It's a theorem that if $f\colon U\subset\Bbb R^n\to \Bbb R^m$ has the property that each of the partial derivatives $\partial_if_j$ exist and are continuous $p\in U$, then $f$ is differentiable at $p$. When I was trying to prove this, I came up with the following "proof" which doesn't use the continuity hypothesis. Can someone tell me what's wrong with this proof?
Since $f_j$ is differentiable at $p$, we can write
$$
f_j(p+v) = f_j(p) + \sum_i \partial_if_j(p)v_i + R_j(v),
$$
where $|R_j(v)|/|v| \to 0$ as $v\to 0$. Hence, we can write
\begin{align*}
f(p+v) &= f(p) + \big(\sum_i \partial_if_1(p)v_i + R_1(v),\dots,\sum_i \partial_if_m(p)v_i + R_m(v)\big) \\
&= f(p) + \sum_j\big(\sum_i\partial_if_j(p)v_i\big)e_j + R_j(v)e_j \\
&= f(p) + [Df_p][v] + (R_1,\dots,R_m)(v),
\end{align*}
where $[Df_p] = [\partial_if_j(p)]$ is the usual Jacobian matrix, and $[v]$ is the column vector $[v_1\ \dotsb\ v_n]^T$. Now,
$$
\frac{|(R_1,\dots,R_m)(v)|^2}{|v|^2} = \frac{R_1(v)^2 + \dots + R_m(v)^2}{|v|^2} \to 0,
$$
where the last expression goes to $0$ as $v\to 0$ since it is a sum of finitely many terms, each of which goes to $0$. Hence we have written $f(p+v)$ as a sum of a constant term, a linear part, and a sublinear piece, so $f$ is differentiable at $p$. At no point did I explicitly use the continuity hypothesis, so what exactly is wrong with this proof? Best.
Best Answer
The very first step is wrong. You are only given that partial derivatives exist and are continuous, not that $f_j$ is a differentiable function on $\mathbb R^{n}$.