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\begin{document}

\title[A critical regularity algebra for the wave equation]{
An algebra for critical regularity solutions to the free wave equation}

\author{Terence Tao}
\address{Department of Mathematics, UCLA, Los Angeles, CA 90024}
\email{tao@@math.ucla.edu}

\subjclass{35L70}
\begin{abstract}
We present a Banach space $V \cap \dot H^{n/2,1/2} \cap L^\infty$
of functions in $\R^{n+1}$ which is
closed under multiplication and which contains all bounded free
$\dot H^{n/2}$ solutions to the free wave equation, when $n>2$.
\end{abstract}

\maketitle


\section{Introduction}\label{intro}

(Note: This is NOT an expository article, but rather a technical calculation which seems to have some mild interest to the critical regularity wave map problem.  I haven't been able to take this idea much further than what's in these notes, so feel free to do something with it.)  

Let $n > 2$.  Let $F$ denote the Banach 
space of all free $\dot H^{n/2}$ solutions
to the wave equation, and let $F \cap L^\infty$ denote the subspace of
those solutions which remain bounded.  In a recent article of Klainerman and 
Machedon, it is shown (among other things) that polynomial expressions of
$F \cap L^\infty$ always stay in the space $\dot H^{n/2,1/2}$ of
functions $\phi$ on $\R^{n+1}$ such that
$$ D_+^{n/2} D_-^{1/2} \phi \in L^2,$$
where $D_+$ and $D_-$ are the Fourier multipliers
corresponding to $|\tau|+|\xi|$ and $\left| |\tau| - |\xi|\right|$
respectively.  
We adopt the convention that functions in $F$ are also
in $\dot H^{n/2,1/2}$; such functions are in the weak closure 
of $\dot H^{n/2,1/2}$, and in practice one can perform a smooth time
localization to some large compact time interval $[-T,T]$,
using the ideas first used by Bourgain, to make the imbedding more precise.

This result would be easy to show if $\dot H^{n/2,1/2} \cap L^\infty$
were an algebra, but this is unlikely to be the case; certainly for
$n=1$ this space is not closed under multiplication.  However
in this note we show that there exists a subspace 
$V \cap \dot H^{n/2,1/2} \cap L^\infty$ of this space which contains
$F$ and which does
turn out to be an algebra, thus recovering Klainerman and
Machedon's result.  The algebra involves Besov-like
decompositions into frequency-localized pieces, and uses pointwise
majorization in physical space as well as in frequency space.

Unfortunately there are technical difficulties when $n=2$, which probably
require a modification of the definition of the space $V$, but
they do not appear insurmountable.  It would of course also
be very nice if the space
is closed under the bilinear operation $(\phi, \psi) \to
\Box^{-1}(\phi \Box \psi)$.  Although this operation does seem to
map to $\dot H^{n/2,1/2} \cap V$ if the dimension is sufficiently high,
it does not map into $L^\infty$, so this space is not quite powerful
enough to resolve a critical well-posedness problem.

We begin the construction of the algebra.  For any integer $j$,
let $Q^+_j$ denote
the Littlewood-Paley projection to the region of frequency space
where $D_+ \sim 2^j$, in such 
a way that $\sum_j Q^+_j$ is the identity.
Let $F_j = Q^+_j F$ denote the Banach space of functions of functions of the form
$Q^+_j \phi$, where $\phi$ is in $F$; this is the space of all
$\dot H^{n/2}$ free solutions with frequencies $\sim 2^j$.

Consider the set $F_j \cdot L^\infty$, the set of all functions which
are pointwise dominated by a function in $F_j$.  This set is
not closed under addition and is therefore not a Banach space; however
we can complete it by taking the elements of that set to be atoms.
More precisely, we define $V_j$ to be the closure of the normed space
whose unit ball is
$$
\{ \sum_\alpha \lambda_\alpha Q^+_j(\phi_\alpha) \chi_\alpha: 
\sum_\alpha |\lambda_\alpha| \leq 1, \|\phi_\alpha\|_{F} \leq 1,
\| \chi_\alpha\|_\infty \leq 1 \}$$
and $\alpha$ can range over any finite index set.  It is easy to see
that $V_j$ is a Banach space.

