# $C^k$ version of Hadamard's Lemma. Differentiability of the remainder

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 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 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 $n>1.$

Remarks. Few days ago I posted similar question on Math.SE, 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, Duke Math. J. 10, 153-158 (1943). ZBL0063.08234.

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 $$R= x \int_0^1 2|tx+ty|\, dt = x|x+y| .$$