Let $M$ be a smooth manifold. Show that $f\in l^2_p$ if and only if in any smooth local coordinates, its first-order Taylor polynomial at p is zero.

Question

Let $M$ be a smooth manifold and p be a point of M. Let $l_P$ denote the subspace of $C^\infty(M)$ consisting of smooth functions that vanish at p, and let $l_p^2$ be the subspace of $l_p$ spanned by functions of the form $fg$ for some $f,g\in l_p$. Show that $f\in l^2_p$ if and only if in any smooth local coordinates, its first-order Taylor polynomial at p is zero.

Here's what I've done:

Let $f,g\in l_p$. The first degree Taylor polynomial of $fg$ at $p$ is given by:

$$P_1(q)=f(p)g(p)+\frac{\partial fg(p)}{\partial x^i}x^i(q-p)$$

As $f(p)=g(p)=0$ and $\displaystyle \frac{\partial fg(p)}{\partial x^i}=f(p)\frac{\partial g(p)}{\partial x^i}+g(p)\frac{\partial f(p)}{\partial x^i}$, we have that $$P_1\equiv 0.$$

But I cannot show that $f\in l_p$ if $f(p)+\frac{\partial f(p)}{\partial x^i}x^i(q-p)=0$. I'm even trying the contrapositve and nothing.

Answer 1

This is a consequence of the remainder formula of the Taylor polynomial of a map. Take any map $f$. Then, we can write $$ f(x)=f(p)+\sum_{i=1}^n \frac{\partial f}{\partial x^i}(x)(x^i-p^i)+R_1(x)=T_1(x)+R_1(x), $$ where $T_1(x)$ is the first order Taylor polynomial of $f$ and $R_1(x)$ is the remainder, which is given by: $$ R_1(x)=\sum_{i,j=1}^n(x^i-p^i)(x^j-p^j)\int_0^1(1-t)\frac{\partial f}{\partial x^i\partial x^j}(p+t(x-p))dt. $$ Note that $R_2(x)\in l_p^2$, therefore if $f$ is such that its first order Taylor polynomial vanishes the claim follows.

Answer 2

Working in $\mathbb R^n,$ suppose $f(0) = 0, \nabla f (0) =0$ for convenience. Let $x\in \mathbb R^n.$ For $t\in \mathbb R,$ set $g(t) = f(tx).$ Then

$$f(x) = f(x) - f(0) = g(1) - g(0) = \int_0^1 g'(t)\, dt = \int_0^1 (\nabla f (tx) \cdot x)\,dt = \sum_{k=1}^{n}x_k\int_0^1 D_kf(tx)\, dt.$$

Since each integral on the right is a smooth function of $x$ and vanishes at $0,$ we have $f\in l^2_0$ as desired.