A version of Borel's Lemma states that, given a sequence $(f_0, f_1, f_2, \ldots)$ of functions in $C^\infty(\mathbb{R}^{n-1})$, there always exists a function $F$ in $C^\infty(\mathbb{R}^{n})$ such that $$ \frac{\partial^k F}{\partial t^k}(0,\cdot) = f_k \quad \forall \ k \geq 0.$$ Note that, clearly, $\{ 0 \} \times \mathbb{R}^{n-1}$ is a closed subset of $\mathbb{R}^{n}$.
On the other hand, the much more general Whitney extension theorem deals with arbitrary closed subsets $X$ of $\mathbb{R}^n$. In its $C^\infty$ formulation, it gives necessary and sufficient conditions for the existence of a function $f$ in $C^\infty(\mathbb{R}^n)$ whose all partial derivatives take prescribed values $f^\alpha$ on $X$, with $\alpha \in \mathbb{N}_0^n$ an arbitrary multi-index. According to the theorem, such an extension $f$ of the candidate derivative data $\{f^\alpha\}$ exists if and only if the $\{f^\alpha\}$ (are continuous and) satisfy the same estimates, restricted to the set $X$, which the collection $\{\partial^\alpha g\}$ of partial derivatives of any $g \in C^\infty(\mathbb{R}^n)$ would satisfy as a result of Taylor's theorem. Let me call a collection $\{f^\alpha\}$, of functions defined and continuous on $X$ and satisfying such estimates, a $C^\infty$-Whitney field on $X$. More precisely, what is required of the collection is that, for any $m \geq 0$ and any compact subset $K$ of $X$, the "remainders" $$f^\alpha(\boldsymbol{x})-\sum_{|\beta|\le m}\frac{f^{\alpha+\beta}(\boldsymbol{y})}{\beta!}\cdot(\boldsymbol{x}-\boldsymbol{y})^\beta,\qquad \boldsymbol{x},\boldsymbol{y}\in K \tag I$$ be $o\big(||\boldsymbol{x}-\boldsymbol{y}||^{m}\big)$ uniformly as $\boldsymbol{x},\boldsymbol{y} \in K,$ $||\boldsymbol{x}-\boldsymbol{y}||\to 0$.
It is obvious (and frequently pointed out in the literature) that if $X = \{a\}$ then Whitney's extension theorem states that given any collection $\{c^\alpha\}_{\alpha \in \mathbb{N}_0^n}$ of real constants we can find a $f \in C^\infty(\mathbb{R}^n)$ such that $\partial^\alpha f (a) = c^{\alpha}$ for all multi-indices $\alpha$. In particular, if $n=1$ we get a particular case of the version of Borel's Lemma I mentioned in the first paragraph.
Now, to my question: How does one show that Borel's Lemma as above is a simple corollary of Whitney's extension theorem when $n>1$?
Let me try to show the origin of my confusion by illustrating the case $n=2$. I.e., we are given a sequence $(f_0, f_1, f_2, \ldots)$ of functions in $C^\infty(\mathbb{R})$, which we ultimately want to promote to be the derivatives of all orders and transverse to $X = \{ 0\} \times \mathbb{R}$ of an $F \in C^\infty(\mathbb{R}^2)$. Given that we'd like to use the Whitney theorem to find $F$, we must first complete the collection $$\{ f^{(0,0)} := f_0 , \, f^{(1,0)} := f_1, \, f^{(2,0)} := f_2, \ldots \}$$ (having identified the $f_i$'s with functions on $X$ in the obvious way). In view of what we want to achieve, the way to do so is by setting, for any $\alpha=(i,j) \in \mathbb{N}_0^2$, $$ f^{\alpha} = f_i^{(j)}, $$ where the superscripts denotes the ordinary $j$-th order derivative. We must now verify that the thus defined collection $\{ f^\alpha\}$ is a $C^\infty$-Whitney field on $X$, for then Borel's Lemma would immediately follow.
So let me test (I) for $m=1$, and with $\alpha=(0,1)$. Then $f^\alpha(0,x) = f_0'(x)$, while since the multi-indices of length less than or equal to $1$ are $(0,0), (1,0), (0,1)$, we have, setting $\boldsymbol{x}=(0,x)$ and $\boldsymbol{y}=(0,y)$: $$ f^\alpha(\boldsymbol{x}) - \sum_{|\beta|\le 1}\frac{f^{\alpha+\beta}(\boldsymbol{y})}{\beta!}\cdot(\boldsymbol{x}-\boldsymbol{y})^\beta = f_0'(x) - \sum_{k+l\le 1}\frac{f^{(k,l+1)}(0,y)}{k! l!}\cdot(x-y)^l \\= f_0'(x) - f^{(0,1)}(0,y) - f^{(1,1)}(0,y) - f^{(0,2)}(0,y)(x-y) \\ = f_0'(x) - f_0'(y) - f_1'(y) - f_0''(y)(x-y). $$ Now, $f_0'(x) - f_0'(y) - f_0''(y)(x-y)$ is clearly $o(||\boldsymbol{x} - \boldsymbol{y}||^m) = o(|x-y|)$ uniformly on compact sets as $|x-y| \to 0$, because $f_0'$ is smooth. My problem is with the $- f_1'(y)$ term above. As $|x-y| \to 0$ it is just $-f_1'(x)$, so unless $f_1$ is zero we seem to have a problem! Also, since $f_1$ was prescribed completely independently from the other $f_i$'s, we don't seem to have estimates linking it to $f_0$ in a way which could obviously save the day.
I have been stuck on this quite a while. Clearly I must have either messed up my calculations, or have badly misunderstood the statement of the (I'd guess) Whitney extension theorem. Either way, I'd really appreciate your help.
The answer is simple.
For $\mathbf{x} = (0,x)$ and $\mathbf{y} = (0,y)$, and $\beta = (1,0)$, the term
$$ (\mathbf{x} - \mathbf{y})^{\beta} = (0-0)^1 \cdot(x-y)^0 = 0 $$
and not $1$, which you assumed. So when you are doing your expansion, the $f'_1(y)$ term should be multiplied by $0$ and disappear.