Big $O$ vs Big $\Theta$ I am aware of the big theta notation $f = \Theta(g)$ if and only if there are positive constants $A, B$ and $x_0 > 0$ such that for all $x > x_0$ we have
$$
A|g(x)| \leq |f(x)| \leq B |g(x)|.
$$
What if the condition is the following:
$$
C_1 + A|g(x)| \leq |f(x)| \leq C_2 + B |g(x)|
$$
where $C_1, C_2$ are possibly negative? Certainly more can be said than just $f = O(g)$. Is there a generalized $\Theta$ notation which allows shifts (by, say $C_1, C_2$)? In particular, I'm interested in the special case:
\begin{eqnarray}
-C  \leq f(x) - g(x) \leq C
\end{eqnarray}
for some positive $C$. How does $f$ compare to $g$ in this case? If $f$ and $g$ are positive functions of $x$ which both diverge to $\infty$, is it true that $f(x) = -C + g(x) + \Theta(1)$? What is the appropriate asymptotic notation in this case?
Update Thanks for the clarifying answers. Now here is a slightly harder question. Suppose $f$ is discrete and $g$ is continuous. Suppose further that as $x \to \infty$, the difference $f(x) - g(x)$ is asymptotically bounded in the interval $[-C,C]$ but does not necessarily converge to $0$. Does $f \sim g$ still make sense? Would it be more appropriate to use $\liminf_{x \to \infty} f(x) - g(x) = - C$ and $\limsup_{x \to \infty} f(x) - g(x) = C$?

If $\lim |f(x)| = \lim |g(x)| = \infty$ then there is no difference between your two concepts.
If $f$ is a $\Theta(g)$, then it is a "shifted" $\Theta(g)$ with $C_1 = C_2 = 0$.
If $f$ is a "shifted $\Theta(g)$, then it is a $\Theta(g)$ :
Since $\lim |g(x)| = \infty$, there exists $x_1$ from which $C_2/B \le |g(x)|$ and $ -2C_1/A \le |g(x)|$. This shows that for $x \ge \max(x_0,x_1)$, $A|g(x)|/2 \le C_1 + A|g(x)| \le |f(x)| \le C_2 + B|g(x)| \le 2B|g(x)|$.
If $-C \le f(x)-g(x) \le C$, then this is exactly the same as saying $f-g = O(1)$.
In this case, you have $f = g+O(1)$, and if their limit is $\pm \infty$, $f \sim g$
