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I am wondering how the remainder in multivariate Taylor's formula could be uniformly bounded in a small ball around a point.

For instance, let $f:\mathbb R^n\rightarrow\mathbb R^m$ be a function of class $C^2$ around a point $x\in \mathbb R^n$. For $\varepsilon>0$, let $B$ (resp. $\overline B$) denote the open (resp. closed) ball of center $x$ and radius $\varepsilon$. We assume $\varepsilon$ sufficiently small so that $f$ is $C^2$ in $B$.

Then, for $y\in B$, define $$R(y)=\frac{\lVert f(y)-f(x)-D_x f(y-x)\rVert}{\lVert y-x\rVert^2}.$$

My question is: can $R(y)$ be uniformly bounded on $B$?

For instance, if $n=m=1$ then the mean value theorem asserts that $R(y)\leq \frac{\max_{z\in \overline B}\mid f''(z)\mid}{2}$.

Is there a similar bound in higher dimensions? (for instance involving $\max_{z\in \overline B}\frac{\lVert D^2_z f\rVert}{2}$ where $\lVert D^2_z f\rVert=\sup\frac{\lVert D^2_z f(u,v)\rVert}{\lVert u\rVert\lVert v\rVert}$)

Thank you!

emeu
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  • maybe someone will give you a precise estimate. My thought is this: the proof of the existence of the multivariate Taylor series is built from the usual Taylor series. If you study that proof then you ought to be able to piggy-back a formula from one of the usual error estimates from the single-variate case. – James S. Cook Jun 21 '13 at 16:25

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