The space $F$ can be recovered from its components $F_j$ by the
formula
$$ \| \phi \|_F \sim (\sum_j \| Q^+_j \phi \|_{F_j}^2)^{1/2}.$$
In analogy with this formula, we define the Banach space $V$ by
$$ \| \phi \|_V \sim (\sum_j \| Q^+_j \phi \|_{V_j}^2)^{1/2}.$$
Since $V_j$ clearly contains $F_j$, we see that $V$ contains $F$.
Thus to complete the claim of this note, we need to show

\begin{proposition}  The space $V \cap \dot H^{n/2,1/2} \cap L^\infty$
is an algebra.
\end{proposition}

We now begin the proof.  Let $\phi$, $\psi$ be functions such that
$$ \| \phi\|_V, \| \phi \|_{\dot H^{n/2,1/2}}, \| \phi\|_\infty,
\| \psi\|_V, \| \psi \|_{\dot H^{n/2,1/2}}, \| \psi\|_\infty \lesssim 1.$$
We have to show that
\ba
\| \phi \psi \|_\infty &\lesssim 1. \label{l-est}\\
\| \phi \psi \|_{\dot H^{n/2,1/2}} &\lesssim 1 \label{h-est}\\
\| \phi \psi \|_V &\lesssim 1 \label{v-est}
\end{align}
The estimate \eqref{l-est} is trivial from the $L^\infty$ control
on $\phi$, $\psi$, so it remains to show \eqref{h-est} and \eqref{v-est}.

\section{Proof of \eqref{h-est}.}

We use the formula
\be{form}
 \| \phi \|_{\dot H^{n/2,1/2}} \sim
\left(\sum_{j,k} (2^{nj/2} 2^{k/2} \| Q^+_j Q^-_k \phi \|_2)^2\right)^{1/2},
\end{equation}
where $Q^-_k$ is the Fourier projection to the region of frequency space
in which $D_- \sim 2^k$.  We thus consider the problem of controlling
\be{qjk}
 \| Q^+_j Q^-_k (\phi \psi) \|_2
\end{equation}
for some integers $j,k$, which we now fix.  
We may of course assume $j \geq k - C$ since
the expression vanishes otherwise.

We break up $\phi$, $\psi$ dyadically as
$$ \phi = P^+_{k-C} \phi + \sum_{j_1 > k-C} Q^+_{j_1} \phi$$
$$ \psi = P^+_{k-C} \psi + \sum_{j_2 > k-C} Q^+_{j_2} \psi$$
where $P^+_k = \sum_{j \leq k} Q^+_j$,
and consider the various contributions of the summands to \eqref{qjk}
and hence to \eqref{form}.

We first consider the contribution of $P^+_{k-C} \phi$.  The only
terms in the expansion of $\psi$ which give a nonvanishing contribution
to \eqref{qjk} are the ones in which $2^{j_2} \sim 2^j$.  In fact
one can see from an examination of frequency supports that
$$ Q^+_j Q^-_k ((P^+_{k-C} \phi) \psi) =
\sum_{j' = j + O(1)} \sum_{k' = k + O(1)}  Q^+_j Q^-_k ((P^+_{k-C} \phi) 
Q^+_{j'} Q^-_{k'} \psi).$$
We can thus essentially reduce to estimating
$$ Q^+_j Q^-_k ((P^+_{k-C} \phi) Q^+_j Q^-_k \psi).$$
Since $Q^+_j Q^-_k$ is bounded on $L^2$, the 
contribution of $P^+_{k-C} \phi$ to \eqref{qjk} is thus majorized 
by
$$ \| (P^+_{k-C} \phi) 
Q^+_{j} Q^-_{k} \psi\|_2.$$
Since $\phi$ is in $L^\infty$ by hypothesis and $P^+_{k-C}$ is an approximation
to the identity, $P^+_{k-C} \phi$ is in $L^\infty$ and we can therefore
estimate the above by
$$\| 
Q^+_j Q^-_{k} \psi\|_2.$$
By \eqref{form}, the contribution of the above to 
\eqref{h-est} is thus majorized by
$$
\left(\sum_{j,k} (2^{nj/2} 2^{k/2} \| Q^+_{j_2} Q^-_{k_2} \psi\|_2)^2\right)^{1/2}.
$$
But by \eqref{form} this is comparable to $\| \psi \|_{\dot H^{n/2,1/2}}$,
which is acceptable by hypothesis.

