Let X,Y be two, iid random variables. I want to demonstrate that the characteristic function of their difference is non-negative ($\phi_{X-Y}(t)\geq 0$). Below is what I've worked, does this proof hold to demonstrate non-negativity?
$$ \phi_{X-Y}(t) = E(e^{itX}e^{-itY}) = E( [cos(tX) + isin(tx)][cos(-tY) + isin(-tY)]) $$
$$ = E[cos(tX)cos(tY) - isin(tY)cos(tX) + isin(tX)cos(tY) + sin(tX)sin(tY)] $$
remembering that $cos(x)=cos(-x)$ and $sin(x)=-sin(-x)$. Then by linearity of expectations and given X,Y are $iid$:
$$ = E[cos(tX)]E[cos(tX)] - E[isin(tX)]E[cos(tX)] + E[isin(tX)]E[cos(tX)] + E[sin(tX)]E[sin(tX)] $$
$$ = E^2(cos(tX)) + E^2(sin(tX)) \geq 0 $$
Additional question, our professor has asserted that more simply $\phi_X(t)\phi_X(-t) = |\phi_x(t)|^2$, which is then obviously non-negative, but I am unsure how to demonstrate this equivalence to the $L^2$ norm.
$$ \phi_{X-Y}(t) \stackrel{\perp}{=} \phi_X(t)\phi_Y(-t) \stackrel{iid}{=} \phi_X(t)\phi_X(-t) $$
Let's make a bit of clarity.
1) If we have $X,Y$ i.i.d. real random variables than it holds:
$\phi_X(-t)=E[e^{-itX}]=E[\overline{e^{itX}}]=\overline{E[{e^{itX}}]}=\overline{\phi_X(t)}$
So we have:
$\phi_{X-Y}(t)=E[e^{it(X-Y)}]=E[e^{itX}e^{-itY}]=E[e^{itX}]E[e^{-itY}]=$
$=E[e^{itX}]E[e^{-itX}]=\phi_X(t)\phi_X(-t)=\phi_X(t)\overline{\phi_X(t)}=|\phi_X(t)|^2$
so that the charachteristic function is real and non-negative. Note that $L^2$ norm here is not involved, just the standard norm of a complex number.
2) If we use the Euler expansion $e^{itX}=cos(tX)+isin(tX)$:
$\phi_X(t)=E[cos(tX)+isin(tX)]=E[cos(tX)]+iE[sin(tX)]$
and therefore:
$|\phi_X(t)|^2=(E[cos(tX)])^2+(E[sin(tX)])^2$
One could have derived this expression also the way the OP did (and his derivation looks correct to me).