Critical Homogeneous Sobolev Embedding For $s\in\mathbb{R}$, $1<p<\infty$, define the homogeneous Sobolev space $\dot{W}^{s,p}(\mathbb{R}^{n})$ as follows: For $f\in\mathcal{S}_{0}(\mathbb{R}^{n})$ (Schwartz functions with Fourier transforms supported away from origin), we define the homogeneous Sobolev norm $\|f\|_{\dot{W}^{s,p}(\mathbb{R}^{n})}$ by
    $$\|(2\pi|\xi|^{s}\widehat{f})^{\vee}\|_{L^{p}(\mathbb{R}^{n})}=\||\nabla|^{s}f\|_{L^{p}(\mathbb{R}^{n})}$$
We define $\dot{W}^{s,p}(\mathbb{R}^{n})$ as the closure of all $f$ under this norm. I am interested in the following critical Sobolev embedding:

For $1<p<\infty$ and $s=n/p$, if $f\in\dot{W}^{s,p}(\mathbb{R}^{n})$,
  then $f\in BMO(\mathbb{R}^{n})$ with    $$\|f\|_{BMO(\mathbb{R}^{n})}\lesssim_{n,p}\|f\|_{\dot{W}^{s,p}(\mathbb{R}^{n})}$$

One proof of this result is via Riesz potentials and goes something as follows. Using the fact that for $0<s<n$, the Fourier transform of the distribution $|x|^{-n+s}$ is a scalar multiple of $(2\pi|\xi|)^{-s}=\widehat{|\nabla|^{-s}}$, it suffices to show that the Riesz potential satisfies the inequality
    $$\|I_{s}f\|_{BMO}\leq\|f\|_{L^{n/s}},\quad\forall f\in\mathcal{S}_{0}(\mathbb{R}^{n})$$
Indeed, if $f\in\mathcal{S}_{0}$, then $|\nabla|^{s}f\in\mathcal{S}_{0}$, whence
    $$\|f\|_{BMO}=\|I_{s}|\nabla|^{s}f\|_{BMO}\lesssim_{n,s}\||\nabla|^{s}f\|_{L^{n/s}}=\|f\|_{\dot{W}^{s,n/s}}$$
By translation invariance, it suffices to show that there exists a constant $A>0$ such that for all cubes $Q$ centered at the origin, there exists a constant $c_{Q}\in\mathbb{C}$
    $$\int_{Q}|I_{s}(f)(y)-c_{q}|dy\leq A|Q|$$
We then have that the infimum of all such $A$ is comparable to $\|I_{s}f\|_{BMO}$. To do this, we let $Q^{*}$ denote the enlargement of a cube $Q$ by a factor of $2\sqrt{n}$ (i.e. $l(Q^{*})=2\sqrt{n}l(Q)$). We write $f=f_{1}+f_{2}$, where $f_{1}=f\chi_{Q^{**}}$.
Choose $1<p_{0}<p$ and let $q_{0}$ be such that $q_{0}^{-1}=p_{0}^{-1}-(s/n)$. By the Hardy-Littlewood-Sobolev inequality together with Holder's inequality, we have the estimate
\begin{align*}
\dfrac{1}{|Q|}\int_{Q}|I_{s}f_{1}|dx\leq\left(\dfrac{1}{|Q|}\int_{Q}|I_{s}f_{1}|^{q_{0}}dx\right)^{1/q_{0}}&\leq|Q|^{-1/q_{0}}\left(\int_{\mathbb{R}^{n}}|I_{s}f_{1}|^{q_{0}}dx\right)^{1/q_{0}}\\
&\lesssim_{n,p_{0},s}|Q|^{-1/q_{0}}\left(\int_{Q^{*}}|f_{1}(x)|^{p_{0}}dx\right)^{1/p_{0}}\\
 &\lesssim_{n,p,s}|Q^{*}|^{\frac{1}{p_{0}}-\frac{1}{q_{0}}}\left(\dfrac{1}{|Q^{**}|}\int_{Q^{*}}|f(x)|^{p}dx\right)^{1/p}\\
 &\lesssim_{n,p,s}|Q|^{\frac{1}{p_{0}}-\frac{1}{q_{0}}-\frac{1}{p}}\|f\|_{L^{p}}
\end{align*}
where we make use of Holder's inequality. Since $\frac{1}{p_{0}}-\frac{1}{q_{0}}-\frac{1}{p}=\frac{s}{n}-\frac{1}{p}=0$, we obtain the desired estimate.
