Multivariate version of Taylor's theorem in several but not all variables resulting in mixed derivative order

Question

Let $U\subseteq \Bbb R^n$ open and $f:U \to \Bbb R$ be a continuously differentiable function, such that additionally the second partial derivatives exist and are continuous in the variables $(x_r , \ldots , x_n)$.

Is there a form of mixed Taylor formula such that

$$f(x+h) = f(x) + \sum_{k=1}^{n} \frac{\partial f}{\partial x_k} (x) h_k + \frac 1 2\sum_{k,\ell = r}^n \frac{\partial ^2 f}{\partial x_k \partial x_\ell} (x) h_k h_\ell + r(h;x)$$

such that $$\frac{r(h;x)}{\sum_{k=1}^{r-1} \vert h_k\vert +\sum_{k=r}^n h_k^2} \to 0$$ for $h\to0$?

Especially I want to have some more explicit information on $r(h;x)$ such I can deduce something about the behaviour on compacta, namely that the convergence to $0$ above holds uniform in $x$ on compacta.

I think denominator can be in this form because of the intuition that the error in the direction of the first $r-1$ variables behaves as $o(\vert h\vert $) while the error in the last variables behaves as $o(\vert h\vert ^2)$.

Matching to this I tried to apply the usual Taylor approximation for the two cases separately, but for example in the case $(x_r , \ldots , x_n)$ one can either approximate around $x + (h_1 , \ldots , h_{r-1}, 0, \ldots , 0)$ or $x$ with adding $(0, \ldots , 0 , h_r , \ldots , h_n)$ but the mixed term does not lead to the desired.

Answer 1

Ok here is half of an answer: if $f \in C^2(U)$ your formula is correct. From the usual Taylor formula you get $r(h, x)=p(x,h) + o (\|h\|_2^2)$ where $$p(h, x) = \sum_{\substack{1 \le i < r\\ r \le j<n}}\frac{\partial^2 f}{\partial x_i \partial x_j} (x) h_ih_j + \frac{1}{2} \sum_{\substack{1 \le i < r\\ 1 \le j < r}}\frac{\partial^2 f}{\partial x_i \partial x_j} (x) h_ih_j$$ Doing some standard calculation you get
$$ |p(x,h)| \le \left(\sum_{i=1}^{r-1} |h_i| \right) \left(\sum_{j=r}^n |h_j| \max_{1 \le i <r} \left|\frac{\partial^2 f}{\partial x_i \partial x_j} (x)\right| + \frac{1}{2} \sum_{j=1}^{r-1} |h_j| \max_{1 \le i \le r} \left|\frac{\partial^2 f}{\partial x_i \partial x_j} (x)\right| \right)$$

from which you get $$\frac{r(x,h)}{\sum_{i=1}^{r-1} |h_i| + \sum_{i=r}^n |h_i|^2} \rightarrow 0 \quad \text{for} \quad h \to 0$$ as desired.

Answer 2

$$ r(x,z):= f(x+z)-f(x) - \sum^n_{k=1} \frac{\partial f}{\partial x_k} (x) z_k - \frac{1}{2} \sum^n_{k,l=r} \frac{\partial^2f}{\partial x_k \partial x_l} (x) z_k z_l. $$ Write $$ z=(z',z'') \text{ mit } z'=(z_1, \ldots , z_{r-1}), z''=(z_r, \ldots , z_n). $$ The function \quad $z' \mapsto f(x+(z', z''))$ \quad is continuously differentiable,

the function \quad $z'' \mapsto f(x +(0,z''))$ \quad is twice continuously differentiable.\ Application of Taylors formula yields \begin{align*} f(x+z)-f(x+(0,z'')) &= f(x+(z',z''))-f(x+(0,z'')) \\ &= \sum^{r-1}_{k=1} \frac{\partial f}{\partial x_k} (\zeta_{x,z})z_k \\ f(x+(0,z''))-f(x) &= \sum^n_{k=r} \frac{\partial f}{\partial x_k} (x)z_k + \frac{1}{2} \sum^n_{k,l=r} \frac{\partial^2f}{\partial x_k \partial x_l} (\bar{\zeta}_{x,z})z_kz_l, \end{align*} where $\zeta_{x,z}$ is on the line between $x+z$ und $x+(0,z'')$ and ,, $\bar{\zeta}_{x,z}$ is on the line betweeen $x+(0,z'')$ and $x$.

