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\topmatter
\title
Limiting embedding theorems for $W^{s, p}$ when $s\uparrow 1$
and applications
\endtitle
\author
Jean Bourgain, Haim Brezis
 and Petru Mironescu
\endauthor
\NoRunningHeads
\endtopmatter
\document

\bigskip
\centerline{Dedicated to the memory of T.~Wolff}
\bigskip
\subhead
1. Introduction
\endsubhead

This is a follow-up of our paper [3] where we establish that
$$
\lim\limits_{s\uparrow 1} (1-s) \int_\Omega\int_\Omega
\frac{|f(x)-f(y)|^p}{|x-y|^{d+sp}}dxdy \sim\Vert\nabla
f\Vert_{L^p(\Omega)},\tag 1
$$
for any $p\in [1, \infty)$, where $\Omega$ is a smooth bounded domain in
$\Bbb R^d, d\geq 1$.

On the other hand, if $0<s<1$, $p>1$ and $sp<d$, the Sobolev inequality for
fractional Sobolev spaces (see e.g. [1], Theorem 7.57 or [6], Section 3.3) 
asserts that
$$
\Vert f\Vert^p_{W^{s, p}(\Omega)} \geq C(s, p, d)\Vert f-\hbox{$\notint$}
f\Vert_{L^q(\Omega)}
\tag 2
$$
where
$$ 
\frac 1q=\frac 1p -\frac sd.\tag 3
$$
Here we use the standard semi-norm on $W^{s, p}$
$$
\Vert f\Vert^p_{W^{s, p}(\Omega)}=\int_\Omega\int_\Omega
\frac{|f(x) -f(y)|^p }{|x-y|^{d+sp}} dxdy.\tag 4
$$

When $s=1$ the analog of (2) is the classical Sobolev inequality
$$
\Vert\nabla f\Vert^p_{L^p(\Omega)} \geq C(p, d)\Vert f-\hbox{$\notint$}
f\Vert^p_{L^{p^*}(\Omega)}\tag 5
$$
where
$$
\frac 1{p^*} =\frac 1p -\frac 1d \ \text { and } \ 1\leq p< d.
$$
The behaviour of the best constant $C(p, d)$ in (5) as $p\uparrow d$ is known
(see e.g. [5], Section 7.7 and also Remark 1 below); more precisely one has
$$
\Vert\nabla f\Vert^p_{L^p(\Omega)} \geq C(d)(d-p)^{p-1} \Vert f-\hbox{$
\notint$} f\Vert_{L^{p^*}(\Omega)}.\tag 6
$$
Putting together (1), (4) and (6) suggests that (2) holds with
$$
C(s, p, d)= C(d)(d-sp)^{p-1}/(1-s),\tag 7
$$
for all $s<1$, $s$ close to 1 and $sp<d$.

This is indeed our main result.
For simplicity we work with $\Omega =$  the unit cube $Q$ in $\Bbb R^d$.

\proclaim
{Theorem 1}
Assume $d\geq 1, p\geq 1, 1/2 \leq s<1$ and $sp<1$.
Then
$$
\int_Q\int_Q \frac{|f(x)-f(y)|^p}{|x-y|^{d+sp}}dxdy\geq C(d)
\frac{(d-sp)^{p-1}}{1-s} \Vert f-\hbox{$\notint$} f\Vert^p_{L^{q}(Q)}\tag 8
$$
where $q$ is given by (3) and $C(d)$ depends only on $d$.
\endproclaim

As can be seen from (8) there are two phenomena that govern the behaviour of
the constant in (8).
As $s\uparrow 1$ the constant gets bigger, while as $s\uparrow d/p$ the
constant deteriorates.
This explains why the we consider several cases in the proof.

As an application of Theorem 1 with $p=1$ and $f=\chi_A$, the characteristic
function of a measurable set $A\subset Q$ we  easily obtain

\proclaim
{Corollary 1}
For all $0<\ve\leq 1/2$,
$$
|A| \, |^c\!A|\leq \bigg(C(d)\ve\int_A\int_{^c\!A} \frac{dxdy}{|x-y|^
{d+1-\ve}}\bigg) ^{d/(d-1+\ve)}.
\tag 9
$$
\endproclaim

Note that in the special case $d=1$, (9) takes the simple form
$$
|A| \, |^c\!A|\leq
\bigg(C^*\ve\int_A\int_{^c\!A}\frac{dxdy}{|x-y|^{2-\ve}}\bigg)^{1/\ve}\tag 10
$$
for some absolute constant $C^*$.
Estimate (10) is sharp as can be easily seen when $A$ is an interval.

The conclusion of Corollary 1 is related to a result stated in [3] (Remark
4).
There is however an important difference.
In [3] the set $A$ was {\it fixed} (independent of $\ve$) and the statement
there
provides a bound for $|A| \ |^c\!A|$ in terms of the limit, as $\ve\to 0$, of
the RHS in (9).
The improved version - which requires a more delicate argument- is used in
Section 7; we apply Corollary 1 (with $d=1$) to give a proof of a result
announced in [2] (Remark E.1).
Namely, on $\Omega =(-1, +1)$ consider the function
$$
\vp_\ve(x)= \cases 0 \qquad &{\text { for }} \qquad -1<x<0,\\
2\pi x/\delta\qquad &{\text { for }} \qquad 0<x<\delta,\\
2\pi \qquad &{\text { for }}\qquad \delta<x<1,
\endcases
$$
where $\delta=e^{-1/\ve}, \ve>0$ small.

Set $u_\ve =e^{i\vp_\ve}$.
It is easy to check (by scaling) that
$$
\Vert u_\ve\Vert_{H^{1/2}} =\Vert u_\ve -1\Vert_{H^{1/2}}\leq C
$$
as $\ve\to 0$ and consequently $\Vert u_\ve\Vert_{H^{(1-\ve)/2}}\leq C$ as
$\ve \rightarrow 0$.
On the other hand, a straightforward computation shows that
$\Vert\vp_\ve\Vert_{H^{(1-\ve)/2}}\sim\ve^{-1/2}$.


The result announced in [2] asserts that {\it any } lifting $\vp_\ve$ of $u_\ve$ blows up
in $H^{(1-\ve)/2}$ (at least) in the same rate as $\vp_\ve$:

\proclaim
{Theorem 2}
Let $\psi_\ve:\Omega\to \Bbb R$ be any measurable function such that $u_\ve
=e^{i\psi_\ve}$.
Then
$$
\Vert\psi_\ve\Vert_{H^{(1-\ve)/2}}\geq c\ve^{-1/2}, \forall \ve\in (0,
1/2), 
$$
for some absolute constant $c> 0$.
\endproclaim

\noindent
{\bf Remark 1.}
There are various versions of the Sobolev inequality (5).
All these forms hold with equivalent constants:

\noindent
{\bf Form 1:}
$\Vert\nabla f\Vert_{L^p(Q)}\geq A_1 \Vert f-\notint_Qf\Vert_{L^q(Q)} \quad
 \forall f\in W^{1, p}(Q)$.

\noindent
{\bf Form 2:} $\Vert \nabla f\Vert_{L^p(Q)} \geq A_2 \Vert f-\notint_Q
f\Vert_{L^q(Q)}\ $ for all $Q$-periodic functions $f\in W^{1, p}_{loc} (\Bbb
R^d)$.

\noindent
{\bf Form 3:}
$\Vert \nabla f\Vert_{L^p(\Bbb R^d)}\geq A_3 \Vert f\Vert_{L^q(\Bbb R^d)}
 \quad \forall f\in C_0^\infty(\Bbb R^d)$.

\noindent
{\bf Form 1 $\Rightarrow$ Form 2.} Obvious with $A_2=A_1$.

\noindent
{\bf Form 2 $\Rightarrow$ Form 1.}
Given any function $f\in W^{1, p}(Q)$, it can be extended by reflections to a
periodic function on a larger cube $\tilde Q$ so that Form 2 implies Form 1
with $A_1\geq CA_2$, and $C$ depends only on $d$.

\noindent
{\bf Form 1 $\Rightarrow$ Form 3.}
By scale invariance,
Form 1 holds with the same constant $A_1$ on the cube $Q_R$ of side $R$.
Fix a function $f\in C^\infty_0(\Bbb R^d)$ and let $R>$ diam (Supp $f$).
We have
$$
\Vert\nabla f\Vert_{L^p(Q_R)} \geq A_1\Vert f-\hbox{$\notint_{Q_R}$}
f\Vert_{L^q(Q_R)}.
$$
As $R\rightarrow \infty$ we obtain Form 3 with $A_3=A_1$.

