Proof of the derivative of a map into R^n by components

Proof

We compare the Euclidean norm of a vector with the absolute values of its entries.

1. Let $U$, $f$, $a$ and $L$ be as in the statement. The component $L_j$ is the map $h \mapsto (L h)_j$. It is linear, since $L$ is linear and taking the $j$-th entry is linear.
2. For $h$ with $a + h \in U$, put $r(h) = f(a + h) - f(a) - L h \in \mathbb{R}^m$. Its entries are $r_j(h) = f_j(a + h) - f_j(a) - L_j h$.
3. For each $v \in \mathbb{R}^m$ and each $j$, we have $|v_j| \le \|v\|$, since $v_j^2 \le v_1^2 + \dots + v_m^2$.
4. For each $v \in \mathbb{R}^m$, we have $\|v\| \le |v_1| + \dots + |v_m|$. Indeed, the square of the right side is $\sum_j v_j^2$ plus nonnegative cross terms.
5. Suppose that $f$ is differentiable at $a$ with $Df(a) = L$. So $\|r(h)\| / \|h\| \to 0$ as $h \to 0$.
6. By step 3, $0 \le |r_j(h)| / \|h\| \le \|r(h)\| / \|h\|$ for each $h \ne 0$. So $|r_j(h)| / \|h\| \to 0$.
7. On $\mathbb{R}$, the Euclidean norm is the absolute value. So step 6 says that $f_j$ is differentiable at $a$ with $Df_j(a) = L_j$, for each $j$.
8. Conversely, suppose that each $f_j$ is differentiable at $a$ with $Df_j(a) = L_j$. So $|r_j(h)| / \|h\| \to 0$ for each $j$.
9. By step 4, $0 \le \|r(h)\| / \|h\| \le \sum_{j=1}^m |r_j(h)| / \|h\|$ for each $h \ne 0$. The right side is a sum of $m$ terms that tend to $0$.
10. So $\|r(h)\| / \|h\| \to 0$, and $f$ is differentiable at $a$ with $Df(a) = L$.