Proof that C^k functions are characterised by continuous derivatives

Proof

We identify each iterated derivative with the list of its values on basis vectors, and we argue by induction on the order $r$.

1. Let $U$, $f$ and $k$ be as in the statement. For $r \ge 0$, let $E_r = L^r(\mathbb{R}^n; \mathbb{R}^m)$, and let $e_1, \dots, e_n$ be the standard basis of $\mathbb{R}^n$. For a tuple $l = (l_1, \dots, l_r)$ in $\{1, \dots, n\}^r$, write $e_l = (e_{l_1}, \dots, e_{l_r})$.
2. Define $\Phi_r \colon E_r \to (\mathbb{R}^m)^{n^r} = \mathbb{R}^{N}$, with $N = m n^r$, by $\Phi_r(T) = (T(e_l))_l$. For $r = 0$, $\Phi_0$ is the identity of $\mathbb{R}^m$. The map $\Phi_r$ is linear.
3. The map $\Phi_r$ is a bijection. Write each argument $v_s = \sum_t v_{s,t} e_t$. By multilinearity, $T(v_1, \dots, v_r) = \sum_l v_{1,l_1} \cdots v_{r,l_r} T(e_l)$, so $T$ is determined by $\Phi_r(T)$. Conversely, for any family $(y_l)$ this formula with $y_l$ in place of $T(e_l)$ defines an $r$-linear map with values $y_l$.
4. Give $E_r$ the norm $\|T\| = \|\Phi_r(T)\|$, with the Euclidean norm on $\mathbb{R}^N$. All norms on $E_r$ are equivalent. Equivalent norms give the same continuous maps, and they change the quotient defining a Fréchet derivative by bounded factors only. So we may use this norm.
5. Let $G \colon U \to E_r$ and $x \in U$. Each linear map $\mathbb{R}^n \to E_r$ is bounded, by linear maps on finite-dimensional spaces are continuous and bounded linear maps are continuous. Since $\Phi_r$ is a linear bijection preserving norms, $\|G(x + h) - G(x) - A h\| = \|\Phi_r G(x + h) - \Phi_r G(x) - \Phi_r A h\|$ for each linear $A \colon \mathbb{R}^n \to E_r$.
6. By step 5, $G$ is Fréchet differentiable at $x$ if and only if $\Phi_r \circ G$ is differentiable at $x$. For the converse direction, a derivative $L$ of $\Phi_r \circ G$ gives the linear map $A = \Phi_r^{-1} \circ L$ in step 5. Then $D(\Phi_r \circ G)(x) = \Phi_r \circ DG(x)$. Also $G$ is continuous if and only if $\Phi_r \circ G$ is continuous.
7. A map into $\mathbb{R}^N$ is continuous if and only if each coordinate is continuous. This follows from $|y_q| \le \|y\| \le |y_1| + \dots + |y_N|$ for $y \in \mathbb{R}^N$, and the triangle inequality.
8. When $D^r f$ is defined on $U$, let $c^r_{j,l}(x)$ be the $j$-th entry of $D^r f(x)(e_l)$. These functions are the coordinates of $\Phi_r \circ D^r f$.
9. For $l$ of length $r$, let $p_{j,l} = \partial^r f_j / \partial x_{l_1} \cdots \partial x_{l_r}$, where we differentiate first in $x_{l_r}$ and last in $x_{l_1}$. For $r = 0$, $p_{j,()} = f_j$. By the definition of higher partial derivatives, $p_{j,(i, l)} = \partial_i p_{j,l}$ whenever the right side exists. As $l$ varies, the $p_{j,l}$ are all the partial derivatives of order $r$ of $f_j$.
10. Claim: suppose that $D^r f$ is differentiable on $U$, for some $r \ge 0$. Then for all $i$, $j$, $l$ and $x \in U$, the partial derivative $\partial_i c^r_{j,l}(x)$ exists and equals $c^{r+1}_{j,(i,l)}(x)$.