\noindent
{\bf Form 3 $\Rightarrow$ Form 2.} Given a smooth periodic function $f$
on $\Bbb R^d$, let $\rho$ be a smooth cut-off function with $\rho=1$ on 
$Q$ and $\rho =0$ outside $2Q$.
Then
$$
\Vert\nabla(\rho f)\Vert_{L^p(\Bbb R^d)} \geq A_3 \Vert\rho f\Vert_{L^q(\Bbb
R^d)}
$$
and thus
$$
A_3\Vert f\Vert_{L^q(Q)} \leq C(\Vert\nabla f\Vert_{L^p(Q)}+\Vert
f\Vert_{L^p(Q)})
$$
where $C$ depends only on $d$.
Replacing $f$ by $(f-\notint_Q f)$ and applying Poincar\'e's inequality (see e.g.
[5], Section 7.8) yields
$$
A_3\Vert f-\hbox{$\notint$} f\Vert_{L^q{(Q)}} 
\leq C\Vert\nabla f\Vert_{L^q(Q)}.
$$
The reader will check easily that the same considerations hold for the
fractional Sobolev norms such as in (8).
The proof of the last implication (Form 3 $\Rightarrow$ Form 2) involves a
Poincar\'e-type inequality.
What we use here is the following

\noindent
{\bf Fact:} Let $1\leq p<\infty$, $1/2\leq s<1$, then
$$
(1-s)\int_Q\int_Q \frac{|f(x)-f(y)|^p}{|x-y|^{d+sp}} \geq c(d) \Vert
f-\hbox{$\notint_Q$} f\Vert_{L^p(Q)}^p.
$$
The proof of this fact is left to the reader. (It is an adaptation of the
argument in the beginning of Section 5.
In (3) of Section 5 one uses an obvious lower bound:
$$
(3) \geq c\bigg(\sum_r\Vert f_r\Vert_{L^p}
\bigg)^p \geq c \Vert f-\hbox{$\notint$}
f\Vert^p_{L^p}.)
$$

For the convenience of the reader we have divided the proof of Theorem 1 into
several cases.
The plan of the paper is the following:

1. Introduction.

2. Proof of Theorem 1 when $p=1$ and $d=1$.

3. Proof of Theorem 1 when $p=1$ and $d\geq 2$.

4. Square function inequalities.

5. Proof of Theorem 1 when $1<p<2$.

6. Proof of Theorem 1 when $p\geq 2$.

7. Proof of Theorem 2.

Appendix: Proof of square function inequality. 


\subhead
2. Proof of Theorem 1 when $p=1$ and $d=1$
\endsubhead

For simplicity, we work with periodic functions of period $2\pi$ (for
non-periodic functions see Remark 1 in the Introduction).
All integrals, $L^p$ norms, etc...., are understood on the interval $(0,
2\pi)$.
We must prove that, (with $\ve=1-s$), for all $\ve\in(0, 1/2]$,
$$
C \ve\iint \frac{|f(x)-f(y)|}{|x-y|^{2-\ve}}dxdy \geq \Vert
f-\hbox{$\notint$} f\Vert_{L^{1/\ve}}.
\tag 1
$$

Write the left side as
$$
\align
&\ve \ \int \frac 1{|h|^{2-\ve}} \Vert f-f_h\Vert_1 dh \sim\\
& \ve \ \sum_{k\geq 0} 2^{k(2-\ve)} \int_{|h|\sim 2^{-k}}
\Vert f-f_h\Vert_1 dh.\tag 2
\endalign
$$
For $|h|\sim 2^{-k}$
$$
\align
&\Vert f-f_h\Vert_1\geq\\
&\Vert (f-f_h)*F_{N_k}\Vert_1=\bigg(N_k=2^{k-100}, F_N(x) =\sum_{|n|\leq
N}
\frac {N-|n|}{N} e^{inx} = \text { F\'ejer kernel}\bigg)\\
&\bigg\Vert \sum_{|n|<N_k} \frac {N_k-|n|}{N_k} \hat f(n) (e^{inh}-1)
e^{inx}\bigg\Vert_1\sim\\
&2^{-k} \bigg\Vert\sum_{|n|<N_k} \frac{N_k-|n|}{N_k} n\hat f(n)
e^{inx}\bigg\Vert_1 \qquad \text {(by the choice of $N_k$)}.
\endalign
$$
This last equivalence is justified via a smooth truncation as in
the following

\proclaim
{Lemma 1} $\left\Vert\sum_{|n|<N} \hat f(n) (e^{inh}-1) e^{inx}
\right\Vert_1\gtrsim \frac 1N\left\Vert\sum_{|n|<N} n\hat f(n)
e^{inx}\right\Vert_1$\hfill\break
for $|h|<\frac 1{100N}$.
\endproclaim

\noindent
{\bf Proof.} Write
$$
\bigg\Vert\sum_{|n|<N} n\hat f(n) e^{inx}\bigg\Vert_1\leq
\bigg\Vert\sum_{|n|<N} \hat f(n) (e^{inh}-1)e^{inx}\bigg\Vert_1.
\bigg \Vert\sum \vp\left(\frac nN\right)\frac n{e^{inh}-1}
e^{inx}\bigg\Vert_1
$$
where $0\leq \vp\leq 1$ is a smooth function with
$$
\vp(t)=
\cases 1 \ \text { for } \ |t|\leq 1\\
0 \ \text { for } \ |t|
\geq 2
\endcases
$$
We have from assumption
$$
\bigg\Vert\sum\vp\left(\frac nN\right) \frac n{e^{inh}-1}
e^{inx}\bigg\Vert_1\sim N\bigg\Vert\sum\vp\bigg(\frac n N\bigg)
\frac{nh}{e^{inh}-1}e^{inx}\bigg\Vert_1
$$
and the second factor remains uniformly bounded.
This may be seen by expanding
$$
\frac y{e^{iy}-1} \sim \frac 1i+0(y)
$$
for $|y| <\frac 1{50}$ and using standard multiplier bounds.
\bigskip

We now return to the proof of Theorem 1 $(p=1, d=1)$.

Substitution in (2) gives thus
$$
\ve \ \sum_{k\geq 0} 2^{-\ve k}\bigg\Vert\sum_{|n|<N_k} \, \frac
{N_k-|n|}{N_k} n \hat f (n) e^{inx}\bigg\Vert_1.\tag 3
$$
Define
$$
k_0=\frac{10} \ve.
$$

For $k_0<k<2k_0$, minorate (using Lemma 1)
$$
\bigg\Vert \sum_{|n|<N_k}\frac{N_k-|n|}{N_k} n\hat f(n)
e^{inx}\bigg\Vert_1\gtrsim \bigg\Vert \sum_{|n|<N_{k_0}} \ \frac
{N_{k_0}-|n|}{N_{k_0}} n\hat f (n) e^{inx}\bigg\Vert_1
$$
and therefore
$$
\align
(3) \gtrsim &\bigg\Vert \sum_{|n|<N_{k_0}} \frac{N_{k_0}-|n|}{N_{k_0}}
n\hat f(n) e^{inx}\bigg\Vert_1=\\
&\bigg\Vert\sum_{|n|<N_{k_0}} \frac{N_{k_0}-|n|}{N_{k_0}} \hat f(n)
e^{inx}\bigg\Vert_{W^{1, 1}} \geq\\
&\bigg\Vert \sum_{0<|n|<N_{k_0}} \frac{N_{k_0}-|n|}{N_{k_0}} \hat f
(n) e^{inx}\bigg\Vert_\infty.\tag 4
\endalign
$$
Next write also
$$
\align
(3)&\gtrsim \ve \ \sum_{r\geq 1} 2^{-r}\sum_{[\frac {r+2}\ve]\leq k<
[\frac{r+3}\ve]}\bigg\Vert \sum_{|n|<N_k} \ \frac{N_k-|n|}{N_k}
n\hat f(n) e^{inx}\bigg\Vert_1\\
&\gtrsim \sum_{r\geq 1} 2^{-r} \bigg\Vert\sum_{|n|<2^{[\frac {r+1}{\ve}]}}
\ \frac {2^{[\frac{r+1}{\ve}]} -|n|}{2^{[\frac{r+1}{\ve}]}}
e^{inx}\bigg\Vert_1.\tag 5
\endalign
$$
Denote for each $r$ by $\lambda_r=\{\lambda_r(n)|n\in\Bbb Z\}$ the
following multiplier
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Thus
$$
\lambda_r(n) =\lambda_r(-n)
$$
$$
\bigg\Vert\sum \lambda_r(n) e^{inx}\bigg\Vert_1< C.
$$
(This multiplier may  be reconstructed from F\'ejer-kernels $F_N$ with $N=
2^{[\frac{r+1}{\ve}]}, 2^{[\frac r\ve]}, 2^{[\frac{r-1}{\ve}]})$.

