This question is from my assignment on smooth manifolds and I need help in proving this result.
Let $M$ be a smooth manifold and $p\in M$. Let $f,g \in C^{\infty} (M)$ and $U \subset M$ be an open subset such that $p\in U$ and $f(x)=g(x)$ for all $x\in U$. Show that for any $X\in T_p(M)$, we have $X(f)=X(g)$.
So, I think that it is clear that how $X(f) =X(g)$ on $U$ ( as it is given that $f(x)=g(x)$ are equal on $U$ ) but how to approach this when a point say $x\in M \setminus U$. I am at loss of ideas.
Kindly shed some light on this.
If $T_pM$ denotes the tangent space of $M$ at point $p$ that is the set of all derivations $X: C^{\infty}(M) \rightarrow\mathbb{R}$ ($X$ is a linear map) satisfying the Leibniz rule at $p$, that is, $\forall f,g \in C^{\infty}(M)$ $$X(fg)=f(p)Xg+g(p)Xf,$$ then, to prove that $Xf=Xg$ if $f$ and $g$ agree, setting $h=f-g$, by linearity is suffices to show that $Xh=0$ whenever $h$ vanishes in a neighborhood $W$ of $p.$
Let $A$ be a closed subset $M \setminus W$. Then exists a smooth function $u \in C^{\infty}(M)$ equal to $1$ on $A$ and is supported in $M \setminus \{p\}$. Because $u=1$ where $h$ is nonzero, $hu=h$. Since $h(p) = f(p)-g(p) = 0 = u(p) $, by Leibniz rule $$Xh = X(hu) =h(p)Xu+u(p)Xh =0 $$