Derivations on the spaces of continuous functions form an infinite dimensional vector space(generalized tangent space) This question basically asks why the notion of tangent space can't be well-generalized to topological manifolds without coming across the issue of dimension.
Let $X$ be a smooth manifold of finite dimension and $C(X)$ denote the space of continuous functions on $X$. Let $p\in X$.
A linear derivation $v$ w.r.t.$p$ is a linear functional on $C(X)$ satisfying $v(fg)=f(p)v(g)+g(p)v(f), \forall f,g\in C(X)$.
My question is, is the space of linear derivations an infinite dimensional space? I will be happy enough to see a proof for $X=\mathbb R, p=0$(or other special cases that could lead to infinite dimension), but general results will be great.
As we know, if $C(X)$ is replaced by $C^\infty(X)$, then this space is the ordinary tangent space, which is finite dimensional.
 A: A quick google search finds that the question had been answered in planetmath.org (The post there concerns with derivations $C(X)\to C(X)$ instead of $C(X) \to \mathbb R$, but the proof is similar). The result is:

Theorem Let $X$ be a topological space and let $C(X)$ be the space of continuous functions to $\mathbb R$. Let $x\in X$ and let $D$ be a derivation at $x$: that is, $D : C(X)\to \mathbb R$ is linear and
$$ D(fg) = f(x) Dg + g(x) Df, \ \ \ \forall f, g\in C(X).$$
Then $D(f) = 0$ for all $f\in C(X)$.

The proof goes as follows:

Lemma 1: For any constant function $c$, $D(c) = 0$. Proof: exercise...

From lemma 1, it suffices to check $D(f) = 0$ for all $f\in C(X)$ with $f(x) = 0$. Write $f = f_+ - f_-$, where $f_\pm$ are the positive and negative parts of $f$ respectively. Then
$$ D(f) = D(f_+) - D(f_-)$$
and it suffices to check that $D(g) = 0$ for all nonnegative functions $g\in C(X)$ with $g(x) = 0$.
Write $D(g) = D(\sqrt g \sqrt g)$ and note $\sqrt g \in C(X)$. Using that $g(x) = 0$ and the definition of a derivation at $x$,
$$ D(g) = \sqrt g(x) D(\sqrt g) + \sqrt g(x) D(\sqrt g) = 0$$
and that finishes the proof.
Remark When $f\in C^\infty (X)$, $f_\pm$ might not be smooth and thus the argument does not work for $C^\infty(X)$.