Also
$$
\align
&\bigg\Vert \sum_{|n| < 2^{[\frac {r+1}{\ve}]}}
\frac{2^{[\frac{r+1}\ve]}-|n|}{2^{[\frac{r+1}\ve]}} n\hat f (n)
e^{inx}\bigg\Vert_1\gtrsim\\
&\bigg\Vert\sum_{2^{[\frac {r-1}{\ve}]} <|n|<2^{[\frac{r+1}{\ve}]}}
\lambda_r (n) \, n\hat f(n) e^{inx}\bigg\Vert_1\tag 6
\endalign
$$
and
$$
(5)\gtrsim \sum_{r\geq 1}
2^{-r}\bigg\Vert\sum_{2^{[\frac{r-1}{\ve}]}<|n|<2^{[\frac{r+1}{\ve}]}}
\lambda_r(n) (\text{sign } n)|n| \ \hat f (n)e^{inx}\bigg\Vert_1.\tag 7
$$
We claim that for $q>2$
$$
\bigg\Vert\sum_{N_1<|n|<N_2}\hat g(n) e^{inx}\bigg\Vert_q \leq
CN_1^{-\frac 1q}\bigg\Vert\sum_{N_1<|n|<N_2}|n| (\text{sign } n)
\hat g(n) e^{inx}\bigg\Vert_1\tag 8
$$
with the constant $C$ independent of $q$.

Applying (8) with
$$
q=\frac 1\ve, \quad \hat g(n) =\lambda_r (n) \hat f(n), \quad
N_1=2^{[\frac{r-1}{\ve}]},  \ N_2=2^{[\frac{r+1}{\ve}]}
$$
we obtain the minoration
$$
(7) \gtrsim \sum_{r\geq 1} \bigg\Vert \sum_{2^{[\frac{r-1}{\ve}]}
<|n|<2^{[\frac{r+1}{\ve}]}} \, \lambda_r(n) \hat f(n)
e^{inx}\bigg\Vert_q.\tag 9
$$
By construction
$$
\sum_{r\geq 1} \lambda_r (n) =1 \text { for } |n|> 2^{[\frac 1\ve]}.
$$
Using also minoration (4) together with the triangle-inequality yields
$$
\text {LHS in }(1)\gtrsim (3) +(8)\gtrsim \bigg\Vert\sum_{n\not= 0} \hat f(n)
e^{inx}\bigg\Vert_q
$$
which proves the inequality.

\noindent
{\bf Proof of (8).}

Estimate
$$
\bigg\Vert\sum _{N_1<|n|<N_2} \, \hat g(n) e^{inx}\bigg\Vert_q
\leq \bigg\Vert \sum_{N_1<|n|<N_2} |n|^{-1} (\text{sign } n)
e^{inx}\bigg\Vert_q \ \bigg\Vert\sum_{N_1<|n|<N_2} |n| (\text {sign } n)
\hat g(n)e^{inx}\bigg\Vert_1
$$
where the first factor equals
$$
\bigg\Vert\sum_{N_1<n<N_2} \frac 1n\sin \ n x\bigg\Vert_q\lesssim
$$
$$
\align
&\bigg\Vert\sum_{\log N_1<k<\log N_2}\bigg\vert \sum_{n\sim 2^k} \frac 1n
\sin \, nx\bigg\vert \ \bigg\Vert_q \qquad (\text{assume $N_1, N_2$ powers
of 2)}\\
&\lesssim \bigg\Vert\sum_{\log N_1< k<\log N_2} \min (2^k|x|,
2^{-k}|x|^{-1})\bigg\Vert_q\\
&\lesssim \bigg\Vert\frac 1{1+N_1|x|}\bigg\Vert_q\lesssim N_1^{-1/q}.
\tag 10
\endalign
$$
This proves (8) and completes the proof of Theorem 1 when $p=1$ and $d=1$.

\subhead
3. Proof of Theorem 1 when $p=1$ and $d\geq 2$
\endsubhead

We have to prove that
$$
\iint \frac{|f(x)-f(y)|}{|x-y|^{d+s}} dxdy \geq \frac{C(d)}{1-s} \Vert
f-\hbox{$\notint$} f\Vert_q\tag 1
$$
where $q=d/(d-s)$.
We assume $d=2$.
The case $d>2$ is similar.
Write
$$
\align
\iint \frac{|f(x)-f(y)|}{|x-y|^{d+s}}dxdy &\sim \sum_{0\leq k} 2^{k(d+s)}
\int_{|h|\sim 2^{-k-10}} \Vert f(x+h)-f(x-h)\Vert_1 dh\\
&\geq \sum 2^{k(d+s)} \int_{\Sb |h_1|\sim 2^{-k-10}\\
|h_2|\sim 2^{-k-10}\endSb} \bigg\Vert\sum_{n\in\Bbb Z^d} \hat f(n) (\sin
n.h) e^{in.x}\bigg\Vert_1 dh_1 dh_2\tag 2
\endalign
$$
Let $\vp$ be a smooth function on $\Bbb R$ s.t. $0\leq \vp\leq 1$ and
$$
\vp(t)=  \cases 1 \ \text{ for } |t|\leq 1\\
0 \ \text{ for } |t|\geq 2
\endcases
$$
As for $d=1$, consider (radial) multipliers $\lambda_0$ and $\lambda_r,
r\geq 1$
$$
\align
&\lambda_0(n) =\vp(2^{-\frac 1\ve}|n|)\\
&\lambda_r(n) =\vp(2^{-\frac {r+1}\ve}|n|)-\vp(2^{-\frac r\ve}|n|)
\tag 3
\endalign
$$
where $\ve=1-s$ and $\ve\in(0, 1/2)$.

Hence
$$
\align
\sum\lambda_r(n)& =1\\
\Vert\lambda_r\Vert_{M(L^1, L^1)} &\leq C \qquad {\text {(multiplier
norm)}}\tag 4\\
\supp \lambda_0&\subset B(0, 2^{\frac 1\ve+1})\tag 5\\
\supp \lambda_r&\subset B(0, 2^{\frac {r+1}\ve+1})
\backslash B(0, 2^{\frac r\ve}).\tag 6
\endalign
$$
Write
$$
(2) =\sum_{\frac 1\ve<k<\frac 2\ve}+ \sum_{r\geq 1} \  \sum_{\frac{r+1}\ve <
k< \frac{r+2}\ve}.\tag 7
$$
For $\frac 2\ve >k>\frac 1\ve$ and $|h|< 2^{-k-10}$, (4), (5) permit us to
write
$$
\align
\bigg\Vert\sum_n\hat f(n) e^{in. x}\sin n.h\bigg\Vert_1&\gtrsim
\bigg\Vert\sum_n\lambda_0(n) \hat f(n) e^{in.x} \sin n.h\bigg\Vert_1\\
&\sim\bigg\Vert\sum_n\lambda_0(n) (n.h) \hat f(n)e^{in.x}\bigg\Vert_1
\endalign
$$
and thus
$$
\align
&2^{k(d+1-\ve)}\int\limits_{|h_1|, |h_2|\sim 2^{-k-10}}\bigg\Vert\sum \hat f(n)
(\sin n.h) e^{in.x}\bigg\Vert_1 dh_1 dh_2 \gtrsim\\
&2^{k(3-\ve)}8^{-k} \bigg(\bigg\Vert \sum\lambda_0(n)n_1\hat f(n)
e^{in.x}\bigg\Vert_1+
\bigg\Vert\sum\lambda_0(n)n_2\hat f(n) e^{in.x}\bigg\Vert_1\bigg)
\\
&=2^{-k\ve}\bigg(\bigg\Vert\partial_{x_1}\bigg(
\sum\lambda_0(n)\hat f(n) e^{in.x}\bigg)\bigg\Vert_1+\Vert\partial_
{x_2} (\cdots) \Vert\bigg )\sim\\
& \bigg\Vert\sum\lambda_0(n)\hat f(n) e^{in.x}\bigg\Vert_{W^{1, 1}}.\tag 8
\endalign
$$
Similarly, for
$$
\frac{r+1}\ve<k<\frac{r+2}\ve
$$
we have
$$
2^{k(d+1-\ve)}\int\limits_{|h_1|, |h_2|\sim 2^{-k-10}}
\bigg\Vert \sum\hat f(n)(\sin nh) e^{in.x}\bigg\Vert_1\gtrsim
2^{-r}\bigg\Vert\sum\lambda_r(n) \hat f(n) e^{in.x}\bigg\Vert_{W^{1, 1}}.
\tag 9
$$
Since in the summation (7), each of the terms (8), (9) appear at least
$\frac 1\ve $ times, we have
$$
\ve.(2)\gtrsim \bigg\Vert\sum\lambda_0(n)\hat f(n) e^{in.x}\bigg\Vert_{W^{1, 1}}+ \sum_r2^{-r}\bigg\Vert\sum\lambda_r(n)\hat
f(n)e^{in.x}\bigg\Vert_{W^{1, 1}}.\tag 10
$$
Write
$$
\frac{2-s}2=1-s+\frac s2
$$
and by H\"older's inequality
$$
\bigg\Vert\sum \lambda_r(n)\hat f(n) e^{in.x}\bigg\Vert_{\frac 2{2-s}} \leq
\bigg\Vert\sum \lambda_r(n)\hat f(n) e^{in.x}\bigg\Vert_2^s \
\bigg\Vert\sum\lambda_r(n)\hat f(n) e^{in.x}\bigg\Vert_1^{1-s}.\tag 11
$$
By the Sobolev embedding theorem $(d=2)$
$$
\bigg\Vert\sum\lambda_r(n)\hat f(n) e^{in.x}\bigg\Vert_2\leq C
\bigg\Vert\sum\lambda_r(n)\hat f(n) e^{in.x}\bigg\Vert_{W^{1, 1}}.\tag 12
$$
We estimate the last factor in (11).