Now take
$$c_{Q}:=I_{s}(f_{2})(0)=\int_{(Q^{*})^{c}}f(y)|y|^{-s}dy$$
Since for $x\in Q$ and $y\in (Q^{*})^{c}$, we have that $|x-y|\geq $, we obtain from the mean value theorem that
\begin{align*}
|I_{s}(f_{2})(x)-c_{Q}|&\leq\int_{(Q^{*})^{c}}|f(y)|\left|\dfrac{1}{|x-y|^{s}}-\dfrac{1}{|y|^{s}}\right|dy\\
&\lesssim_{s}\int_{(Q^{*})^{c}}|f(y)|\cdot\dfrac{|x|}{|y|^{n-s+1}}dy\\
&\leq l(Q)\left(\int_{(Q^{*})^{c}}|f(y)|^{\frac{n}{s}}dy\right)^{s/n}\left(\int_{(Q^{*})^{c}}|y|^{-p'(n-s+1)}dy\right)^{1/p'}\\
&\leq l(Q)\|f\|_{L^{n/s}}\cdot l(Q)^{\frac{n}{p'}}\cdot l(Q)^{-n+s-1}\left(\int_{|y|\geq 2\sqrt{n}}|y|^{-p'(n-s+1)}dy\right)^{1/p'}\\
&=C_{n,s}\|f\|_{L^{n/s}}
\end{align*}
where we use Holder's inequality and dilation invariance above. By the triangle inequality, we conclude that
$$\dfrac{1}{|Q|}\int_{Q}|I_{s}(f)(x)-c_{Q}|dx\lesssim_{n,s}\dfrac{1}{|Q|}\int_{Q}|I_{s}(f_{1})(x)|dx+\dfrac{1}{|Q|}\int_{Q}|I_{s}(f_{2})(x)-c_{Q}|dx\lesssim_{n,s}\|f\|_{L^{n/s}},$$
which completes the proof.
In some comments which can be found here, Terence Tao says that it's possible to prove this critical Sobolev embedding via Littlewood-Paley theory and references some notes of his which may be helpful. I took a look at them, and the only result which seemed to be useful was that if a Schwartz function has compactly supported Fourier transform, then $\|f\|_{L^{\infty}}\approx\|f\|_{BMO}$. One can apply this result to individual Littlewood-Paley projections $P_{k}f$, but I don't right now how to put this individual estimates together to obtain an estimate for $\|f\|_{BMO}$ in terms of $\|f\|_{L^{n/s}}$.
 A: So, I believe that I have a proof; I'm not sure if it's what Tao has in mind, though. I would appreciate it if someone would take a look, as I made several false attempts before, and my eyes may be too tired to catch my error this attempt.
Let $\phi$ be smooth function which is $\equiv 1$ on the ball $B_{1}(0)$ and supported in the ball $B_{2}(0)$. Set $\psi(\xi):=\psi(\xi)-\psi(2\xi)$ and $\psi_{j}(\xi)=\psi(\xi/2^{j})$, for $j\in\mathbb{Z}$ to obtain a Littlewood-Paley partition of unity.
For $f\in\mathcal{S}(\mathbb{R}^{n})$, we write $f=\sum_{j}\psi_{j}f=\sum_{j}P_{j}f$. Let $B$ be a ball of radius $r>0$. By translation invariance, it suffices to consider the case where $B$ is centered at the origin.