11. Proof of the claim. By step 6, $\Phi_r \circ D^r f$ is differentiable at $x$ with derivative $\Phi_r \circ D(D^r f)(x)$. Since the derivative gives the directional derivatives, its difference quotients in the direction $e_i$ tend to $\Phi_r\big(D(D^r f)(x)(e_i)\big)$.
12. Each coordinate of a convergent family in $\mathbb{R}^N$ converges, by step 7. Taking the coordinate with index $(j, l)$, the partial derivative $\partial_i c^r_{j,l}(x)$ exists and equals the $j$-th entry of $D(D^r f)(x)(e_i)(e_l)$.
13. By the definition of the iterated derivative, $D(D^r f)(x)(e_i)(e_l) = D^{r+1} f(x)(e_i, e_{l_1}, \dots, e_{l_r})$. Its $j$-th entry is $c^{r+1}_{j,(i,l)}(x)$. This proves the claim.
14. Now suppose that $f$ is of class $C^k$ in the sense of partial derivatives. We show by induction on $r \le k$ that $D^r f$ is defined on $U$ and $c^r_{j,l} = p_{j,l}$ on $U$ for all $j$ and $l$. For $r = 0$ both sides are $f_j$.
15. Suppose that this holds for some $r < k$. The coordinates of $\Phi_r \circ D^r f$ are the $p_{j,l}$. Their partial derivatives $\partial_i p_{j,l} = p_{j,(i,l)}$ have order $r + 1 \le k$, so they exist and are continuous on $U$.
16. By the theorem that continuous partial derivatives imply differentiability, $\Phi_r \circ D^r f$ is differentiable at each point of $U$. By step 6, $D^r f$ is differentiable on $U$. So $D^{r+1} f$ is defined on $U$.
17. By the claim and the induction hypothesis, $c^{r+1}_{j,(i,l)} = \partial_i c^r_{j,l} = \partial_i p_{j,l} = p_{j,(i,l)}$ on $U$. Each tuple of length $r + 1$ has the form $(i, l)$, so the induction step is complete.
18. Step 16 shows condition 2 for each $r < k$. For $1 \le r \le k$, the coordinates $c^r_{j,l} = p_{j,l}$ of $\Phi_r \circ D^r f$ are continuous by hypothesis. By steps 6 and 7, $D^r f$ is continuous on $U$.
19. For $r = 0$: by step 16 with $r = 0$, $f$ is differentiable at each $x \in U$. As in step 5, there is $C \ge 0$ with $\|Df(x) h\| \le C \|h\|$. The definition gives $\|f(x + h) - f(x)\| \le (C + 1) \|h\|$ for small $h$. So $f$ is continuous, and condition 1 holds.
20. Conversely, suppose that conditions 1 and 2 hold. We show by induction on $r \le k$ that each $p_{j,l}$ of order $r$ exists on $U$ and equals $c^r_{j,l}$. For $r = 0$ both sides are $f_j$.
21. Suppose that this holds for some $r < k$. By condition 2, $D^r f$ is differentiable on $U$. By the claim, $\partial_i c^r_{j,l}$ exists on $U$ and equals $c^{r+1}_{j,(i,l)}$.
22. Since $c^r_{j,l} = p_{j,l}$ on $U$, the partial derivative $p_{j,(i,l)} = \partial_i p_{j,l}$ exists on $U$ and equals $c^{r+1}_{j,(i,l)}$. This completes the induction step.
23. Let $r \le k$. By condition 1, $D^r f$ is continuous on $U$. By steps 6 and 7, each coordinate $c^r_{j,l}$ of $\Phi_r \circ D^r f$ is continuous. By step 20, each partial derivative $p_{j,l}$ of order $r$ is continuous on $U$.
24. So $f$ is of class $C^k$ on $U$. Together with steps 14 to 19, this proves the equivalence.