Let $f:\Omega\to\mathbb{R}$ be of $C^k$ class, let $0\in\Omega\subset\mathbb{R}^n$ and let $\Omega$ be star shaped at $0.$


From [Hadamard's Lemma](https://en.wikipedia.org/wiki/Hadamard%27s_lemma) we know that we can write function $f$ as


$$f(x)=f(0)+\sum\_{i=1}^n \overbrace{x\_i\int\_0^1\frac{\partial f}{\partial x\_i}(tx)dt}^{R\_i(x)}.$$


This is in fact [multivariate case of Taylor's Theorem](https://en.wikipedia.org/wiki/Taylor%27s_theorem#Taylor.27s_theorem_for_multivariate_functions) with the remainder of first order.


**Question.** Is $R\_i$ of $C^k$ class for $i=1,\dots,n$?


For $n=1$ the answer is yes, cause then $R(x)=f(x)-f(0).$ Problem is with $n>1.$




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**Remarks.**
Few days ago I posted [similar question on Math.SE](https://math.stackexchange.com/questions/2286305/differentiability-of-the-remainder-in-taylors-theorem-1d-vs-2d), but it didn't draw much attention. 


After reading Whitney's article I thought that the the answer may be nontrivial so this is why I post here as well. 


I just can add that from Whitney's paper we immediately know that


$$ x\int\_0^1\frac{\partial f}{\partial x}(tx,y)dt$$


is of $C^k$ class, but in our case we have 


$$ x\int\_0^1\frac{\partial f}{\partial x}(tx,ty)dt.$$


I can't find a way to transform Whitney's argument to my case.


*Whitney, Hassler*, [**Differentiability of the remainder term in Taylor's formula**](http://dx.doi.org/10.1215/S0012-7094-43-01014-2), Duke Math. J. 10, 153-158 (1943). [ZBL0063.08234](https://zbmath.org/?q=an:0063.08234).



$C^k$ version of Hadamard's Lemma. Differentiability of the remainder
No, not in general. Consider $f(x,y)=(x+y)|x+y|$. This is $C^1$ with partial derivatives $f\_x=f\_y=2|x+y|$, but you lose one derivative when you form
$$
R= x \int\_0^1 2|tx+ty|\, dt = x|x+y| .
$$


