Function of class $\mathcal{C}^{\alpha}$ I thought that functions of class $\mathcal{C}^{\alpha}$ were those whose first $\alpha$ derivatives exists and are continuous.
In this paper (page 7), however, Hairer defines $\mathcal{C}^{\alpha}$ as the class of functions $f$ such that, for every $x$ in the domain, it is possible to find a polynomial $P_x$ such that
$$|f(y) - P_x(y)| \lesssim |x-y|^{\alpha},$$
where $\lesssim$ means that there exists a constant such that the left hand side is less than or equal to the product of the right hand side and that constant.
Are these definitions equivalent?
 A: Have a look at page 23 where this is clarified at least to some extent.
In general $C^{k, \alpha}$ with $0<\alpha < 1 $ is the space of $k$ times continuously differentiable functions such that the $k-$th derivative is still Hölder continuous with exponent $\alpha$ (see, e.g., here: en.wikipedia.org if you want to find out what that means). These spaces are used in the regularity theory of (elliptic) partial differential equations and are better suited for this, since Laplace like operators have interesting continuity properties on these spaces.
Sometimes people just write $C^\alpha $ for this, in particular when $k=0$, but sometimes also for non-integer $\alpha$, and this is one of these cases. On page 23 of the paper you are referring to the author explains that his definition actually differs from the usual definition in case $\alpha$ is an integer (you get a Lipshitz condition for the derivatives of order $\alpha-1$, which is not enough to imply differentiability), and he actually gives a slightly different looking definition on that page. He also admits that his notation is an abuse of the notation in case of integer $\alpha$. I personally think he should have used a different notation and should have explained a bit more about this on page 7, but that's probably just me.
A: To make it clear, that $\alpha$ is a real number in $(0,1]$ and not a multiindex:
Consider $f(x) = x(\sin(1/x))^2$ for $x\ne 0$ and $f(0)=0$. Then, for $x\ne 0$ you can take the Taylor polynomial as $P_x$. For $x=0\ne y$ we have
$$ \left| \frac{f(y) - 0}{y} \right| \le 1 $$
but
$$ \left| \frac{f(y) - 0}{y} \right| \not\to 0. $$
That is, $f$ is locally approximately a linear polynomial, but not even differentiable (at 0). 
