smoothness of the functions in Hadamard lemma For every smooth function $f\in C^\infty (\mathbb{R}^n)$ there are smooth functions $g_i$ such that $f(x)=f(0)+\Sigma x_ig_i(x).$
This is proved by defining $g_i(x) = \int_{0}^{1}\frac{\partial f}{\partial {x_i}}(tx)dt$. I think these $g_i$ are exactly smooth but can't prove it.
How do you prove this? Is there a useful lemma?
 A: Note that
$$\frac{\partial}{\partial x_j} \int_0^1 \frac{\partial}{\partial x_i} f(t\mathbf{x})dt = \int_0^1 \frac{\partial}{\partial x_j} (\frac{\partial}{\partial x_i} f(t\mathbf{x})) dt$$
Since $f$ is $C^{2}$, the integrand on the right is continuous. To generalize, let $h(\mathbf{x},t)$ be continuous; then we claim that
$$\int_0^1 h(\mathbf{x}, t) dt$$
is continuous (as a function of $\mathbf{x}$). This is easy to show just by using the limit definition of continuity. 
Therefore, $\frac{\partial g_i}{\partial x_j}$ exists and is continuous for each $j$, and so $g_i$ is $C^1$. Similar methods can be used to show that $g_i$ is $C^{\infty}$. 
A: Consider the function $p(s,t)=st$. The $p:\mathbb{R}^2\rightarrow \mathbb{R}$ is a smooth function. Hence the composition $f\circ p$ is also a smooth function from $\mathbb{R}^2$ to $\mathbb{R}$. Now a theorem states that you can take the derivative inside the integration if the function inside integration is smooth (try to prove it or see Proposition 2.1, Chapter III of "Function of one complex variable" by Conway). The only thing that may confuse you is the uniform continuity. But remember smoothness is a local property and in a compact neighbourhood every continuous function is uniformly continuous.  This proves your claim.
