Adding a "point at infinity" to $\mathcal{C}^{\alpha}$

Question

In this paper, on page 5, it says

We denote by $\bar{\mathcal{C}^{\alpha}}$ the space $\mathcal{C}^{\alpha}$ to which we add a “point at infinity” $\infty$ with neighbourhoods of the form $\{h : \|h\|_{\alpha} > R\} ∪ {∞}$, which turns $\bar{\mathcal{C}^{\alpha}}$ into a Polish space.

The $\alpha$ semi-norm is defined as $$\|f\|_{\alpha}:=\sup_{x\neq y} \frac{|f(x)-f(y)|}{|x-y|},$$ for $x,y$ in the domain (which here is $S^1$).

The $\mathcal{C}^{\alpha}$ norm is defined to be $$\|f\|_{\mathcal{C}^{\alpha}}=\|f\|_{\infty}+\|f\|_{\alpha}.$$

$\bar{\mathcal{C}^{\alpha}}$ is a space of functions, so what does it mean to add a point at infinity? Is it a function whose ${\alpha}$ norm is infinite? Also, what metric is $\bar{\mathcal{C}^{\alpha}}$ endowed with?

Answer 1

I presume that you mean $\alpha$-Hölder with the norm: $$ \|f\|_\alpha := \sup_x |f(x)| + \sup_{x\neq y} \frac{|f(x)-f(y)|}{|x-y|^\alpha}.$$

You may define the following metric on $\bar{C^\alpha}=C^\alpha\cup \{\infty\}$. When both $f$ and $g$ are finite:

$$ d(f,g) = \left\| \frac{f}{1+\|f\|^2} - \frac{g}{1+\|g\|^2} \right\| + \left| \frac{\|f\|}{1+\|f\|} - \frac{\|g\|}{1+\|g\|}\right|. $$

While if $f=\infty$ and $g$ is finite:

$$ d(\infty,g) = \frac{\|g\|}{1+\|g\|^2} +1 - \frac{\|g\|}{1+\|g\|} .$$ In the expression for $d(f,g)$ the first term distinguishes directions (but not the norms) while the second term distinguishes the norms. You have that $d(\infty,g)\rightarrow 0$ as $\|g\|\rightarrow +\infty$. The space is complete, and the topology is the same as the norm topology on the $C^\alpha$ part.

For separability, it is false without interpretation: $C^\alpha(S^1)$, $0< \alpha\leq 1$ is not separable. For e.g. $\alpha=1/2$ you may take the uncountable family, $a\in S^1$: $$ f_a (x) = \sqrt { \strut\;\inf_{k\in {\Bbb Z}} \;\left|x+a-k\right|} .$$ Then $\|f_a-f_b\|_\alpha \geq 1$ whenever $a\neq b$.

Edit: I asked the author of the above article who said that indeed, it wasn't true for the usual Hölder space functions. In the article, $C^\alpha(S^1)$ should be understood as the completion of the smooth functions under the Hölder norm. In that case it is separable (it avoids the above type of functions) and you may take e.g. a countable dense set in e.g. $C^1(S^1)$ as your dense set in the $\alpha$-completion.

Answer 2

I would regard this as a technical trick: The construction gives you a topological space, where unbounded sequences converge to the 'point' infinity. So this trick gives you a unified notion for convergence in norm $f_n\to f$ or divergence to infinity $\|f_n\|\to\infty$.

It also enables you to consider functions with values in such a space taking the value infinity, which is technically nice for the paper at hand that includes the case of blow-up in finite time.