Proof of Taylor's theorem in several variables by restriction to a line

Proof

We restrict $f$ to the line through $a$ in the direction $h$ and use the one-variable Taylor theorem.

1. Let $U$, $k$, $f$ and $a$ be as in the statement. For $h \in \R^n$ with $a + h \in U$, let $R_k(h) = f(a + h) - \sum_{\abs{\alpha} \le k} \frac{D^\alpha f(a)}{\alpha!} h^\alpha$. We must show the two claims about $R_k$.
2. Fix $h \in \R^n$ and let $O = \set{t \in \R : a + t h \in U}$. This set is open, since it is the preimage of the open set $U$ under the continuous map $t \mapsto a + t h$. On $O$ define $g(t) = f(a + t h)$.
3. Let $u$ be a partial derivative of $f$ of order $j < k$. The first partial derivatives of $u$ are partial derivatives of $f$ of order $j + 1 \le k$. By the definition of class C^k, they exist and are continuous on $U$.
4. So $u$ is differentiable on $U$. Each point of $O$ lies in an open interval contained in $O$. On such an interval, the chain rule for a curve applies. So $t \mapsto u(a + t h)$ is differentiable on $O$ with derivative $\sum_{i=1}^n \partial_i u(a + t h) \, h_i$.
5. Claim: for $0 \le j \le k$, the derivative $g^{(j)}$ exists on $O$ and $g^{(j)}(t) = \sum_{i_1, \dots, i_j} \partial_{i_j} \cdots \partial_{i_1} f(a + t h) \, h_{i_1} \cdots h_{i_j}$, where each index runs from $1$ to $n$. For $j = 0$ this is the definition of $g$.
6. If the claim holds for some $j < k$, apply step 4 to each term $u = \partial_{i_j} \cdots \partial_{i_1} f$. This gives the claim for $j + 1$. By induction the claim holds for all $j \le k$. Each $g^{(j)}$ is continuous, since the partial derivatives are continuous.
7. Let $j \le k$ and let $(i_1, \dots, i_j)$ be a sequence of indices. Any rearrangement of this sequence is reached by finitely many swaps of two adjacent entries.
8. Swapping $i_l$ and $i_{l+1}$ does not change the partial derivative. Indeed, $w = \partial_{i_{l-1}} \cdots \partial_{i_1} f$ is of class $C^2$, since $l + 1 \le j \le k$. By Schwarz's theorem, $\partial_{i_{l+1}} \partial_{i_l} w = \partial_{i_l} \partial_{i_{l+1}} w$, and applying the remaining partial derivatives keeps this equality.
9. For the sequence $(i_1, \dots, i_j)$ let $\alpha_l$ be the number of $r$ with $i_r = l$. Then $\abs{\alpha} = j$. By steps 7 and 8, $\partial_{i_j} \cdots \partial_{i_1} f = D^\alpha f$, in the multi-index notation of the statement. Also $h_{i_1} \cdots h_{i_j} = h^\alpha$.
10. For a fixed $\alpha$ with $\abs{\alpha} = j$, the number of sequences that give $\alpha$ is the multinomial coefficient $j! / \alpha!$.
11. Steps 5, 9 and 10 give, for $0 \le j \le k$ and $t \in O$: $$g^{(j)}(t) = \sum_{\abs{\alpha} = j} \frac{j!}{\alpha!} D^\alpha f(a + t h) \, h^\alpha.$$
12. In particular $\sum_{j=0}^{k} g^{(j)}(0) / j! = \sum_{\abs{\alpha} \le k} \frac{D^\alpha f(a)}{\alpha!} h^\alpha$. So $R_k(h) = g(1) - \sum_{j=0}^{k} g^{(j)}(0)/j!$ whenever $1 \in O$.
13. We now prove the limit. Since $U$ is open, choose $r > 0$ with $\set{x : \norm{x - a} < r} \subseteq U$. Let $0 < \norm{h} < r$. Then $a + t h \in U$ for $t \in [0, 1]$, so $[0, 1] \subseteq O$.
14. By step 6, $g^{(k)}$ exists and is continuous on $[0, 1]$. Apply Taylor's theorem with the integral remainder to $g$ on $[0, 1]$, with order $k - 1 \ge 0$: $$g(1) = \sum_{j=0}^{k-1} \frac{g^{(j)}(0)}{j!} + \int_0^1 \frac{(1 - t)^{k-1}}{(k-1)!} \, g^{(k)}(t) \, dt.$$
15. We have $\int_0^1 \frac{(1 - t)^{k-1}}{(k-1)!} \, dt = \frac{1}{k!}$. So, by linearity of the integral, steps 12 and 14 give $$R_k(h) = \int_0^1 \frac{(1 - t)^{k-1}}{(k-1)!} \big( g^{(k)}(t) - g^{(k)}(0) \big) \, dt.$$
16. Each $\abs{h_l} \le \norm{h}$, so $\abs{h^\alpha} \le \norm{h}^k$ when $\abs{\alpha} = k$. Let $N = \sum_{\abs{\alpha} = k} k!/\alpha!$. Then step 11 gives, for $t \in [0, 1]$, $$\abs{g^{(k)}(t) - g^{(k)}(0)} \le N \norm{h}^k \max_{\abs{\alpha} = k} \abs{D^\alpha f(a + t h) - D^\alpha f(a)}.$$
17. Let $\varepsilon > 0$. There are finitely many $\alpha$ with $\abs{\alpha} = k$, and each $D^\alpha f$ is continuous at $a$. So there is $\rho$ with $0 < \rho \le r$ and $\abs{D^\alpha f(x) - D^\alpha f(a)} < \varepsilon$ for all such $\alpha$ and all $x$ with $\norm{x - a} < \rho$.
18. Let $0 < \norm{h} < \rho$. Then $\norm{(a + t h) - a} < \rho$ for $t \in [0, 1]$. By steps 15 to 17 and monotonicity of the integral, $\abs{R_k(h)} \le \int_0^1 \frac{(1 - t)^{k-1}}{(k-1)!} N \varepsilon \norm{h}^k \, dt = \frac{N \varepsilon}{k!} \norm{h}^k$.
19. Since $\varepsilon > 0$ was arbitrary and $N / k!$ does not depend on $\varepsilon$, this shows $R_k(h) / \norm{h}^k \to 0$ as $h \to 0$.
20. Now let $f$ be of class $C^{k+1}$, and let the segment from $a$ to $a + h$ lie in $U$. Then $[0, 1] \subseteq O$.
21. Steps 3 to 11 hold with $k + 1$ in place of $k$. So $g^{(k+1)}$ exists and is continuous on $[0, 1]$, and it is given by step 11 with $j = k + 1$.
22. Taylor's theorem with the integral remainder, now with order $k$, gives $$g(1) = \sum_{j=0}^{k} \frac{g^{(j)}(0)}{j!} + \int_0^1 \frac{(1 - t)^{k}}{k!} \, g^{(k+1)}(t) \, dt.$$
23. By step 12, the integral is $R_k(h)$. Insert step 11 with $j = k + 1$ and use linearity of the integral. Since $(k+1)!/(k! \, \beta!) = (k+1)/\beta!$, this gives $$R_k(h) = \sum_{\abs{\beta} = k + 1} \frac{k + 1}{\beta!} \, h^\beta \int_0^1 (1 - t)^k \, D^\beta f(a + t h) \, dt.$$