I'm going through the last steps of Bourgain and Demeter's proof of the $l^2$ decoupling conjecture, but I'm unable to see how the first inequality in (43) goes through. I'll water down the question a little to make it more transparent.

Let $f_1, \dots, f_n$ be $n$ functions whose Fourier transform is supported on $B(1)$. Bourgain claimed that by the heuristics he and Guth used in their paper we can think that $f_i$ is "essentially constant" on $B_1$. This seems to allow the following "reverse Holder" inequality

$$\prod_{i=1}^n \|f_i\|_{L^p(B_1)}^{1/n}\ll_{n,p} \|\prod_{i=1}^n f_i^{1/n}\|_{L^p(B_1)}. $$

But I skimmed that paper and only found an estimate morally like the following:

$$ f(x)=f*\eta(x)\le \|f\|_{L^1}\|\eta\|_{L^\infty}, $$

where $\hat\eta$ is compactly supported and equals 1 on $B_1$. How do I deduce the reverse Holder inequality from that?

**Added:** Just for reference, the inequality (43) that I referred to is

$$ \prod_{i=1}^n \|\widehat{g_id\sigma}\|_{p,\delta,B_{\delta^{-1/2}}}^{1/n}\ll_{n,p} \|\prod_{i=1}^n (\sum_{\theta\subset\tau_i:\delta^{1/2}\text{ cap}} |\widehat{g_{i,\theta}d\sigma}|^2)^{1/2n}\|_{L^p(w_{B_{\delta^{-1/2}}})}. $$

In the above, the $f_i$ are supposed to be $\widehat{g_{i,\theta}d\sigma}$, and I'm assuming $\tau_i=\text{supp }g_i$ contains only one $\delta^{1/2}$ cap, which is a slab of size $\delta^{1/2}\times\cdots\times\delta^{1/2}\times\delta$. $\tau_i$ are assumed to be $\nu-$transverse, but I feel this is unimportant in this inequality. $w_{B_{\delta^{-1/2}}}$ is just a smooth cutoff weight that equals 1 on $B_{\delta^{-1/2}}$. I have normalized $\delta$ to 1 in the above.