My question is about the proof of the corollary 3.3 page 21 of this paper.
Let $(E,d,\mu)$ be a measured metric space with $\mu$ doubling. We write $\mathcal{M}$ the Hardy-Littlewood maximal operator. Let $1<q\leq\infty$ and $a>1$ be fixed. Let $F,G\in L^1_{loc}(E)$ be non-negative.
Definition : We says $(F,G)\in\mathcal{E}_{q,a}$ if for every ball $B$ we can find non-negative measurable functions $G_B, H_B$ defined on $B$ with $$ F \leq G_B + H_B ~ \text{a.e. on}\, B $$ such that $$ \begin{split} \left( \frac{1}{\mu(B)}\int_B (H_B)^q\, d\mu \right)^{1/q} &\leq a \inf_{x\in B} \mathcal{M} F(x) + \inf_{x\in B} G(x), \\ \frac{1}{\mu(B)}\int_B G_B\, d\mu &\leq \inf_{x\in B} G(x). \end{split} $$
Let $1<\rho$ and let $F,G\in\mathcal{E}_{q,a}$ such that $\|F\|_1<\infty$ and $\|G\|_\rho<\infty$. In this proof we define a function $$ \Phi(t) = \rho\int_0^t \lambda^{\rho-1}\mu\{\mathcal{M}F>\lambda\}\,d\lambda $$ for $t\geq 0$. Also after using a precedent proposition we show the innequality $$ \Phi(Kt) \leq \frac{1}{2} \Phi(t) + \left( \frac{K}{\gamma} \right)^\rho \lVert G \rVert_\rho^\rho $$ where $K$ and $\gamma\leq 1$ are some constants. And then the author says that "An easy iteration shows that $\Phi$ is bounded". I don't see what kind of iteration we have to do.
The iteration that the paper is referring to is that you can apply the inequality iteratively. You can start with $\Phi(K^n t)$ and apply the inequality over and over until you get to $\Phi(t)$, and the form of the inequality makes it so that this is bounded even as $n \rightarrow \infty$. Since $K^nt$ gets as big as you want, this gives boundedness. A more precise answer is below.
To make the inequality simpler, we can just take $C$ to be a constant and say that $\Phi(Kt) \leq \tfrac{1}{2} \Phi(t) + C$ for any $t \geq 0$. Now, consider the sequence $\Phi(1), \Phi(K), \Phi(K^2)$, etc. You can approximate $\Phi(K^2)$ by applying the inequality twice: $$ \Phi(K^2) \leq \tfrac{1}{2}\Phi(K) + C \leq \tfrac{1}{2} (\tfrac{1}{2} \Phi(1) + C) + C = \tfrac{1}{4} \Phi(1) + \tfrac{3}{2}C $$ By continuing like this, you can approximate each $\Phi(K^n)$. By iteration on $n$, you can show that $$ \Phi(K^n) \leq \tfrac{1}{2^n} \Phi(1) + C(2-\tfrac{1}{2^n}). $$ Approximating more loosely, we have $\Phi(K^n) \leq \Phi(1) + 2C$ for each $n$. Since presumably $K>1$ (the paper says we can choose $K$ "large enough"), and since $\Phi$ is increasing (if I understand correctly), this is enough to conclude that $\Phi$ is bounded. (If I've misunderstood and $\Phi$ is not increasing, it is also enough to show that $\lim_{t \rightarrow \infty} \Phi(t)$ exists.)