By definition $\vartheta(x) \leq \psi(x). \hspace{20mm}(1) $
and also we know
$\frac{\psi(x)}{x} \sim \frac{\vartheta(x)}{x},\hspace{45mm}(2) $
Littlewood's result that, for large x, successively
$\pi(x) - Li(x) < -K\frac{\sqrt{x}\log\log\log x}{\log x},$
$\pi(x) - Li(x) > K \frac{\sqrt{x} \log\log\log x}{\log x} $
is also expressed with suitable changes in terms of
$\psi(x) -x.$
My confusion is the following. If $\psi(x) < x $ then certainly $\vartheta(x) < x$ by (1). If $\psi(x) > x$ (1) still holds as does (2), but it seems that $\vartheta(x)$ could satisfy both (1),(2) and still be less than x. Meaning that we could say $\vartheta(x) < x $ infinitely often but not necessarily that it was ever greater than x.
So does $\vartheta(x) - x $ also change sign infinitely often or not necessarily?
Thanks for any clarification.
It is known that $\theta(x)-x$ does change sign infinitely often. I agree that it's not an immediate consequence of the facts you've listed above. However, Littlewood's proof has forms that apply to $\psi$, $\theta$, and $\pi$.
In fact, the oscillations of $\psi(x)-x$ are as large as a constant times $\sqrt x \log\log\log x$ (as you allude to above); and it's easy to see that $\psi(x)-\theta(x)$ is always at most a constant times $\sqrt x$. Combining these two facts gives us oscillations of $\theta(x)-x$ that are also as large as a constant times $\sqrt x \log\log\log x$.