Does there exist a dense set of functions $f \in L^{2}([0, 1])$ such that $x^{-1/2}f(x) \in L^{1}([0, 1])$ and $\int_{0}^{1}x^{-1/2}f(x)\, dx = 0$?
I've noticed that $\int_{0}^{1}x^{-1/2}f(x)\, dx = 2\int_{0}^{1}f(x^{2})\, dx$ and so perhaps I should work with $f(x^{2})$ instead. So I tried $f(x) = e^{2\pi i n\sqrt{x}}$, but then I noticed that $x^{-1/2}f(x) \not\in L^{1}$.
If $g\in L^2[0,1]$, with $x^{-1/2}g(x)\in L^1[0,1]$ and $\int_0^1x^{-1/2}g(x)\,dx=1$,
Take $f\in L^2[0,1]$, then for every $\varepsilon>0$, there exists a $g\in C[0,1]$, with $$ \|f-g\|_{L^2}<\varepsilon. $$ Next, there is a $\tau\in (0,1)$, such that, is $h=g\cdot\chi_{[\tau,1]}$, $$ \|g-h\|_{L^2}<\varepsilon. $$ Let $a=\int_\tau^1 x^{-1/2}h(x)\,dx$. Set $w(x)=\beta\chi_{[\delta,\tau]}x^{-1/2+\eta}$.
Then, for suitably small $\eta$ and $\delta$ and suitable $\beta$, we obtain that $$ \int_0^1 x^{-1/2}\big(h(x)+w(x)\big)\,dx=0, $$
while $\|w\|_{L^2}<\varepsilon$.