Recalling (6),
$$
2^{\frac{r+1} \ve+1} >\max (|n_1|, |n_2|) > 2^{\frac r\ve -1}
$$
if $\lambda_r(n)\not=0, r\geq 1$.

Hence, with $\vp$ as above
$$
\lambda_r(n) = \lambda_r(n).(1-\vp)(2^{-\frac{r-1}\ve}
n_1)+\lambda_r(n).\vp(2^{-\frac{r-1}\ve}n_1).(1-\vp)(2^{-\frac{r-1}\ve}n_2)
$$
and thus
$$
\align
&\bigg\Vert\sum\lambda_r(n)\hat f(n) e^{in.x}\bigg\Vert_1\leq
\\
&\bigg\Vert\sum\lambda_r(n)n_1\hat f(n) e^{in.x}\bigg\Vert_1 \
\bigg\Vert\sum\frac 1{n_1}
(1-\vp)(2^{-\frac{r-1}\ve}n_1)e^{in.x}\bigg\Vert_1\\
&+\bigg\Vert\sum\lambda_r(n)n_2\hat f(n) e^{in.x}\bigg\Vert_1 \
\bigg\Vert\sum\frac 1{n_2}
\vp(2^{-\frac{r-1}\ve}n_1)(1-\vp) (2^{-\frac{r-1}\ve} n_2)
e^{in.x}\bigg\Vert_1\le\\
&\bigg(\bigg\Vert\sum_{n_1}\frac 1{n_1}
(1-\vp)(2^{-\frac{r-1}\ve}n_1)e^{in_1x_1}\bigg\Vert_{L^1_{x_1}}+
\bigg\Vert\sum_{n_2}\frac 1{n_2}
(1-\vp)(2^{-\frac{r-1}\ve}n_2)e^{in_2x_2}\bigg\Vert_{L^1_{x_2}}\bigg).\\
&\bigg\Vert\sum\lambda_r(n)\hat f(n) e^{in.x}\bigg\Vert_{W^{1, 1}}.\tag 13
\endalign
$$
Since $(1-\vp)(2^{-\frac{r-1}\ve}n_1)=0$ for $|n_1| \leq 2^{\frac {r-1}
\ve}$, one easily checks that
$$
\bigg\Vert\sum_{n_1}\frac 1{n_1}
(1-\vp)(2^{-\frac{r-1}\ve}n_1)\ve^{in_1x_1}\bigg\Vert_{L^1_{x_1}}\lesssim
\sum_{\ell\geq \frac{r-1}\ve}2^{-\ell} <2^{\frac {r-2}\ve}.
$$
Similarly
$$
\bigg\Vert\sum_{n_2} \frac 1{n_2}
(1-\vp)(2^{-\frac{r-1}\ve}n_2)e^{in_2x_2}\bigg\Vert_{L^1_{x_2}}
\leq 2^{-\frac{r-2}\ve}.
$$
Thus (13) implies that
$$
\bigg\Vert\sum\lambda_r(n)\hat f(n)e^{in.x}\bigg\Vert_1\leq
2^{-\frac {r-2}\ve} \bigg\Vert \sum \lambda_r(n)\hat f(n)
e^{in.x}\bigg\Vert_{W^{1, 1}}.\tag 14
$$
Substitution of (12), (14) in (11) gives
$$
\align
\bigg\Vert\sum\lambda_r(n)\hat f(n)e^{in.x}\bigg\Vert_{\frac 2{2-s}}
&\lesssim 2^{-\frac{r-2}\ve(1-s)} \bigg\Vert \sum \lambda_r(n)\hat f(n)
e^{in.x}\bigg\Vert_{W^{1, 1}}\\
&\sim 2^{-r}\bigg\Vert\sum \lambda_r(n)\hat f(n) e^{in.x}\bigg\Vert_{W^{1,
1}}.\tag 15
\endalign
$$
By (12), (15)
$$
\align
\ve.(2)&\geq \bigg\Vert\sum\lambda_0(n)\hat f(n)e^{in.x}\bigg\Vert_2
+ \sum_{r\geq 1} \bigg\Vert\sum \lambda_r(n)\hat f(n)
e^{in.x}\bigg\Vert_{\frac 2{2-s}}\\
&\geq \big\Vert f-\notint f\big\Vert_{{\frac 2{2-s}}}
\endalign
$$
by (3).

This proves (1) and completes the proof of Theorem 1 when $p=1$.

\subhead
{4. Square function inequalities}
\endsubhead

We present here some known inequalities used in the proof of Theorem 1 when
$p>1$.
Let $\{\Delta_j f\}_{j= 1, 2, \ldots}$ be a Littlewood-Paley decomposition
with $\Delta_j f$ obtained from a Fourier multiplier of the form
$\vp(2^{-j}|n|)-\vp(2^{-j+1}|n|)$ with $0\leq \vp\leq 1$ a smooth function
satisfying $\vp(t)=1$ for $|t|\leq 1$ and $\vp(t) =0$ for $|t|>2$.

Recall the square-function inequality for $1<q<\infty$
$$
\frac 1{C(q)} \bigg\Vert\bigg(\sum|\Delta_j f|^2\bigg)^{1/2}\bigg\Vert_q
\leq \Vert f\Vert_q
\leq C(q) \bigg\Vert\bigg(\sum|\Delta_j f|^2\bigg)^{1/2}\bigg\Vert_q.\tag
1
$$
We will also consider square-functions wrt a martingale filtration.
Denote thus $\{\Bbb E_j\}$ the expectation operators wrt a dyadic
partition of $[0, 1]^d$ and
$$
\tilde\Delta_j f=(\Bbb E_j-\Bbb E_{j-1})f\tag 2
$$
the martingale differences.

We will use the square-function inequality
$$
\Vert f\Vert_q \leq C\sqrt q\bigg\Vert\bigg(\sum|\tilde\Delta_j
f|^2\bigg)^{1/2}
\bigg\Vert_q \qquad \text { for } \infty> q\geq 2\tag 3
$$
which is precise in terms of the behaviour of the constant for $q\to
\infty$ \big(see [4] and also the Appendix for a proof of (3)\big).

\proclaim
{Remark 2}
One should expect (3) also to hold if $\tilde\Delta_j$ is replaced by
$\Delta_j$ above but we will not need this fact.
\endproclaim

We do use later on the following inequality.

Let 
$$
 p<q \text { and } s=d\bigg(\frac1p -\frac 1q\bigg)\geq \frac 12.
$$

Then, for $q\geq 2$
$$
\Vert f\Vert_q\leq C \sqrt q\bigg[\sum_k (2^{ks}\Vert\Delta_k
f\Vert_p)^2\bigg]^{1/2}.\tag 4
$$

\noindent
{\bf Proof of (4)}

It follows from (3) that since $q\geq 2$
$$
\Vert f\Vert_q\leq C\sqrt q\bigg(\sum_j\Vert\tilde\Delta_j
f\Vert^2_q\bigg)^{1/2}.\tag 5
$$
Write
$$ 
\align
\tilde \Delta _j f&=\sum_{k\leq j}\tilde\Delta_j \Delta_k
f+\sum_{k>j}\tilde\Delta_j\Delta_k f\\
\Vert \tilde\Delta_j f\Vert_q &\lesssim \sum_{k\leq j}
2^{k-j}\Vert\Delta_k f\Vert_q +\sum_{k>j} 2^{js} \Vert\Delta_k f\Vert_p\\
&\lesssim \sum_{k\leq j} 2^{k-j} (2^{ks}\Vert\Delta_k f\Vert_p)+
\sum_{k>j} 2^{(j-k)s} (2^{ks}\Vert\Delta_k f\Vert_p).\tag 6
\endalign
$$
Substitution of (6) in (5) gives
$$
\align
\Vert f\Vert_q&\leq C\sqrt q \bigg\{\bigg(\sum_{k\leq j}
(j-k)^24^{k-j}(2^{ks}\Vert\Delta_k f\Vert_p)^2\bigg)^{1/2}+
\bigg(\sum_{k>j}(k-j)^24^{(j-k)s} (2^{ks}\Vert\Delta_k
f\Vert_p)^2\bigg)^{1/2}\bigg\}\\
&\leq C\sqrt q\bigg(\sum_k(2^{ks}\Vert\Delta_k f\Vert_p)^2\bigg)^{1/2}.\tag
7\\
\endalign
$$

\subhead
5. Proof of Theorem 1 when $1<p<2$
\endsubhead

Write
$$
\align
\iint \frac {|f(x)-f(y)|^p}{|x-y|^{d+ps}}dxdy&\sim \sum_{k\geq
0}2^{k(d+ps)}\int_{|h|\sim 2^{-k-10}}\Vert f(x+h)-f(x-h)\Vert_p^p dh\\
&\geq \sum_{k\geq 0} 2^{k(d+ps)} \int_{|h|\sim 2^{-k-10}}\bigg\Vert
\sum\hat f(n)(\sin n.h) e^{in.x} \bigg\Vert_p^p dh.\tag 1
\endalign
$$
Following the argument in Section 3 (formula (10)), we get again for
$$
\align
s&=d\bigg(\frac 1p-\frac 1q\bigg), 1-s=\ve\tag 2\\
\ve.(1)&\gtrsim \sum_r\bigg(2^{-r}\bigg\Vert\sum_n\lambda_r(n)\hat f(n)
e^{in.x} \bigg\Vert_{W^{1, p}}\bigg)^p\tag 3
\endalign
$$
where the multipliers $\lambda_r$ are defined as before.

