The question is about the proof of Corollary 2.2.5 in Grafakos's book "Modern Fourier Analysis". And the statement can be writen as follows:
Let $\phi,\psi \in \mathcal{S}(\mathbb{R}^n)$ and the Fourier transform of $\psi$ is supported in some annulus, say $\{1/2\le |\xi|\le 2\}$. Then let $0<r<\infty$. We have for all tempered distribution $f\in \mathcal{S}^{'}(\mathbb{R}^n)$ and all $t>0$, we have
$$ |\phi_t * \Delta_j^{\psi}(f)(x)| \le C(n,r,\phi) M(|\Delta_j^{\psi}(f)|^r)^{1/r}(x)\tag{1} $$ where $M$ is the Hardy-Littlewood maximal operator and $\Delta_j^{\psi}(f)$ is the Littlewood-Paley operator associated to $\psi$ i.e. $\Delta_j^{\psi}(f)(x) = (\psi_{2^{-j}}*f)(x)$.
The proof is based on Lemma 2.2.3 which states that:
Let $0<r<\infty$. Then for all $t>0$ and all $C^1$ functions $u$ on $\mathbb{R}^n$ whose Fourier transform is supported in the ball $B(0,t)$, then we have $$ \sup_{z\in \mathbb{R}^n} \frac{|u(x-z)|}{(1+t|z|)^{n/r}} \le C(n,r) M(|u|^r)^{1/r}(x)\tag{2} $$
Then the idea is that we can bound the LHS of ${(1)}$ by $$ \int |\Delta_j^{\psi}(f)(x-z)|t^{-n}|\phi(z/t)|dz \le C(\phi,N)\int \frac{|\Delta_j^{\psi}(f)(x-z)|}{(1+t^{-1}|z|)^{n/r}}t^{-n}(1+t^{-1}|z|)^{n/r-N}dz $$
$$ \le C(\phi,N) \sup_{z}\frac{|\Delta_j^{\psi}(f)(x-z)|}{(1+t^{-1}|z|)^{n/r}} \int \frac{t^{-n}}{(1+t^{-1}|z|)^{N-n/r}}dz $$ then choose $N=n/r+n+1$ to make the integral convergent and by $(2)$ we get $$ |\phi_t*\Delta_j^{\psi}(f)(x)|\le C(n,r,\phi)\sup_{z}\frac{|\Delta_j^{\psi}(f)(x-z)|}{(1+t^{-1}|z|)^{n/r}}\le C(\phi,n,r)M(|\Delta_j^{\psi}(f)|^r)^{1/r}(x)\tag{3} $$
And my question is that, we know $\Delta_j^{\psi}(f)$ is $C^1$ and has Fourier transform supprted in $B(0,2^{j+1})$ then by $(2)$, we can get $$ \sup_{z}\frac{|\Delta_j^{\psi}(f)(x-z)|}{(1+2^{j+1}|z|)^{n/r}}\le C(n,r)M(|\Delta_j^{\psi}(f)|^r)^{1/r}(x) $$ but how can we pass $2^{j+1}$ to any positive $t^{-1}$ with the constant in $(3)$ still only depends on $n,r,\phi.$ In particular, what if $t^{-1}<2^{j+1}$.
Many thanks in advance.