i have this lemma :
Let $f$ be a $C^{\infty}$ function in a convex neighborhood $V$ of $0$ in $\mathbb{R}$ , with $f(0)=0$ then $f(x)= x g(x)$. for suitable $C^\infty$ function $g$ defined in $V$ with $g(0)=f'(0)$
to prove this they say that :
$\displaystyle f(x)=\int_0^1 \frac{df(tx)}{dt} dt=\int_0^1 x \frac{df(tx)}{dx} dt...*$
Therefore we can let $\displaystyle g(x)=\int_0^1\frac{df(tx)}{dx} dt$
I don't understand why $\displaystyle\int_0^1 \frac{df(tx)}{dt} dt=\int_0^1 x \frac{df(tx)}{dx} dt$
Please
Thank you .
I just checked my copy of Milnor, you have the notation written incorrectly
All he has done there is a basic application of chain rule:
$$\frac{d}{dt} f(t x_1, t x_2, ..., t x_n) = f_1 (t x_1, t x_2, ..., t x_n) x_1 + f_2 (t x_1, t x_2, ..., t x_n) x_2 + \dots + f_n (t x_1, tx_2, ..., t x_n) x_n$$
where $f_i$ is the partial derivative of $f$ with respect to the $i^{th}$ variable.
Let's look at the scenario when $f(x) = \sin x$. Then $$f(x) = \int_0^1 \frac{d f(tx)}{dt} dt = \int_0^1 x \cos(tx) dt $$
Now if we were to change the integrand to $ x \frac{d \sin(tx)}{dx}$, we'd have $x t \cos(tx)$ inside the integral. What is meant however is $\left. x \frac{d \sin(u)}{du}\right|_{u = tx}$ which is $x \cos(tx)$ which is what you want inside the integral.