\noindent
{\bf Case $\underline{d=1}$}

Define
$$
f_r= \sum_n\lambda_r(n)\hat f(n) e^{in.x}.
$$
We will make 2 estimates.

First write
$$
f_r=\bigg(\sum n\lambda_r(n)\hat f(n) e^{in.x})*\bigg(\sum_{2^{\frac r\ve}
<|n|< 2^{\frac{r+1}\ve}} \, \frac 1n e^{in.x}\bigg)
$$
implying
$$
\Vert f_r\Vert_q\leq \Vert f_r\Vert_{W^{1, p}}\bigg\Vert
\sum_{2^\frac r\ve <n<2^{\frac{r+1}\ve}}\frac 1n \sin nx\bigg\Vert_{(\frac
1{p'} +\frac 1q)^{^{-1}}}\tag 4
$$
and by estimate (10) in Section 2,
$$
\Vert f_r\Vert_q \lesssim 2^{-\frac r\ve(\frac 1{p'}+\frac 1q)}
\Vert f_r\Vert_{W^{1, p}} =2^{-\frac r\ve(1-s)}\Vert f_r\Vert_{W^{1, p}}
=2^{-r}\Vert f_r\Vert_{W^{ 1, p}}.\tag 5
$$
Estimate then
$$
\Vert f\Vert_q\leq\sum_r\Vert f_r\Vert_q \leq C\sum_r (2^{-r}\Vert
f_r\Vert_{W^{1, p}}).\tag 6
$$
Next apply inequality (4) of Section 4.
Observe that
$$
|\Delta_{k}f |\leq \sum_r |\Delta_k f_r|
$$
where, by construction, there are, for fixed $k$, at most 2 nonvanishing
terms.

Thus
$$
\Vert \Delta_kf\Vert^2_p\lesssim \sum_r\Vert\Delta_k f_r\Vert^2_p.\tag 7
$$
Also, for fixed $r$
$$
\sum_k(2^{ks}\Vert \Delta_k f_r\Vert_p)^2 =\sum_r 4^{-k\ve} \Vert\Delta_k
f_r\Vert^2_{W^{1, p}}\lesssim \frac 1\ve 4^{-r} \Vert f_r\Vert^2_{W^{1,
p}}.\tag 8
$$
Substituting (7), (8) in (4) of Section 4 gives
$$
\Vert f\Vert_q \lesssim C\sqrt q \bigg[\sum_k\sum_r (2^{ks}\Vert\Delta_k
f_r\Vert_p)^2\bigg]^{1/2} \leq [C\sqrt{\frac q\ve}) \bigg[\sum_r(2^{-r} \Vert
f_r\Vert_{W^{1, p}})^2\bigg]^{1/2}\tag 9
$$
which is the second estimate.

Interpolation between (6) and (9) implies thus
$$
\Vert f\Vert_q \leq C\bigg(\sqrt{\frac q\ve}\bigg)^{2(1-\frac
1p)}\bigg[\sum_r (2^{-r}\Vert f_r\Vert_{W^{1, p}})^p\bigg]^{1/p}.\tag 10
$$
Recalling (3) and also (2) (which implies that $1-\ve =\frac 1p -\frac 1q
<\frac 1p$, hence $\ve>1-\frac 1p$) it follows that
$$
\ve.(1)\gtrsim \bigg(\frac 1q\bigg)^{p-1} \Vert f\Vert_q^p\tag 11
$$
which gives the required inequality.

\noindent
{\bf Case $\underline{d>1}$}

We will distinguish the further 2 cases

{\bf Case A}: \ $0<\frac 1p-\frac 1d$ is not near 0

{\bf Case B}: \ $ \frac 1p-\frac 1d$ is near 0

Observe that case B may only happen for $d=2$ and $p$ near 2 (we assumed
$1<p<2)$.

\noindent
{\bf Case A.}

Define $q_1$ by
$$
1=d\bigg(\frac 1p-\frac 1{q_1}\bigg)\tag 12
$$
so that $q<q_1$ and $q_1$ is bounded from above by assumption.

Thus we have the Sobolev inequality
$$
\Vert g\Vert_{q_1} \leq C\Vert g\Vert_{W^{1, p}}.\tag 13
$$
Next, we make the obvious adjustment of the argument in Section 3,
(11)-(15).

Thus H\"older's inequality gives
$$
\Vert f_r\Vert_q\leq \Vert f_r\Vert^{1-\theta}_{q_1} \Vert
f_r\Vert_q^\theta\tag 14
$$
with
$$
\frac 1q =\frac{1-\theta}{q_1}+\frac \theta p,
\text { hence $\theta =1-s=\ve$ by
(2), (12)}.
$$

Hence, by (13)
$$
\Vert f_r\Vert_q\leq C\Vert f_r\Vert^{1-\ve}_{W^{1, p}}\Vert
f_r\Vert^\ve_p.\tag 15
$$
To estimate $\Vert f_r\Vert_p$, proceed as in (13) of Section 3.
Thus
$$
\align
\Vert f_r\Vert_p&\lesssim\bigg\Vert\sum\frac 1n (1-\vp)(2^{-\frac {r-1}
\ve}n)e^{inx}\bigg\Vert_{L^1_x(\Bbb T)} \ \Vert f_r\Vert_{W^{1, p}}\\
&\lesssim 2^{-\frac{r-1}\ve} \Vert f_r\Vert_{W^{1, p}}.\tag 16
\endalign
$$
Substitution of (16) in (15) gives
$$
\Vert f_r\Vert_q \lesssim 2^{-r} \Vert f_r\Vert_{W^{1, p}}.\tag 17
$$
Substitution of (17) in (3) gives (since $q$ is bounded by case A
hypothesis)
$$
\align
\ve.(1)\gtrsim \sum_r\Vert f_r\Vert^p_q  &\sim\sum_r
\bigg\Vert\bigg(\sum_j |\Delta_j f_r|^2\bigg)^{1/2} \bigg\Vert^p_q\\
&\gtrsim \bigg\Vert\bigg(\sum_{r, j} |\Delta_j
f_r|^2\bigg)^{1/2}\bigg\Vert^p_q\\
&\gtrsim\bigg\Vert\bigg(\sum_j|\Delta_jf|^2\bigg)^{1/2}
\bigg\Vert^p_q\sim \Vert f\Vert^p_q\tag 18\\
\endalign
$$
(the second inequality requires distinction of the cases $q\geq 2$ and
$p< q\leq 2$).

(18) gives the required inequality.

\noindent
{\bf Case B.}

Thus $d=2$ and $p$ is near 2.

Going back to (3) and applying (1), (4) of Section 4, we obtain
$$
\align
\ve.(1)&\gtrsim \sum_r(2^{-r}\Vert f_r\Vert_{W^{1, p}})^p\\
&\gtrsim \bigg(\sum_r 4^{-r} \sum_j\Vert\Delta_j f_r\Vert^2_p
4^j\bigg)^{\frac p2}\\
&\gtrsim \bigg(\sum_j(2^{sj}\Vert\Delta_j f\Vert_p)^2\bigg)^{\frac p2}\\
&\gtrsim q^{-\frac p 2} \Vert f\Vert^p_q\tag 19
\endalign
$$
where
$$
q^{-\frac p2}=\bigg(\frac 1p-\frac s2\bigg)^{\frac p2}\sim
(2-ps)^{p-1}\tag 20
$$
which again gives the required inequality.

\subhead
6. Proof of Theorem 1 when $p\geq 2$
\endsubhead


>From (3) in Section 5, we get now the minoration
$$
\ve.(1)\gtrsim \sum_j (2^{sj}\Vert \Delta_j f\Vert_p)^p\tag 1
$$
which we use to majorize $\Vert f\Vert_q$.

We have already inequality (4) in Section 5, thus
$$
\Vert f\Vert_q \leq C\sqrt q \bigg(\sum_j(2^{sj}\Vert \Delta_j
f\Vert_p)^2\bigg)^{1/2}.\tag 2
$$
Our aim is to prove that
$$
\Vert f\Vert_q\leq C q^{1-\frac 1p} \bigg(\sum_j(2^{sj}\Vert\Delta_j
f\Vert_p)^p\bigg)^{\frac 1p}\tag 3
$$
which will give the required inequality together with (1).

