Dense subset of $L^2(\mathbb R)$ We know that continuous functions with compact support is dense in $L^2(\mathbb R)$, but what if we consider continuous functions with compact support which have the additional property that $\int_{\mathbb R}\exp(izx)f(x)dx=0$, where $z$ is a fixed nonreal complex number? 
Are they dense in $L^2(\mathbb R)$? My book just mentions this fact and offers no explanation. I have tried to prove this, for example, via approximation by trigonometric polynomials on a finite interval, but that does not seem to work.
 A: Take $f\in L^2(\mathbb R)$ and $\delta>0$ fixed. Let $f_1$ continuous with compact support such that $||f-f_1||_{L^2(\mathbb R)}<\delta/2$. We have to find $f_2$, continuous with compact support and such that $||f_1-f_2||\leq \delta/2$ and $\int_{\mathbb R}e^{izx}f_2(x)=0$. Put $\alpha:=\int_{\mathbb R}e^{izx}f_1(x)dx$. If $\alpha=0$ take $f_2=f_1$. If we can find $h$ continuous with compact support such that $\int_{\mathbb R}e^{izx}h(x)dx=\alpha$ and $||h||_{L^2}\leq\delta/2$ we are done, since we will put $f_1-f_2=h$ hence $f_2=f_1+h$. Define $h_1(x)$ by the relation $h(x)=e^{-izx}\alpha h_1(x)$. Then we want $\int_{\mathbb R}h_1(x)dx=1$ and $||h_1||_{L^2}\leq \alpha\delta/2$.
To do that, consider a continuous function with compact support $\phi$ whose integral is $1$, then put $\phi_n(x)=\frac 1n\phi\left(\frac xn\right)$. Then $\int_{\mathbb R}\phi_n(x)dx=1$ and $\int_{\mathbb R}\phi_n(x)^2dx=\frac 1n||\phi||^2_{L^2}$, so just choose $n_0$ such that $\frac 1{n_0}||\phi||^2_{L^2}\leq \sqrt{|\alpha|}\frac{\delta^2}4$ and put $h_1=\phi_{n_0}$.
