I want to prove that if $(u_j)_{j=1}^\infty$ is a complete orthonormal system in $L_2(a,b)$, then $(\overline{u_j})_{j=1}^\infty$ is also a complete orthonormal system in $L_2(a,b)$. Any ideas on how to approach this proof?
Proving $(\overline{u_j},\overline{u_i})=\delta_{ij}$ seems to be simple: $$(\overline{u_j},\overline{u_i})=\int_a^b\overline{u_j}(t)\overline{\overline{u_i}}(t)dt=\int_a^bu_i(t)\overline{u_j}(t)dt=(u_i,u_j)=\delta_{ij}.$$ Any ideas on how to prove completeness?
Suppose that $(f,\bar{u}_j)=0$ for all $j$. Then by definition $$ \int_a^bf(x)u_j(x)\;dx=0 $$ for all $j$, and taking the complex conjugate of the integral we get $$ 0=\overline{\int_a^bf(x)u_j(x)\;dx}=\int_a^b\overline{f(x)}\overline{u_j(x)}\;dx=(\bar{f},u_j) $$ for all $j$.
Since $\{u_j\}$ is by hypothesis an orthonormal basis, this implies that $\bar{f}=0$ almost everywhere, hence $f=0$ almost everywhere. Therefore $\{\bar{u}_j\}$ is an orthonormal basis as well.