Can cross-c.(x,y) be positive if signals x and y do not share frequencies? The question is if the cross-correlation of two signals $x$ and $y$ can be non-zero even if $x$ and $y$ do not share frequencies.

My approach is the following:
The cross-correlation  $\rho_{xy}\left(\tau\right):=x\left(\tau\right)\star y\left(\tau\right)$  can be expressed in terms of a convolution:

$$
\begin{array}{rcl}
\rho_{xy}\left(\tau\right) &amp; = &amp; x\left(\tau\right)\star y\left(\tau\right)\\
 &amp; = &amp; \displaystyle\sum_{t=-\infty}^{^{\infty}}x\left(t\right)\bar{y}\left(t+\tau\right))\\
 &amp; = &amp; \left(\bar{x}\right)\left(-\tau\right)*y\left(\tau\right)
\end{array}$$

i.e. the cross-correlation of $ x\left(\tau\right) $ and $ y\left(\tau\right) $ equals the convolution of $ \bar{x}\left[-\tau\right] $ and $ y\left(\tau\right).$ 

The convolution theorem  $\mathcal{F}\left(x*y\right)=\mathcal{F}\left(x\right)\mathcal{F}\left(y\right) $ allows us to express the cross-correlation in terms of it's frequency content and replace the convolution by multiplication: $$ \mathcal{F}\left(\rho_{xy}\right)\left[\omega\right]=\mathcal{F}\left(\left(\bar{x}\right)\left(-\tau\right)*y\right) \left[\omega\right]=\bar{\mathcal{F}\left(x\right)}\left[\omega\right]\mathcal{F}\left(y\right)\left[\omega\right] $$
 Due to Parseval's Theorem, signal energy in frequency and time domain corresponds to: $$ \sum_{\omega=-\infty}^{\infty}\left|\mathcal{F}\left(\rho_{xy}\right)\left[\omega\right]\right|^{2}=\sum_{\tau=-\infty}^{\infty}\left|\rho_{xy}\left(\tau\right)\right|^{2}.$$

Hence the energy of the cross-correlation of the signals corresponds in time and frequency dimension. Thus, unless two different signals share frequencies, they cannot interfere. Two orthogonal signals, i.e. $\left\langle x,y\right\rangle =0 $, can have shared frequencies without having a non-zero cross-correlation.
The solution to the self-proposed problem was correct, as it stands.