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Where one could find the proof of the following version of Taylor theorem for functions of several variables?

Assume that $f$ is a function of class $C^{n+k}$ defined in a neighbourhood $W$ of zero in $\mathbb{R}^m$ with values in $\mathbb{R}.$ Then there exist a functions $S_\alpha$, where $\alpha$ is multiindex with length $n$, defined in $W$ with values in $\mathbb{R}$ such that

  1. $S_\alpha$ is of clas $C^k$ in $W$, $S_\alpha(0)=0$, $D^i S_\alpha(0)=\left(\frac{i!}{(n+i)!}\right) D^{n+i}f(0)$ for every multiindex $i$ with $0 \leq |i| \leq k$.

  2. $S_\alpha$ is of class $C^{n+k}$ in $W\setminus \{0\}$.

  3. $\lim_{x\rightarrow 0} \|x\|^{|i|} D^{i+k}S_\alpha(x)=0$ for each multiindex $i$ with $1\leq |i|\leq n$ and for each $\alpha$ (with $| \alpha|=n$).

  4. $f(x)=\sum_{|\alpha|\leq n}\left(\frac{D^\alpha f(0)}{\alpha !}\right)x^\alpha+\sum_{|\alpha| =n} S_\alpha(x) x^\alpha$ for $x \in W$.


For $m=1$ representation 4. follows from Taylor's formula with integral's remainder: $$f(x)=f(0)+\frac{f'(0)}{1!}x+...+\frac{f^{(n-1)}(0)}{(n-1)!}x^{n-1}+\frac{f^{(n)}(0)}{n!}x^n+S_n(x) x^n,$$ where $$S_n(x):=\frac{1}{(n-1)!}\int_0^1 (f^{(n)}(sx)-f^{(n)}(0)) (1-s)^{n-1}ds.$$ By differentiation under integral we obtain statement 1. The 2. is obvious, because $S_n(x)=\frac{f(x)-P_n(x)}{x^n}$ for $x \neq 0$, where $P_n$ is a polynomial function.

By substitution $y=sx$ formula for $S_n$ can be written in the following form $$S_n(x):=\frac{1}{(n-1)!} \frac{1}{x^n} \int_0^x (f^{(n)}(y)-f^{(n)}(0)) (x-y)^{n-1}dy.$$ By this and Leibniz formula on higher-order derivatives of multiplication follows 3.

For arbitrary $m$ 4. follows from the case $m=1$ applied to the function $g(t):=f(tx)$ for $t \in [0,1].$ Here $$S_\alpha(x)=\frac{1}{(n-1)!}\int_0^1 (D^\alpha f(sx)-D^\alpha f(0))(1-s)^{n-1}ds$$ (where $ \alpha$ is multiindex with length $n$.)

The proof of statement 1. goes as the case $m=1$. But I'm most interested how to obtain the 2 and 3.

share|cite|improve this question
C.H. Edwards, Advanced Calculus of Several Variables. page 129. – Potato Dec 29 '11 at 23:20
possible duplicate of Taylor's theorem in Banach spaces – t.b. Dec 30 '11 at 2:42
I think that they concern rather standard version of Taylor formula with remainder in the form of Peano or Lagrange. But I look for a special version. It probably follows from that classical one but in what way? – L.T Dec 30 '11 at 8:41

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