Thus it holds \begin{align*} r(x,z) &= f(x+z)-f(x+(0,z'')) + f(x+(0,z''))-f(x) \\ & \qquad - \sum^n_{k=1} \frac{\partial f}{\partial x_k} (x) z_k - \frac{1}{2} \sum^n_{k,l=r} \frac{\partial^2f}{\partial x_k \partial x_l}(x) z_k z_l \\ &= \sum^{r-1}_{k=1} \left( \frac{\partial f}{\partial x_k} (\zeta_{x,z}) - \frac{\partial f}{\partial x_k}(x) \right) z_k + \frac{1}{2} \sum^n_{k,l=r} \left( \frac{\partial^2f}{\partial x_k \partial x_l} (\bar{\zeta}_{x,z}) - \frac{\partial^2f}{\partial x_k \partial x_l}(x) \right) z_k z_l. \end{align*} Hence it holds with the triangle inequality \begin{align*} |r(x,z)| & \le \sum^{r-1}_{k=1} \bigg| \frac{\partial f}{\partial x_k} (\zeta_{x,z}) - \frac{\partial f}{\partial x_k} (x)\bigg| \, \, |z_k| + \frac{1}{2} \sum^n_{k,l=r} \bigg| \frac{\partial^2f}{\partial x_k \partial x_l} (\bar{\zeta}_{x,z}) - \frac{\partial^2f}{\partial x_k \partial x_l} (x) \bigg| \,\, |z_k z_l|. \end{align*} We have $$ \sum^{r-1}_{k=1} \bigg| \frac{\partial f}{\partial x_k} (\zeta_{x,z}) - \frac{\partial f}{\partial x_k} (x) \bigg| \,\, |z_k| \le \max\limits_{1 \le k \le r-1} \bigg| \frac{\partial f}{\partial x_k} (\zeta_{x,z}) - \frac{\partial f}{\partial x_k}(x) \bigg| \sum^{r-1}_{k=1} |z_k|. $$ Note that, since $2|z_k| \, |z_l| \le |z_k|^2 + |z_l|^2$: $$ \sum^n_{k,l=r} |z_k| \, |z_l| \le \sum^n_{k=r} \sum^n_{l=r} \frac{1}{2} (|z_k|^2 + |z_l|^2) = (n+1-r) \sum^n_{k=r} |z_k|^2. $$ Hence: \begin{align*} & \frac{1}{2} \sum^n_{k,l=r} \bigg| \frac{\partial^2 f}{\partial x_k \partial x_l} (\bar{\zeta}_{x,z}) - \frac{\partial^2 f}{\partial x_k \partial x_l} (x) \bigg| |z_k z_l| \\ & \le \frac{1}{2} \cdot \max\limits_{r \le k,l \le n} \bigg| \frac{\partial^2 f}{\partial x_k \partial x_l} (\bar{\zeta}_ {x,z}) - \frac{\partial^2 f}{\partial x_k \partial x_l} (x) \bigg| \cdot \sum^n_{k,l=r} |z_k| \,\, | z_l| \\ & \le \frac{(n+1-r)}{2} \cdot \max\limits_{r \le k,l \le n} \bigg| \frac{\partial^2 f}{\partial x_k \partial x_l} (\bar{\zeta}_{x,z}) - \frac{\partial^2 f}{\partial x_k \partial x_l} (x) \bigg| \cdot \sum^n_{k=r} |z_k|^2. \end{align*} All in all: $$ |r(x,z)| \le {\cal K} (x,z) \left( \sum^{r-1}_{k=1} |z_k| + \sum^n_{k=r} |z_k|^2 \right) $$ with $$ {\cal K} (x,z) := \max\limits_{k=1, \ldots , r-1} \bigg| \frac{\partial f}{\partial x_k} (\zeta_{x,z}) - \frac{\partial f}{\partial x_k} (x) \bigg| + \left( \frac{n+1-r}{2} \right) \max\limits_{r \le k,l \le n} \bigg| \frac{\partial^2 f}{\partial x_k \partial x_l} (\bar{\zeta}_{x,z}) - \frac{\partial^2 f}{\partial x_k \partial x_l} (x) \bigg|, $$ therefore we have also $$ | \varepsilon (x,z)| := \frac{|r(x,z)|}{\sum^{r-1}_{k=1} |z_k| + \sum^n_{k=r} |z_k|^2} \le {\cal K} (x,z). $$ Für $K \subset U$ compact we have $\exists \varepsilon > 0$, ssuch that the compact set \ $K_{\varepsilon} := \{ x \in U: \inf\limits_{y \in K} \| x-y \| \le \varepsilon \}$ also satisfies $K_{\varepsilon} \subset U$, since $U$ is open.

Note that $\| z \| < \varepsilon$ implies that $\zeta_{x,z}, \bar{\zeta}_{x,z} \in K_{\varepsilon}$ for $x \in K$.

Further the continuous functions
$\left( \frac{\partial f}{\partial x_k} \right)_{k=1, \ldots , r-1}$ und $\left( \frac{\partial^2 f}{\partial x_k \partial x_l} \right)_{k,l=r, \ldots , n}$ are uniformly continuos on the compact set $K_{\varepsilon}$. This means that for $\eta > 0$ exists $\delta > 0$, such that $\| z \| < \delta$ implies, that $\forall x \in K |{\cal K} (x,z)| < \eta$, since $( \| z \| < \delta \Rightarrow \| \zeta_{x,z} - x \| < \delta$ and $\| \bar{\zeta}_{x,z} - x \| < \delta)$.

Therefore $$ \lim\limits_{\| z \| \rightarrow 0} \,\, \sup_{x \in K}\,\, | \varepsilon (x,z)| = 0. $$