Using interpolation for $2\leq p<\frac ds$, it clearly suffices to
establish (3) for large values of $q$.
To prove (3), we assume $2\leq p\leq 4$ (other cases may be treated by
adaption of the argument presented below).
Assume further (taking previous comment into account)
$$
q\geq 2p.\tag 4
$$
Again by interpolation, (3) will follow from (2) and the inequality
$$
\Vert f\Vert_q \leq Cq^{\frac 34} \bigg(\sum_j(2^{sj}\Vert\Delta_j
f\Vert_p)^4\bigg)^{1/4}.\tag 5
$$

We use the notation from Section 4 and start from the martingale
square function inequality (3) in Section 4; thus
$$
\Vert f\Vert_q \leq C\sqrt q\bigg\Vert\bigg(\sum |\tilde\Delta_j
f|^2\bigg)^{1/2}\bigg\Vert_q.\tag 6
$$
Write
$$
|\tilde\Delta_j f|\leq \sum_k|\tilde\Delta_j\Delta_k f|=\sum_{m\in\Bbb
Z}|\tilde\Delta_j\Delta_{j+m} f|
$$
(putting $\Delta_k=0$ for $k<0$).

Writing
$$
\bigg\Vert\bigg(\sum_j|\tilde\Delta_jf|^2)^{1/2}\bigg\Vert_q \leq
\sum_{m\in\Bbb Z} \, \bigg\Vert\bigg(\sum_j|\tilde\Delta_j
\Delta_{j+m}f|^2\bigg)^{1/2}\bigg\Vert_q\tag 7
$$
we estimate each summand.

Fix $m$.
Write
$$
\align
\bigg\Vert\bigg(\sum_j|\tilde\Delta_j\Delta_{j+m}
f|^2\bigg)^{1/2}\bigg\Vert^4_q &=
\bigg\Vert\bigg(\sum_j|\tilde\Delta_j\Delta_{j+m}f|^2\bigg)^2\bigg
\Vert_{\frac q4}\\
& \leq 2 \, \sum_{j_1\leq j_2}\Vert\, |\tilde\Delta_{j_1}
\Delta_{j_1+m} f|^2 \, |\tilde\Delta_{j_2} \Delta_{j_2+m} f|^2\Vert_{\frac
q4}\tag 8 
\endalign
$$
and
$$
\align
\Vert \ | \tilde\Delta_{j_1}\Delta_{j_1+m}f|^2 \ &|\tilde\Delta_{j_2}
\Delta_{j_2+m} f|^2\Vert_{\frac q4} =\bigg[\int
|\tilde\Delta_{j_1}\Delta_{j_1+m} f|^{\frac q2} . \Bbb
E_{j_1}\big[|\tilde\Delta_{j_2}\Delta_{j_2+m}f|^{\frac q2}]\big]\bigg]
^{\frac4q}\\
&\leq\Vert\tilde \Delta_{j_1} \Delta_{j_1+m} f\Vert^2_q \
\big\Vert\big(\Bbb E_{j_1}[|\tilde \Delta_{j_2}\Delta_{j_2+m} f|^{\frac
q2}]\big)^{\frac 2q}\big\Vert^2_q\\
&\leq 4^{d(j_2-j_1)(\frac 1p-\frac 2q)} \Vert\tilde
\Delta_{j_1}\Delta_{j_1+m}f\Vert^2_q \ \big\Vert(\Bbb
E_{j_1}[|\tilde\Delta_{j_2}\Delta_{j_2+m} f|^p]\big)^{1/p}\Vert^2_q\\
&\leq 4^{d(j_2-j_1)(\frac 1p-\frac 2q)} \, 4^{d{j_1}(\frac 1p-\frac 1q)}
\Vert\tilde\Delta_{j_1}\Delta_{j_1+m} f\Vert^2_q \ \Vert\tilde
\Delta_{j_2}\Delta_{j_2+m} f\Vert^2_p.\tag 9 
\endalign
$$
Assume $\underline{m\leq 0}$

Estimate
$$
\Vert \tilde\Delta_{j_1} \Delta_{j_1+m} f\Vert_q \lesssim
2^m\Vert\Delta_{j_1+m} f\Vert_q \leq 2^m 2^{d(j_1+m)(\frac 1p-\frac 1q)} 
\Vert \Delta_{j_1+m} f\Vert_p\tag 10
$$
$$
\Vert \tilde\Delta_{j_2} \Delta_{j_2+m} f\Vert_p \lesssim
2^m\Vert\Delta_{j_2+m} f\Vert_p.\tag 11
$$
Substitution of (10), (11) in (9) gives
$$
4^{(1-d(\frac 1p-\frac 1q))m+m} \ 4^{-\frac dq(j_2-j_1)} [2^{d(\frac
1p-\frac 1q)(j_1+m)}\Vert
\Delta_{j_1+m}f\Vert_p]^2 \ [ 2^{d(\frac 1p-\frac
1q)(j_2+m)}\Vert\Delta_{j_2+m} f\Vert_p]^2\tag 12
$$
where
$$
d\bigg(\frac 1p-\frac 1q\bigg)=  s.
$$
Summing (12) for $j_1<j_2$ and applying Cauchy-Schwartz implies for $m<0$
$$
\align
(8)&< 4^{(2-s)m} \bigg(\sum_{\ell\geq 0} 4^{-\frac dq \ell}\bigg) 
\bigg[\sum_j(2^{sj}\Vert\Delta_j f\Vert_p)^4\bigg]\\
&\lesssim 4^{(2-s)m} q
\bigg[\sum_j(2^{sj}\Vert\Delta_j f\Vert_p)^4\bigg].
\tag 13
\endalign
$$

Assume next $\underline{m>0}$.

Estimate
$$
\Vert \tilde\Delta_{j_1}\Delta_{j_1+m} f\Vert_q \lesssim
2^{d_{j_1}(\frac 1p-\frac 1q)}
\Vert\Delta_{j_1+m} f\Vert_p
$$
and
$$
\align
(9) &\leq 4^{d(j_2-j_1)(\frac 1p-\frac 2q)} \ 16^{d{j_1}(\frac 1p-\frac
1q)} \Vert\Delta_{j_1+m} f\Vert^2_p \ \Vert\Delta_{j_2+m} f\Vert^2_p\\
&\leq 16^{-ms} 4^{-(j_2-j_1)\frac dq} \Vert 2^{s(j_1+m)}
\Delta_{j_1+m}f\Vert^2_p \, 
\Vert 2^{s(j_2+m)} \Delta_{j_2+m} f\Vert^2_p.\tag 14
\endalign
$$
Summing over $j_1< j_2$ implies that for $m>0$
$$
(8) \lesssim 16^{-ms} q \bigg[\sum_j(2^{sj}\Vert\Delta_j
f\Vert_p)^4\bigg].\tag 15
$$
Summing (13), (15) in $m$ implies that
$$
\align
(7)&\leq \bigg(\sum_{m\leq 0} 2^{(1-\frac s2)m}+\sum_{m>0}
2^{-sm}\bigg) q^{1/4} \bigg[\sum_j(2^{sj}\Vert\Delta_j
f\Vert_p)^4\bigg]^{1/4}\\
&\leq q^{1/4}\bigg[\sum_j(2^{sj}\Vert\Delta_j f\Vert_p)^4\bigg]^{1/4}.\tag
16
\endalign
$$
To bound $\Vert f\Vert_q$, apply (6) which introduces an additional
$q^{1/2}$-factor.
This establishes (5) and completes the argument and the proof of
Theorem 1.

\subhead
7. Proof of Theorem 2
\endsubhead

We will make use of the following two lemmas

\proclaim
{Lemma 2}
Let $I\subset\Bbb R$ be an interval and let $\psi: I\to\Bbb Z$ be any
measurable function.
Then, there is some $k\in\Bbb Z$ such that
$$
|\{x\in I; \psi(x)\not= k\}|\leq
2\bigg
(C^*\ve\int_I\int_I\frac{|\psi(x)-\psi(y)|^2}{|x-y|^{2-\ve}}dxdy\bigg)^
{1/\ve},
$$
for all $\ve\in(0, 1/2]$. where $C^*$ is the absolute constant in Corollary
1 (inequality (10) in Section 1).
\endproclaim

