Smooth functions

Let $f:\mathbb{R}^n \rightarrow \mathbb{R}$ be a smooth function (i.e a function that has continuous partial derivatives of any order). Prove that there exist functions $g_i, \; i=1, 2, \dots, n$  such that:

$$f(x_1, x_2, \dots, x_n) -f (0, 0, \dots, 0) = \sum_{i=1}^{n} x_i g_i (x_1, x_2, \dots, x_n)$$

Solution

 Let $\mathbf{x}=\left ( x_1, x_2, \dots, x_n \right ) \in\mathbb{R}^n$ and $t \in \mathbb{R}$. We set: $\varphi_\mathbf{x}(t)=f\left ( t \mathbf{x} \right )$ and we note that:

$$\int_{0}^{1}\varphi'_{\mathbf{x}}(t) \, {\rm d}t = \varphi_{\mathbf{x}}(1) - \varphi_{\mathbf{x}} (0)= f (\mathbf{x}) - f(0)$$

Using the chain rule however, we get that:

\begin{align*}
\int_0^1 {{{\varphi '}_{\bf{x}}}\left( t \right)} \, {\rm d}t &= \int_0^1 {\left( {\sum\limits_{i = 1}^n {\frac{\partial }{{\partial {x_i}}}} f\left( {t{\bf{x}}} \right) \cdot \frac{{\rm d}}{{{\rm d}t}}\left( {t{x_i}} \right)} \right)\, {\rm d}t}  \\
 &= \int_0^1 {\left( {\sum\limits_{i = 1}^n {\frac{\partial }{{\partial {x_i}}}} f\left( {t{\bf{x}}} \right) \cdot {x_i}} \right)\, {\rm d}t} \\
 &= \sum\limits_{i = 1}^n {{x_i}\left( {\int_0^1 {\frac{\partial }{{\partial {x_i}}}f\left( {t{\bf{x}}} \right)\, {\rm d}t} } \right)} 
\end{align*}

Now the result follows immediately , by setting:

$$g_i(\mathbf{x})= \int_{0}^{1}\frac{\partial }{\partial x_i} f (t \mathbf{x}) \, {\rm d}t , \;\;\; i=1, 2, \dots, n$$

The exercise can also be found at mathematica.gr