How to prove this integration is not zero Let {$f_{n}$}$_{n=1}^{\infty}$ be a sequence of non-zero elements of $L^{2}[0,1]$. Prove that there is a function $g\in L^{2}[0,1]$ such that for all $n\ge1$ we have $\int_{0}^{1}g(x)f_{n}(x)dx\neq0$.
I try to assume there isn't such function $g$ and get contradiction. But I couldn't find what's the contradiction. I can show all $f_{n}=0 \quad a.e$. 
Maybe my idea is not correct and maybe we can construct a function $g$ satisfying $\int_{0}^{1}g(x)f_{n}(x)dx\neq0$.

Note that  
$$U_n = \left\{ f\in L^2([0,1]) : \int_0^1 f f_n dx\neq 0 \right\}$$
is an nonempty (since $f_n$ is nonzero) open sets which is dense in $L^2 ([0,1])$. The Baire Category theorem says that 
$$ \bigcap U_n$$
is nonempty. Thus there is $g\in L^2([0,1])$ so that 
$$ \int_0^1 g f_n dx\neq 0$$
for all $n$. 