\noindent
{\bf Proof of Lemma 2.}
After scaling and shifting we may assume that $I=(-1, +1)$.
For each $k\in \Bbb Z$, set
$$
A_k =\{x\in I; \psi(x) < k\}.
$$
Note that $A_k$ is nondecreasing, $\lim\limits_{k\to -\infty} |A_k|=0$ and
$\lim\limits_{k\to +\infty}|A_k|=2$.
Thus, there exists some $k\in \Bbb Z$ such that
$$
|A_k|\leq 1 \text { and } |A_{k+1}|>1.\tag 1
$$
Applying Corollary 1 with $A=A_k$ and with $A=A_{k+1}$ we find (using (1))
$$
|A_k|\leq |A_k| \ |^c\!\!A_{k}|\leq \bigg(C^*\ve
\int_A\int_{^c\!A_k}\frac{dxdy}{|x-y|^{2-\ve}}\bigg)^{1/\ve}\tag 2
$$
and
$$
|^c\!A_{k+1}|\leq |A_{k+1}| \ |^c\!A_{k+1}|\leq
\bigg(C^*\ve\int_{A_{k+1}}\int_{^c\!A_{k+1}}\frac{dxdy}
{|x-y|^{2-\ve}}\bigg)^{1/\ve}.\tag 3
$$
On the other hand
$$
|\psi(x)-\psi(y)|\geq 1 \text { for a.e. } x\in A_k, y\in{}^c\!A_k
$$
and
$$
|\psi(x)-\psi(y)|\geq 1\text { for a.e. }
x\in A_{k+1}, y\in {}^c\!A_{k+1}.
$$
Therefore
$$
\align
|\{x\in I; \psi(x)\not=k\}|&= |A_k|+|^c\!A_{k+1}|\\
&\leq 2
\bigg(C^*\ve\int_I\int_I\frac{|\psi(x)-\psi(y)|^2}{|x-y|^{2-\ve}}
dxdy\bigg)^{1/\ve}.
\endalign
$$

\proclaim
{Lemma 3}
If $\alpha>0, a<b<x, A\subset (a, b)$ is measurable, then
$$
\int_{(a, b)\backslash A} \frac{dy}{(x-y)^\alpha} \geq \int_a^{b-|A|}
\frac{dy}{(x-y)^\alpha}
$$
and similarly, if $x<a<b$, then
$$
\int_{(a, b)\backslash A} \frac{dy}{(y-x)^\alpha} \geq \int^b_{a+|A|}
\frac{dy}{(y-x)^\alpha}.
$$
\endproclaim

The proof of Lemma 3 is elementary and left to the reader.

\noindent
{\bf Proof of Theorem 2.}
Let $\psi_\ve:\Omega=(-1, +1)\to \Bbb R$ be any measurable function such that
$u_\ve=e^{i\psi_\ve}$.
We have to prove that for all $\ve< 1/2$,
$$
\Vert\psi_\ve\Vert_{H^{(1-\ve)/2}(\Omega)}\geq c\ve^{-1/2}\tag 4
$$
for some absolute constant $c$ to be determined.

We argue by contradiction and assume that for some $\ve< 1/2$
$$
\Vert\psi_\ve\Vert_{H^{(1-\ve)/2}(\Omega)}<\eta\ve^{-1/2}.\tag 5
$$
We will reach a contradiction if $\eta$ is less than some absolute constant.
Set
$$
\psi=\frac 1{2\pi} (\psi_\ve -\vp_\ve)
$$
so that $\psi:\Omega\to\Bbb Z$; recall that $u_\ve =e^{i\vp_\ve}$ and the
function $\vp_\ve$ is defined by
$$
\vp_\ve(x)= \cases 0 \qquad &  \text { for } -1<x<0,\\
2\pi x/\delta & \text { for } 0<x<\delta,\\
2\pi & \text { for } \delta<x<1,\endcases
$$
where $\delta=e^{-1/\ve}$.

A straightforward computation (using the fact that $\psi$ takes its values
into $\Bbb Z$) shows that
$$
|\psi(x)-\psi(y)|\leq |\psi_\ve(x)-\psi_\ve(y)| \text { for a.e. } x, y \in
\left(-1, \frac{2\delta}3\right)\tag 7
$$
and
$$
|\psi(x)-\psi(y)|\leq |\psi_\ve(x)-\psi_\ve(y)| \text { for a.e. } x, y\in
\left(\frac \delta 3, 1\right).\tag 8
$$
Applying Lemma 2 with $I=(-1, \frac{2\delta}3)$ and $I=(\frac\delta 3, 1)$,
together with (5), (7) and (8) yields the existence of $\ell, m\in\Bbb Z$
such that
$$
\bigg|\bigg\{x\in \left(-1, \frac{2\delta}3\right); \psi (x)\not= \ell
\bigg\} \bigg|\leq 2 (C^*\eta^2)^{1/\ve}
$$
and
$$
\bigg|\bigg\{x\in \left(x\in \frac{\delta}3, 1\right); \psi (x)\not= m
\bigg\} \bigg|\leq 2 (C^*\eta^2)^{1/\ve}.
$$
We choose $\eta$ in such a way that
$$
4(C^*\eta^2)^{1/\ve}< \delta/3, \text { for } \ve<1/2,
$$
for example
$$
\eta^2<1/4eC^*.\tag 9
$$
It follows that $\ell=m$.
Without loss of generality (after adding a constant to $\psi_\ve$) we may
assume that
$$
\ell=m=0.\tag 10
$$
Therefore
$$
\psi_\ve(x)=\vp_\ve(x) \ \text { for } x\in[(-1, 0)\backslash A]\cup
[(\delta, 1)\backslash B]\tag 11
$$
where
$$
A=\{x\in (-1, 0); \psi(x)\not= 0\}
$$
and
$$
B=\{x\in (\delta, 1); \psi(x)\not= 0\}
$$
with
$$
|A|<\delta/6, |B|<\delta/6.\tag 12
$$
>From (11) and the definition of $\vp_\ve$ we have
$$
\align
\ve\int_\Omega&\int_\Omega \frac{|\psi_ \ve(x) -\psi_\ve (y)|^2}{|x-y|^{2-\ve}}
dxdy \geq \ve\int^0_{-1}
dx\int^1_0\frac{|\psi_\ve(x)-\psi_\ve(y)|^2}{|x-y|^{2-\ve}}dy
\\
&\geq \ve \int_{(-1, 0)\backslash A} dx\int_{(\delta, 1)\backslash B} \frac
{|\vp_\ve(x)-\vp_\ve(y)|^2}{|x-y|^{2-\ve}} dy\\
& \geq \ve\int_{(-1, 0)\backslash A} dx\int_{(\delta, 1)\backslash B}\frac
{4\pi^2dy}{|x-y|^{2-\ve}}.
\endalign
$$
Applying Lemma 3 and (5) we find
$$
\align
\eta^2>\ve\int_\Omega&\int_\Omega \frac{|\psi_\ve(x)-\psi_\ve(y)|^2}{|x-y|^{2-\ve}}
dxdy \geq\ve\int_{-1}^{-|A|} dx\int^1_{\delta+|B|} \frac{4\pi^2
dy}{|x-y|^{2-\ve}}\\
&\geq \ve\int_{-1}^{-\delta/6} dx \int^1_{\delta+\delta/6}
\frac{4\pi^2dy}{|x-y|^{2-\ve}}=4\pi^2(1-e^{-1})+o(1)
\endalign
$$
as $\ve\to 0$.
We obtain a contradiction for an appropriate choice of $\eta$.




\bigskip
\noindent
{\bf APPENDIX.}
{ \bf Proof of square function inequality}

Let $\{\Cal F_n\}_{n= 0, 1, 2, \ldots}$ be refining finite partitions
such that
$$
\align
&\# \Cal F_n=K^n\\
&|Q| =K^{-n} \ \text { if $Q$ is an $\Cal F_n$-atom}
\endalign
$$
(If $\Omega =[0, 1]^d, K=2^d$).

Denote $\Bbb E_n$ the $\Cal F_n$-expectation
$$
\alignat2
\Delta_n f &=\Bbb E_n f -\Bbb E_{n-1} f \qquad&{\text {(we used the
notation $\tilde\Delta_n f$ in Section 4)}}\\
Sf&= \bigg(\sum|\Delta_n f|^2\bigg)^{1/2} \qquad &\text {(the square
function)}\\
|f|\underset {a.e.}\to \leq f^*&=\sup |\Bbb E_n f| \qquad &\text{(the
maximal function)}
\endalignat
$$

\proclaim
{Proposition 1} 
$$
{\text{\rm mes\,}}[|f|>\lambda\Vert Sf\Vert_\infty] < e^{-c\lambda^2} \qquad
(\lambda\geq 1)\tag 1
$$
where $c =c(K)>0$ is a constant.
\endproclaim 

\proclaim {Proposition 2} (good-$\lambda$ inequality)
$$
\text{\rm{mes\,}} [f^*>2\lambda, Sf<\ve\lambda, \sup\Bbb E_{n-1} [|\Delta_n
f|]<\ve\lambda] < e^{^{-\frac c{\ve^2}}}{\text{\rm mes\,}}[f^*>\lambda]\qquad
(0<\ve<1)\tag 2
$$
\endproclaim

\proclaim
{Proposition 3}
$$
\Vert f^* \Vert_q\leq C\sqrt q\Vert Sf\Vert_q \quad \text  { for $q\geq
2$}\tag 3
$$
\endproclaim

We follow essentially [4].

