Prove that if $\int_0^1 f(yx) dx = g(y)$, so $f(x)=g(x)+xg'(x)$ for any $x$.

Question

I'm trying to solve this question:

$f: \mathbb{R} \to \mathbb{R}$ is a continuous function and $g: \mathbb{R} \to \mathbb{R}$ is of $C^1$ class, such that $g(y) = \int_0^1 f(yx) dx$. Prove that $f(x)=g(x)+xg'(x)$ for all $x \in \mathbb{R}$.

But my progress until now was really slow. I've tried to use the Leibniz integral rule and the first mean value theorem for definite integrals, but I had no success. Can somebody help me?

Answer 1

If you change variables in the integral ($u = xy$), you get $$ \int_0^1 f(yx) dx = \frac 1y \int_0^yf(u) du, $$

so, $$\int_0^y f(u) dy = yg(y) \Rightarrow f(y)=(yg(y))' = g(y) + yg'(y).$$

Note: This method requires that $y\ne 0$, but in that case the equality can be checked directly.

Answer 2

welcome toi MSE . a hint can be $$f(x)=g(x)+xg'(x)\\ f(x)=x'g(x)+xg'(x)\\f(x)=(xg(x))'\\\to \\\int f(x)=xg(x)+const$$