No, that would be incorrect.

Perhaps it is the (arguably) sloppy condition $\varepsilon > 0$ which is the root of your confusion. It is supposed to mean that $\varepsilon$ is a real positive number and constant (with respect to $n$). If we allow $f \in \omega(n^{\log_b a})$, for instance, then for $a = b = 2$ we would have $f(n) = n \log n \in \Theta(n)$ as a consequence of the theorem, which is false.

On a side note, you write "$f$ must be asymptotically less than or equal to $n^{1.999\ldots}$". This does mean that $f$ must be bounded by $n^2$, though I believe not for the reasons you think it does. Recall that $1.999\ldots = 2$, so that statement is rather trivial...

No, that would be incorrect.

Perhaps it is the condition $\varepsilon > 0$ which is the root of your confusion. It is supposed to mean that $\varepsilon$ is a real positive number and constant (with respect to $n$). If we allow $f \in \omega(n^{\log_b a})$, for instance, then for $a = b = 2$ we would have $f(n) = n \log n \in \Theta(n)$ as a consequence of the theorem, which is false.

On a side note, you write "$f$ must be asymptotically less than or equal to $n^{1.999\ldots}$". This does mean that $f$ must be bounded by $n^2$, though I believe not for the reasons you think it does. Recall that $1.999\ldots = 2$, so that statement is rather trivial...

No, that would be incorrect.

Perhaps it is the (arguably) sloppy condition $\varepsilon > 0$ which is the root of your confusion. It is supposed to mean that $\varepsilon$ is positive and constant (with respect to $n$). If we allow $f \in \omega(n^{\log_b a})$, for instance, then for $a = b = 2$ we would have $f(n) = n \log n \in \Theta(n)$ as a consequence of the theorem, which is false.

On a side note, you write "$f$ must be asymptotically less than or equal to $n^{1.999\ldots}$". This does mean that $f$ must be bounded by $n^2$, though I believe not for the reasons you think it does. Recall that $1.999\ldots = 2$, so that statement is rather trivial...

No, that would be incorrect.

Perhaps it is the (arguably) sloppy condition $\varepsilon > 0$ which is the root of your confusion. It is supposed to mean that $\varepsilon$ is a real positive number and constant (with respect to $n$). If we allow $f \in \omega(n^{\log_b a})$, for instance, then for $a = b = 2$ we would have $f(n) = n \log n \in \Theta(n)$ as a consequence of the theorem, which is false.

On a side note, you write "$f$ must be asymptotically less than or equal to $n^{1.999\ldots}$". This does mean that $f$ must be bounded by $n^2$, though I believe not for the reasons you think it does. Recall that $1.999\ldots = 2$, so that statement is rather trivial...