\noindent
{\bf Proof of Proposition 1.}

One verifies that there is a constant $A=A(K)$ such that if $\vp$ is
$\Cal F_n$-measurable and $\Bbb E_{n-1} \vp =0$, then
$$
\Bbb E_{n-1} [e^{\vp-A\vp^2}]\leq 1.\tag 4
$$
Hence
$$
\Bbb E_{n-1}[e^{\Delta_n f-A(\Delta_n f)^2}]\leq 1\tag 5
$$
and, denoting $S_n f=\big(\sum_{m\leq n}|\Delta_m f|^2\big)^{1/2}$,
$$
\align
\int e^{\Bbb E_n f-A(S_n f)^2} &=\int e^{\Bbb E_{n-1} f-A(S_{n-1} f)^2}
\Bbb E_{n-1}[e^{\Delta_n f -A(\Delta_n f)^2}]\\
&\leq \int e^{\Bbb E_{n-1}} f -A(S_{n-1} f)^2 \qquad \text { (by (5))}\\
&\leq 1.
\endalign
$$
Thus
$$
\int e^{f-A(Sf)^2} \leq 1.\tag 6
$$
Assume $\Vert Sf\Vert_\infty\leq 1$.
Applying (6) to $tf$ ($t>0$ a parameter), we get
$$
\align
\int e^{tf}&\leq e^{At^2}\\
\mes [f>\lambda]&\leq e^{At^2-t\lambda}
\endalign
$$
and for appropriate choice of $t$
$$
\mes [f>\lambda]< e^{-\frac{\lambda^2}{4A}}.
$$
This proves (1).

\noindent
{\bf Proof of Proposition 2.}

This is a standard stopping time argument.

Consider a collection of maximal atoms $\{Q_\alpha\}\subset\bigcup \Cal
F_n$ s.t. if $Q_\alpha$ is an $\Cal F_n$-atom, then $|\Bbb E_n
f|>\lambda$ on $Q_\alpha$.
Thus $Q_\alpha\cap Q_\beta=\phi$ for $\alpha\not=\beta$.
Fix $\alpha$. From the maximality
$$
|\Bbb E_{n-1} f|\leq \lambda \text { on } Q_\alpha.\tag 7
$$
Therefore
$$
\align
&[f^*>2\lambda, Sf<\ve\lambda, \sup\Bbb E_{m-1} [|\Delta_m f|] <\frac
1K\ve \lambda]\cap Q_\alpha\subset\\
&[(f-\Bbb E_n f)^*>(1-\ve)\lambda, Sf< \ve\lambda, \sup \Bbb E_{m-1}
[|\Delta_m f|]<\frac 1K\ve\lambda]\cap Q_\alpha = (8)
\endalign
$$
For $m>n$, denote $\chi_m$ the indicator function of the set
$$
Q_\alpha\cap \bigg[\bigg(\sum^{m-1}_{\ell=n+1} |\Delta_\ell
f|^2\bigg)^{1/2} <\ve \lambda\bigg]
\cap \bigg[\Bbb E_{m-1} [|\Delta_m f|] <\frac
1K\ve\lambda\bigg] \cap\bigcap_{n\leq\ell < m} [|\Bbb E_\ell f-\Bbb E_n
f|\leq (1-\ve)\lambda] = (9).
$$
Thus
$$
\chi_m=\Bbb E_{m-1}  \ \chi_m
$$
and
$$
g=\sum_{m> n} \chi_m \Delta_m f
$$
is an $\{\Cal F_m|m\geq n\}$-martingale on $Q_\alpha$.

>From the definition of $\chi_m$, we have clearly
$$
S(g) =\bigg(\sum_{m>n} \chi_m|\Delta_m f|^2\bigg)^{1/2}
<\ve\lambda+\ve\lambda\lesssim \ve \lambda \tag 10
$$
and 
$$
|g|>(1-\ve)\lambda \text { on the set (8)}.
$$
>From Proposition 1 and (10)
$$
\mes[x\ \ve Q_\alpha\big| \ |g|>(1-\ve)\lambda]< e^{-\frac
c{\ve^2}}|Q_\alpha|\tag 11
$$
hence
$$
\mes (8) \lesssim e^{-\frac c{\ve^2}}|Q_\alpha|.\tag 12
$$

Summing (12) over $\alpha$ implies
$$
\mes[f^*>2\lambda, Sf<\ve\lambda, \sup \Bbb E_{m-1} [|\Delta_m f|]
<\frac 1K \ve\lambda]< e^{-\frac c{\ve^2}}\sum|Q_\alpha|\leq e^{-\frac
c{\ve^2}}\mes [f^*>\lambda]
$$
which is (2).

\noindent
{\bf Proof of Proposition 3.}
$$
\align
\Vert f^*\Vert^q_q &= q\int\lambda^{q-1} \mes[f^*>\lambda]d\lambda\\
&=2^q q\int \lambda^{q-1}\mes [f^*>2\lambda] d\lambda\\
&\leq 2^q q\int\lambda^{q-1}\{ \mes [Sf\geq \ve\lambda]+ \mes [\sup \Bbb
E_{n-1}[|\Delta_n f|] \geq \frac\ve K\lambda]+ e^{-\frac c{\ve^2}}\mes
[f^*>\lambda]\}\\
&<\bigg(\frac 2\ve\bigg)^q (\Vert Sf\Vert^q_q +K^q\Vert \sup\Bbb
E_{n-1}[|\Delta_n f|] \Vert^q_q) + 2^q e^{-\frac c{\ve^2}}\Vert
f^*\Vert^q_q\tag 13
\endalign
$$
Take $\frac 1\ve\sim\sqrt q$ so that the last term in (13) is at most
$\frac 12 \Vert f^*\Vert^q_q$.
Thus
$$
\Vert f^*\Vert_q < C\sqrt q(\Vert Sf\Vert_q +\Vert\sup\Bbb E_{n-1}
[|\Delta_n f|]\Vert_q).\tag 14
$$
Also
$$
\align
\Vert\sup \Bbb E_{n-1}[|\Delta_n f|]\Vert_q &\leq \bigg(\sum_n\Vert
\Bbb E_{n-1} [|\Delta_n f|] \Vert_q^q\bigg)^{1/q}\\
&\leq \bigg(\sum_n \Vert\Delta_n f\Vert^q_q\bigg)^{1/q}\\
&\leq \Vert Sf\Vert_q.\tag 15
\endalign
$$
and (3) follows from (14), (15).

\noindent
{\bf Acknowledgment}

The first author (J.B.) is partially supported by NSF Grant
DMS-9801013.
The second author (H. B.) is partially sponsored by a European Grant ERB FMRX
CT98 0201. He is also a member of the Institut Universitaire de France.
Part of this work was done during a visit of the third author (P. M.)
at Rutgers University; he thanks the Mathematics Department for its support
and hospitality.

\Refs
\widestnumber\no{XXXXXX}

\item
{[1]}  R.A.~Adams,
{\it Sobolev spaces},
Acad. Press, 1975.

\item
{[2]}  J.~Bourgain, H.~Brezis and P.~Mironescu,
{\it Lifting in Sobolev spaces},
J.~d'Analyse {\bf 80} (2000), 37-86.

\item
{[3]}  J.~Bourgain, H.~Brezis and P.~Mironescu,
{\it Another look at Sobolev spaces},
Volume dedicated to A.~Bensoussan, IOS Press (to appear).

\item
{[4]}  S.~Chang, T.~Wilson, T.~Wolff,
{\it Some weighted norm inequalities concerning the Schr\"odinger operators
}, Comment. Math. Helv. 60 (1985), no 2, 217-246.

\item
{[5]}  D.~Gilbarg and N.S.~Trudinger,
{\it Elliptic partial differential equations of Second order},
Second edition, Springer-Verlag, Berlin, Heidelberg, New-York, 1983.

\item
{[6]}  H.~Triebel,
{\it Theory of function spaces},
Birkh\"auser, Basel, Boston, 1983.

\bigskip
\noindent
$\overline { \qquad\qquad \qquad\qquad\qquad }$

\bigskip
\line{\null\hfil{\vtop{
\hbox{Jean Bourgain}
\hbox{Institute for Advanced Study}
\hbox{Princeton, NJ 08540}
\hbox { email: bourgain\@math.ias.edu}
\bigskip
\hbox{H.~Brezis}
\hbox{Analyse Num\'erique}
\hbox{Universit\'e P. et M.~Curie, B.C. 187}
\hbox{4 Pl. Jussieu}
\hbox{75252 Paris Cedex 05, France}
\hbox { email: brezis\@ccr.jussieu.fr}
\bigskip
\hbox{ and }
\bigskip
\hbox{Department of Mathematics}
\hbox{Rutgers University}
\hbox{Hill Center, Busch Campus}
\hbox{110 Frelinghuysen Rd.}
\hbox{Piscataway, NJ 08854, USA}
\hbox{ email: brezis\@math.rutgers.edu}
\bigskip
\hbox{P.~Mironescu}
\hbox{Departement de Math\'ematiques}
\hbox{Universit\'e Paris-Sud}
\hbox{91405 Orsay, France}
\hbox{ email: Petru.Mironescu\@math.u-psud.fr}}\hfill}}
\endRefs
\bigskip

\enddocument