We split $f$ into a low frequency piece and a high frequency piece by $f=P_{\leq C}f+P_{>C}f$, where $C\in\mathbb{R}$ is a parameter to be determined later, and use the triangle inequality to obtain
    $$\dfrac{1}{|B|}\int_{B}|f-f_{B}|dx\leq2\dfrac{1}{|B|}\int_{B}|P_{> C}f|dx+\dfrac{1}{|B|}\int_{B}|P_{\leq C}f-(P_{\leq C}f)_{B}|dx,$$
where the subscript $B$ denotes the average over $B$. For the first term, Holder's inequality together with Bernstein's lemma gives us the estimate
    \begin{align*}
\dfrac{1}{|B|}\int_{B}|P_{> C}f|dx\leq\|P_{>C}f\|_{L^{p}}|B|^{-1/p}&\lesssim_{n,s}2^{-Cs}\||\nabla|^{s}P_{>C}f\|_{L^{p}}|B|^{-1/p}\\
&\lesssim_{n} 2^{-Cs}|B|^{-1/p}\||\nabla|^{s}f\|_{L^{p}}
\end{align*}
where the last inequality follows from Young's inequality. Since $1/p=s/n$, $|B|^{-1/p}=_{n,s}r^{-s}$. We choose $C$ so that $(2^{C}r)\sim 1$.
For the second piece, we have by the mean value theorem that
\begin{align*}
\dfrac{1}{|B|}\int_{B}|P_{\leq C}f-(P_{\leq C}f)_{B}|dx&\lesssim\dfrac{1}{|B|^{2}}\int_{B}\int_{B}|\nabla(P_{\leq C}f)(y)|\cdot|x|dydx\\
\end{align*}
Now observe that
\begin{align*}
\left|\nabla(P_{\leq C}f)(y)\right|&=\left|\int_{\mathbb{R}^{n}}(2\pi i\xi)\phi(\xi/2^{C})\widehat{f}(\xi)e^{2\pi i \xi\cdot y}d\xi\right|\\
&\leq\int_{\mathbb{R}^{n}}|2\pi \xi||\phi(\xi/2^{C})||\widehat{f}(\xi)|d\xi\\
&=\int_{\mathbb{R}^{n}}|2\pi \xi|^{1-s}|\phi(\xi/2^{C})||\widehat{|\nabla|^{s}f}(\xi)|d\xi\\
&=\int_{|\xi|\leq 2^{C+1}}|2\pi\xi|^{1-s}|\widehat{(P_{\leq C}|\nabla|^{s}f)}(\xi)|d\xi\\
\end{align*}
Choose $q\geq 2$ so that $q'(1-s)>-n$. This is possible since $q'(1-s)>q'(1-n)$. Then by Holder's inequality, the last expression above is $\leq$
\begin{align*}
\left(\int_{|\xi|\leq 2^{C+1}}|2\pi\xi|^{q'(1-s)}d\xi\right)^{1/q'}\|\widehat{P_{\leq C}|\nabla|^{s}f}\|_{L^{q}}&\lesssim_{n,q}\left(\int_{|\xi|\leq 2^{C+1}}|2\pi\xi|^{q'(1-s)}d\xi\right)^{1/q'}\|P_{\leq C}|\nabla|^{s}f\|_{L^{q'}}\\
&=2^{C\frac{n}{q'}}2^{C(1-s)}\left(\int_{|\xi|\leq 2}|2\pi\xi|^{q'(1-s)}d\xi\right)^{1/q'}\|P_{\leq C}|\nabla|^{s}f\|_{L^{q}}
\end{align*}
by Hausdorff-Young inequality and dilation invariance. By Bernstein's lemma and Young's inequality,
$$\|P_{\leq C}|\nabla|^{s}f\|_{L^{q'}}\lesssim_{n,s,p}2^{C(\frac{n}{p}-\frac{n}{q'})}\||\nabla|^{s}f\|_{L^{p}}=2^{Cs-C\frac{n}{q'}}\||\nabla|^{s}f\|_{L^{p}}$$
Substituting this estimate in, we obtain that
$$|\nabla( P_{\leq C}f)(y)|\lesssim_{n,s,p}2^{C}\||\nabla|^{s}f\|_{L^{p}}$$
and therefore
\begin{align*}
\dfrac{1}{|B|}\int_{B}|P_{\leq C}f-(P_{\leq C}f)_{B}|dx\lesssim_{n,s,p}\dfrac{1}{|B|}\int_{B}|x|2^{C}\||\nabla|^{s}f\|_{L^{p}}dx\leq 2^{C}r\||\nabla|^{s}f\|_{L^{p}}&\sim\||\nabla|^{s}f\|_{L^{p}}
\end{align*}
