Let $f:[0,1]\to \mathbb{R}$ be continuous such that $$\int_{0}^{1} f(xt)dt=0$$ for all $x \in [0,1]$.
Show that $f(x)=0$ for all $x \in [0,1]$.
Using the FTC and substitution:
$$F(t)=\frac{1}{x}\int_{0}^{t} f(u)du$$
$$F'(t)=\frac{1}{x}[f(t)]$$
I'm not sure if I am going in the right direction. As of right now, I can't see how to go from my last step to showing that $f(x)=0$ for all $x$ in $[0,1]$.
You put $u = xt$ so that $du = xdt$. Note $$ 0 = \int_0^x f(u)\frac{1}{x}\; du $$ for all $x\in (0,1]$. So $$ 0 = \int_0^xf(u)\; du $$ for all $x\in (0,1]$. Taking the derivative on both sides you get $$ 0 = f(x) $$ for all $x\in (0,1)$. Now the fact that $f$ is continuous gives you that also $f(0) = f(1) = 0$.
$F(x)=\int_{0}^{1}f(xt)dt=\frac{1}{x}\int_{0}^{x}f(u)du=0$ for all $x\in [0, 1].$ Suppose $f(x)\neq 0$ on $[0,1]$ then there exist a interval $[a,b]\subset [0,1]$ such that $f(x)>0$ for all $x\in [a,b]$. Using above equality we get $\int_{0}^{a}f(x)dx=0$ and $\int_{0}^{b}f(x)dx=0.$ But $0=\int_{0}^{b}f(x)dx-\int_{0}^{a}f(x)dx=\int_{a}^{b}f(x)dx>0.$
