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Prove or find a counterexample: if $f:\mathbb{R}^n\rightarrow \mathbb{R}$ is continuously differentiable with $f(0)=0$ then there exist continuous functions $g_1,\dots,g_n:\mathbb{R}^n\rightarrow \mathbb{R}$ with $$f(x)=x_1g_1(x_1,\dots, x_n)+\dots+x_ng_n(x_1,\dots, x_n)$$

I don't know where to start. Should I use the implicit function theorem? Please give me some hints.

Happy new year.

learning_math
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  • How is differentiation defined here? It seems that the usual definition does not work. – Kenny Lau Jan 02 '18 at 05:01
  • @KennyLau How does the usual definition not work? What is your usual definition of differentiable for functions $\Bbb R^n\to\Bbb R$? – Arthur Jan 02 '18 at 07:45
  • @Arthur The usual definition of differentiable is for function $\Bbb R \to \Bbb R$ and is $f'(a) := \displaystyle \lim_{h \to 0} \frac {f(a+h) - f(a)} h$. There is also a definition for partial differentiability for functions $\Bbb R^n \to \Bbb R$, and also a gradient for functions $\Bbb R^n \to \Bbb R$, but the gradient is a vector. – Kenny Lau Jan 02 '18 at 07:47
  • @KennyLau There is also a common definition of differentiable at a point $p$ for any function $\Bbb R^n\to \Bbb R^m$. Such a function $h$ is said to be differentiable at $p$ if there is a linear map $L:\Bbb R^n\to \Bbb R^m$ such that $\lim_{v\to p}\left(\frac{f(v) - f(p) - L(v)}{|v-p|}\right) = 0$. In other words, $h$ has a linear approximation at the point $p$. In the case $\Bbb R^n\to \Bbb R$, this just means a well-defined tangent plane (and $\Bbb R\to \Bbb R$ is a special case where it means a well-defined tangent line). – Arthur Jan 02 '18 at 08:06
  • @Arthur when $m=1$, it is essentially the gradient. – Kenny Lau Jan 02 '18 at 08:07
  • @KennyLau Yes, that is true. When $m = 1$, $L$ is just the gradient, seen as a linear map from $\Bbb R^n\to \Bbb R$ (a row vector in linear algebra). Still, I don't see how the definition fails to work, as you seem to claim. – Arthur Jan 02 '18 at 08:09
  • @Arthur I'm not sure I understand your definition via linear map. Take $n=m=1$ and $f=\sin$. Are you saying that there is a constant $c$ such that $\displaystyle \lim_{x\to0} \frac{f(x)-f(0)-cx}{|x-0|} = 0$? – Kenny Lau Jan 02 '18 at 08:42
  • @KennyLau Yes, and that constant is $1$, and we usually call it "the derivative of the sine function at $x = 0$. At any other point than $0$, there is a different constant $c$. I realize now that I messed up the norm signs slightly, and they should've been around the numerator as well. It should've been $\lim_{v\to p}\frac{|f(v) - f(p) - L(v)|}{|v-p|} = 0$ – Arthur Jan 02 '18 at 08:47
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    FYI: This is Exercise 2-35 from Spivak's "Calculus on Manifolds". – user557 Jan 07 '18 at 07:50

2 Answers2

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Fix $x\in{\mathbb R}^n$ and consider the auxiliary function $$\phi(t):=f(t\,x)\qquad(0\leq t\leq 1)\ .$$ The chain rule then gives $$f(x)=\phi(1)-\phi(0)=\int_0^1\phi'(t)\>dt=\int_0^1\nabla f(t\,x)\cdot x\>dt=\sum_{i=1}^n x_i\,g_i(x)$$ with $$g_i(x):=\int_0^1 f_{.i}(t\, x)\>dt\qquad(1\leq i\leq n)\ .$$

Christian Blatter
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  • And what exactly is the difference between our functions? – Kenny Lau Jan 02 '18 at 09:22
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    @KennyLau: They are not the same. E.g., if $n=3$ your $g_2$ amounts to $\int_0^1f_{.2}(x,ty,0)>dt$, and mine to $\int_0^1 f_{.2}(tx,ty,tz)>dt$. But the $g_i$ are not uniquely determined anyway. The problem with your solution is that the continuity of the $g_i$ is not obvious. Note that your $g_i$ are defined without reference to differentiability. – Christian Blatter Jan 02 '18 at 10:15
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For $1 \le i \le n$, for $x_i \ne 0$, define: $$g_i(x_1,\ldots,x_n) := \frac {f(x_1, \cdots, x_{i-1}, x_i, 0, 0, \cdots, 0) - f(x_1, \cdots, x_{i-1}, 0, 0, 0, \cdots, 0)} {x_i}$$

When $x_i = 0$, let $g_i(x_1,\ldots,x_n)$ be its limit.

Note that $\displaystyle g_1(x_1,\ldots,x_n) = \frac {f(x_1, 0, \cdots, 0) - f(0, 0, \cdots, 0)} {x_1} = \frac {f(x_1, 0, \cdots, 0)} {x_1}$.

It is a trivial exercise to show that $f = \sum x_i g_i$.

The well-definedness of such functions (i.e. the fact that the limit exists) and the continuity of such functions follow from the definition of "continuously differentiable", which unfortunately you have not provided.

For example, when $n=2$, this construction amounts to: $$f(x,y) = x \left( \frac {f(x,0)} x \right) + y \left( \frac {f(x,y) - f(x,0)} y \right)$$

When $n=3$, this construction amounts to: $$f(x,y,z) = x \left( \frac {f(x,0,0)} x \right) + y \left( \frac {f(x,y,0) - f(x,0,0)} y \right) + z \left( \frac {f(x,y,z) - f(x,y,0)} z \right)$$

Kenny Lau
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