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

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.