# Show if $f(n)$ has polynomial growth and $g(n)=\Theta(f(n))$, then $g(n)$ also has polynomial growth

As stated in the question title, if $$f(n)$$ has polynomial growth and $$g(n)=\Theta(f(n))$$, then how can we show $$g(n)$$ also has polynomial growth?

$$g(n)=\Theta(f(n))$$ gives us $$0\leq c_1f(n)\leq g(n)\leq c_2f(n)$$ for some positive constants $$c_1, c_2$$ for some $$n>n_0>0$$. But how can I conclude $$g(n)$$ also satisfies the polynomial growth condition?

A function $$f(n)$$ defined on all sufficiently large positive reals satisfies the polynomial-growth condition if there exists a constant $$\hat{n} > 0$$ such that the following holds: for every constant $$\phi \geq 1$$, there exits a constant $$d>1$$ (depending on $$\phi$$) such that $$f(n)/d \leq f(\psi n) \leq d f(n)$$ for all $$1 \leq \psi \leq \phi$$ and $$n \geq \hat{n}$$.

Let us consider only sufficiently large values of $$n$$. As you have pointed out there are two constants $$c_1, c_2$$ with $$0< c_1 \le c_2$$ such that: $$c_1 f(n) \le g(n) \le c_2f(n).$$
Also, since $$f$$ has polynomial growth, for every $$\phi \ge 1$$ there exists some $$d_\phi > 1$$ such that: $$\frac{1}{d_\phi} f(n) \le f(\psi n) \le d_\phi f(n) \quad \forall 1 \le\psi \le \phi.$$
Then the following holds for all $$1 \le\psi \le \phi$$: $$\frac{c_1}{c_2 d_\phi} g(n) \le \frac{c_1 }{d_\phi} f(n) \le c_1 f(\psi n) \le g(\psi n) \le c_2 f(\psi n) \le c_2 d_\phi f(n) \le \frac{c_2 d_\phi}{c_1} g(n).$$ Which means that $$g$$ also satisfies the polynomial growth condition. In particular, for a given $$\phi$$, we can choose $$d'_\phi = \frac{c_2 d_\phi}{c_1}$$ to obtain: $$\frac{1}{d'_\phi} g(n) \le g(\psi n) \le d'_\phi g(n) \quad \forall 1 \le\psi \le \phi,$$ where $$d'_\phi > 1$$ holds since $$c_2 \ge c_1$$ and $$d_\phi > 1$$.