Do Holder continuous functions preserve rate of uniform convergence sequence?

Question

Suppose $g$ is $\alpha$-Hölder continuous and $f_n$ converges uniformly to $f$ at rate $a_n$, so that

$a_n \sup_{x}|f_n(x) - f(x)| \to 0 $ as $n \to \infty$

Then does $g \circ f_n$ converges uniformly to $g\circ f$ at rate $a_n$?

Answer 1

The claim is false in general. To see it, take $f_n(x) = x + \frac 1n$, $f(x) = x$, and set $a_n = n^{1/2}$ (for instance), let also $g(x)= x^{1/2}$, where $x\geq 0$.

We have that $g$ is Holder-$\frac 12$, but the claim in your question fails for these data.

Indeed, $f_n$ converges to $f$ uniformly on $\mathbb{R}$, in particular $$ n^{1/2} \sup_{x \in [0,1]} |f_n(x) - f(x)| \to 0. $$ On the other hand, $$ n^{1/2} \sup_{x\in [0,1]} |g(f_n(x)) - g(f(x))| = n^{1/2} \sup_{x\in [0,1]} \left| \frac{ f_n(x) - f(x) }{\sqrt{f_n(x)} + \sqrt{f(x)}} \right| = \\ n^{1/2} \frac 1n \sup_{x\in [0,1]} \left| \frac{ 1 }{\sqrt{x+1/n} + \sqrt{x}} \right| \geq (\text{ take } x=0) \\ n^{-1/2} \frac{1}{\sqrt{1/n}} = 1, $$ so the convergence fails.

Answer 2

No.

We need $a_n \sup_x|f_n(x) - f(x)|^\alpha \to 0$

From the Hölder continuity of $g$, there are non-negative constant $C$, $\alpha$ such that

$ a_n|g(x) - g(y)| \le a_n C |x - y |^{\alpha} $

For $\alpha > 1$, the function $g$ is constant and the convergence is automatic. For $\alpha < 1$, we have, from the uniform convergence of $f_n$ at rate $a_n$, that for any $\epsilon > 0$, $\alpha \le 1$, $x$, there exists a $N > N_*$ such that for $n > N$

$a_nC|f_n(x) - f(x)|^\alpha < \epsilon $

Which implies that for any $\epsilon > 0$ and $x$, we have a $n$ so that

$ a_n|g(f_n(x)) - g(f(x))| \le a_n C |f_n(x) - f(x) |^{\alpha} < \epsilon$

which proves the result.