If $$\widehat{\varphi} = \chi_{[-\frac{1}{2},\frac{1}{2}]}$$ then it is not hard to compute via the inverse Fourier transform that $$\varphi(x) = \frac{\sin(\pi x)}{\pi x}$$ so we need to show $$\langle \varphi(2^{u}x), \varphi(2^vx)\rangle = \int_{\mathbb{R}} \varphi(2^u x) \varphi(2^v x) dx = \int_{\mathbb{R}} \frac{\sin(2^u \pi x)}{\pi 2^u x}\cdot \frac{\sin(2^v \pi x)}{\pi 2^v x}\neq 0$$ for some values of $u,v$. This integral is not very easy to compute for any values of $u,v$, but Mathematica can confirm that for $u = 0$ and $v = 1$, it is not zero.
I am wondering if all the work I did, i.e. computing the inverse FT and needing to compute the integral with the sines is necessary or can I conclude this more directly using the fact that $\varphi$ is the characteristic function of a symmetric interval about the origin? Thanks in advance!
I think you should use that with $\phi_v(x) = \phi(2^v x)$ it holds that $$\langle \phi_v,\phi_u\rangle = \langle\widehat{\phi_v},\widehat{\phi_u}\rangle$$ (by Plancherel). This suggests to compute $\widehat{\phi_u}$ from $\widehat{\phi}$...