Schwartz rule in differentiable manifolds.

Question

Let $M$ a differentiable manifold and $(U,\varphi)$ a chart with coordinate functions $(x_1,...,x_n)$. Let $p \in U$. Given $f:U\longrightarrow \mathbb{R}$, $f \in C^\infty(U)$, it is possible to consider $\frac{\partial^2 f}{\partial x_i \partial x_j}|_p$ and $\frac{\partial^2 f}{\partial x_j \partial x_i}|_p$. My goal is to prove that $$\frac{\partial^2 f}{\partial x_i \partial x_j}|_p=\frac{\partial^2 f}{\partial x_j \partial x_i}|_p$$

I know that $\frac{\partial f}{\partial x_i}|_p=\frac{\partial(f \circ \varphi^{-1})}{\partial r_i}|_{\varphi(p)}$ and $\frac{\partial f}{\partial x_j}|_p=\frac{\partial(f \circ \varphi^{-1})}{\partial r_j}|_{\varphi(p)}$, but I'm not sure about how to write correctly the proof. This is my attemp: $$\frac{\partial^2 f}{\partial x_i \partial x_j}|_p= \frac{\partial (\frac{\partial f}{\partial x_i})}{\partial x_j}|_p=\frac{\partial ( \frac{\partial f}{\partial x_i}\circ \varphi^{-1})}{\partial r_j}|_{\varphi(p)}=\frac{\partial ( \frac{\partial f \circ \varphi^{-1}}{\partial r_i}\circ \varphi^{-1})}{\partial r_j}|_{\varphi(p)}$$ In a similar way we have

$$\frac{\partial^2 f}{\partial x_j \partial x_i}|_p=\frac{\partial ( \frac{\partial f \circ \varphi^{-1}}{\partial r_j}\circ \varphi^{-1})}{\partial r_i}|_{\varphi(p)}$$

Answer 1

We have

\begin{align}\frac{\partial^2 f}{\partial x_i \partial x_j}\bigg|_p &= \frac{\partial}{\partial x_i}\left(\frac{\partial f}{\partial x_j}\right)\bigg|_p = \frac{\partial}{\partial r_i}\left(\frac{\partial f}{\partial x_j} \circ \phi^{-1}\right)\bigg|_{\phi(p)}\\ &= \frac{\partial}{\partial r_i}\left(\frac{\partial(f\circ \phi^{-1})}{\partial r_j}\circ \phi \circ \phi^{-1}\right)\bigg|_{\phi(p)}\\ &= \frac{\partial}{\partial r_i}\left(\frac{\partial (f\circ \phi^{-1})}{\partial r_j}\right)\bigg|_{\phi(p)}\\ \end{align}

and similarly $$\frac{\partial^2 f}{\partial x_j \partial x_i}\bigg|_p = \frac{\partial}{\partial r_j}\left(\frac{\partial (f\circ \phi^{-1})}{\partial r_i}\right)\bigg|_{\phi(p)}.$$ Since $$\frac{\partial}{\partial r_j}\left(\frac{\partial(f\circ \phi^{-1})}{\partial r_i}\right)\bigg|_{\phi(p)} = \frac{\partial}{\partial r_i}\left(\frac{\partial (f \circ \phi^{-1})}{\partial r_j}\right)\bigg|_{\phi(p)}$$

the result follows.