Dimension of the space of derivations for finitely differentiable manifolds The $C^\infty$-case: If $M$ is a (paracompact Hausdorff) real $C^\infty$-manifold and $p \in M$ then the tangent space $T_pM$ can be defined in several different but equivalent ways. One possibility is to define a tangent vector via its action on smooth functions $M \to \mathbb{R}$ and set $T_pM := \mathrm{Der}_p(C_p^\infty(M))$. Here $C_p^\infty(M)$ is the algebra of germs at $p$ of smooth real valued functions, and $\mathrm{Der}_p(C^\infty(M))$ is the space of derivations of $C_p^\infty(M)$ relative to evaluation at $p$, i.e. the space linear maps $\varphi: C_p^\infty(M) \to \mathbb{R}$ satisfying $\varphi(fg) = \varphi(f)g(p) + f(p)\varphi(g)$ for $f,g \in C_p^\infty(M)$. Equivalently one can also use $\mathrm{Der}_p(C^\infty(U))$ for an arbitrary open neighbourhood $U$ of $p$. To proof the finite-dimensionality of $T_pM$ we can assume without loss of generality that $M=\mathbb{R}^n$, $p=0$ and that $U$ is a convex set containing $0$. One can then write down a basis for $T_pM$, namely the derivations $\frac{\partial}{\partial x^i}|_0$ where $i \in \{1,\ldots,n\}$. To see that this is a spanning set one first writes any smooth function $g$ on $U$ as $g = g(0) + \sum_i g_ix^i$, where $g_i(x) = \int_{0}^1\frac{\partial g}{\partial x^i}(tx)dt$ (using the fundamantal theorem of calculus) and then applies a derivation $\varphi$ to obtain $\varphi(g) = \sum_i \varphi(x^i)\frac{\partial}{\partial x^i}|_0(g)$. For linear independence one evaluates at the coordinate functions $x^i$.
The $C^r$-case: In contrast, for a $C^r$-manifold where $r$ is a positive integer the situations looks different. Namely here $\mathrm{Der}_p(C^r(U))$ is infinite dimensional. A proof outline of this can for example be found in J.M.Lee's book "Manifolds and differential geometry", Chapter 2, Problem (18) on page 124. First it is established that as vector spaces $\mathrm{Der}_p(C^r(U)) \cong (\mathfrak{m}_r/\mathfrak{m}_r²)^*$, where $\mathfrak{m}_r$ is the maximal ideal of $\mathrm{Der}_p(C^r(U))$ given by functions $f$ that vanish at $p$, $f(p)=0$. In the case $M=\mathbb{R}$ and $p=0$ one then defines functions $g^r_\varepsilon(x) := \begin{cases}x^{r+\varepsilon} & x \geq 0 \\ 0 & x \leq 0\end{cases}$ for $\varepsilon \in (0,1)$ and shows that their equivalence classes are all linearly independent in $\mathfrak{m}_r/\mathfrak{m}_r²$. The general case follows from this as well.
Question: Going through the proof of finite-dimensionality in the $C^\infty$-case I can't see where we need the assumption of smoothness, it all seems to work verbatim for the $C^r$-case. That we can choose an arbitrary open neighbourhood $U$ for the proof follows from the existence of cutoff functions. These are smooth and in particular $C^r$, so this also works in the latter case. In the argument that the $\frac{\partial}{\partial x^i}|_0$ form a basis we only seem to assume $C^1$ really. This of course stands in contradiction to the fact that $\mathrm{Der}_p(C^r(U))$ is infinite-dimensional. So where does the proof go wrong?
 A: Certainly the powerful formula
$$g(x) = g(0) + \sum x^ig_i(x)\tag{$\star$}$$
only requires $g$ to be $C^1$, in which case the $g_i$ are just continuous. If $g$ is $C^k$, then the $g_i$ are just $C^{k-1}$. However, if $g$ is $C^\infty$, then the $g_i$ will also be $C^\infty$, and so we can use this formula to see that for any derivation $\phi$ (at $p=0$) we have
$$\phi(g) = \sum\phi\big(x^i g_i(x)\big) = \sum \big(\phi(x^i)g_i(0) + x^i(0)\phi(g_i)\big) = \sum \phi(x^i)g_i(0).$$
From this we conclude that $\phi(x^i)$ uniquely determine $\phi$.
Note that if $\phi$ is a derivation on $C^k$ functions, we cannot play this game, as $g_i$ will not be $C^k$.
Moreover, in the smooth case the formula ($\star$) can be continued to the next term and we see that
$$g(x)=g(0)+\sum_i x^i\big(g_i(0) + \sum_j x^jg_{ij}(x)\big).$$
If we let $\mathfrak m$ be the set of germs of smooth functions vanishing at $0$, then we see from this that $\{x^i+\mathfrak m^2: i=1,\dots,n\}$ gives a basis for $\mathfrak m/\mathfrak m^2$ and we conclude that the derivations (in the smooth case), namely, $(\mathfrak m/\mathfrak m^2)^*$,  will of course be $n$-dimensional.
In the $C^1$ case, all the functions $g^1_\epsilon$ with $0<\epsilon<1$ will yield linearly independent elements of $\mathfrak m_1/\mathfrak m_1^2$. What goes wrong? Well, of course the formula ($\star$) still works but you find that $g_1(0)$ doesn't distinguish all these elements, hence doesn't distinguish the derivations. You definitely do not get a spanning set.
