Let: $$\Phi_{mn}=g(x-mD)e^{-inWx}$$ $$g(x)=\frac{1}{\sqrt D}P(\frac{2x}{D}h(x))$$ $$P(x)=\begin{cases} 1 & \quad -1<x\leq 1\\ 0 & \quad \text{else}\\ \end{cases}$$ were $D=1,W=2\pi$ and $$h(x)=\begin{cases} 1 & \quad 0<x\\ 0 & \quad \text{else}\\ \end{cases}$$ Why is it easy to see $\Phi_{mn}$ is not orthogonal?
It is supposed to be a quick and visible without a proof answer in a multiple choice test.
It is clear that $g(x)=1$ on $(0,1/2)$ and $0$ outside this interval.
Thus $g(x-m)=1$ on $(m,m+1/2)$ and $0$ outside this interval.
The basis is not orthogonal because - in particular - dot products (in the hermitian sense)
$\langle \Phi_{m,n}, \Phi_{m,n+1} \rangle$ (with the same $m$) are non zero : they are equal to
$$\ \int_{m}^{m+1/2} e^{2i\pi n x}e^{-2i\pi (n+1)x}dx=\int_{m}^{m+1/2} e^{-2i\pi x}dx=\dfrac{i}{\pi}\neq 0$$
Remark: one can be fooled by the fact that many other dot products are zero. In particular, if $m \neq m'$, all dot products $\langle \Phi_{m,n}, \Phi_{m',n'} \rangle$ are zero (no common support between $[m,m+1/2]$ and $[m',m'+1/2]$).