f_1(1)-f_1(0)\over (1-0)^\alpha}=1$$ Thus $\|f_n\|_{\alpha}\ge 1$ and $f_n\rightrightarrows 0.$