User blog:Simply Beautiful Art/A better lower bound to f k in the fast growing hierarchy

In this blog post, I will be proving the following lower bound to the fast growing hierarchy:

$$f_{k+1}(n)\ge f_k(n)\uparrow^kn\quad\quad(*)$$

for $$k\ge3$$, by proving

$$f_k^j(n)\ge f_k(n)\uparrow^kj\quad\quad(0)$$

We start with the known lower bound of

$$f_3(n)\ge n((2^n)\uparrow^2n)\ge n\uparrow^2n\quad\quad(1)$$

We will prove $$(0)$$ by induction over $$k$$.

For our base case, we show this holds for $$k=3$$.

We do this by applying induction over $$j$$ in

$$f_3^j(n)\ge f_3(n)\uparrow^3j\quad\quad(2)$$

Trivially $$(2)$$ holds for $$j=1$$. Assume $$(2)$$ holds for some arbitrary $$j=m\ge1$$. We now show $$(2)$$ must hold for $$j=m+1$$.

$$\begin{align}f_3^{m+1}(n)&=f_3(f_3^m(n))\\&\ge f_3^m(n)\uparrow^2f_3^m(n)\\&\ge f_3(n)\uparrow^2f_3^m(n)\\&\ge f_3(n)\uparrow^2(f_3(n)\uparrow^3m)\\&=f_3(n)\uparrow^3(m+1)\end{align}$$

which concludes our base case for $$(0)$$. Assume $$(0)$$ holds for some arbitrary $$k=u\ge3$$. We now show $$(0)$$ must hold for $$k=u+1$$ by applying induction over $$j$$.

Trivially $$(0)$$ holds for $$j=1$$. Assume $$(0)$$ holds for some arbitrary $$j=v\ge1$$. We now show $$(0)$$ must hold for $$j=v+1$$.

$$\begin{align}f_{u+1}^{v+1}(n)&=f_{u+1}(f_{u+1}^v(n))\\&=f_u^{f_{u+1}^v(n)}(f_{u+1}^v(n))\\&\ge f_u(f_{u+1}^v(n))\uparrow^uf_{u+1}^v(n)\\&\ge f_{u+1}(n)\uparrow^uf_{u+1}^v(n)\\&\ge f_{u+1}(n)\uparrow^u(f_{u+1}(n)\uparrow^{u+1}v)\\&=f_{u+1}(n)\uparrow^{u+1}(v+1)\end{align}$$

Q.E.D.