A similar argument disposes of the contribution of
$P^+_{k-C} \psi$.
Note that the simultaneous contribution of $P^+_{k-C} \phi$ and
$P^+_{k-C} \psi$ to \eqref{qjk} vanishes.

By the triangle inequality,
the remaining contribution to \eqref{qjk} is majorized by
$$ \sum_{j_1 > k-C} \sum_{j_2 > k-C} 
\|Q^+_j Q^-_k(Q^+_{j_1} \phi Q^+_{j_2} \psi) \|_2.$$
By symmetry we can restrict ourselves to the case $j_1 \geq j_2$.
By inspecting the frequency support of $Q^+_{j_1} \phi Q^+_{j_2} \psi$,
we see that the summand vanishes unless the two largest members
in $\{j, j_1, j_2\}$ are within $O(1)$ of each other.
We may therefore estimate the above sum by
\be{test}
 \sum_{j_1 > j-C}  \sum_{j_2 = j_1 +O(1)} + \sum_{j_1 = j+O(1)} \sum_{k-C < j_2 < j+C}
\|Q^+_j Q^-_k(Q^+_{j_1} \phi Q^+_{j_2} \psi) \|_2.
\end{equation}
We may drop the operator $Q^+_j Q^-_k$ since it is bounded in
$L^2$. 

We now use

\begin{proposition}\label{useful}  Let $j \geq k - C$. 
If $\phi \in V_j$, $\psi \in V_k$, and $n>2$, then
$$ \| \phi \psi\|_2 \lesssim 2^{-nj/2} 2^{-k/2} \| \phi\|_{V_j} \|\psi\|_{V_k}.$$
\end{proposition}

\begin{proof}  We may assume that $\phi$, $\psi$ are atoms of $V_j$, $V_k$
respectively.  Since multiplication by $L^\infty$ functions has no
harmful effect on $\|\phi \psi\|_2$, it thus suffices to show that
$$ \| Q^+_j(\phi) Q^+_k(\psi) \|_2 \lesssim 2^{-nj/2} 2^{-k/2} \| \phi \|_{F_j}
\| \psi \|_{F_k}$$
for all $\phi \in F_j$, $\psi \in F_k$.

By Plancherel's theorem, this is equivalent to showing that
$$ \| f d\sigma_j * g d\sigma_k \|_2 \lesssim 2^{-nj/2} 2^{-k/2} 
2^{nj/2} \|f\|_2 2^{nk/2} \|g\|_2$$
for all $f \in L^2(S_j)$, $g \in L^2(S_k)$, where $S_j$ is the
surface $\{ (\tau,\xi): |\tau| = |\xi| \sim 2^j\}$ with surface measure
$d\sigma_j$, and similarly for $S_k$.  By interpolation it suffices
to prove the estimates
$$ \| f d\sigma_j * g d\sigma_k \|_1 \lesssim 
\|f\|_1 \|g\|_1$$
and
$$ \| f d\sigma_j * g d\sigma_k \|_\infty \lesssim 2^{-nj} 2^{-k} 
2^{nj} \|f\|_\infty 2^{nk} \|g\|_\infty.$$
The former estimate is just Young's inequality, whereas the latter
follows from the easily verified calculation
$$ \| d\sigma_j * d\sigma_k \|_\infty \lesssim 2^{(n-1) k}.$$
Note that this last calculation requires $n>2$.
\end{proof}

From this proposition, we can bound \eqref{test} by
 $$
\sum_{j_1 > j-C}  \sum_{j_2 = j_1 +O(1)} + \sum_{j_1 = j+O(1)} \sum_{k-C < j_2 < j+C}
2^{-nj_1/2} 2^{-j_2/2} \| \phi \|_{V_{j_1}} \|\psi \|_{V_{j_2}}
$$
This in turn is easily seen to be majorized by
$$ 2^{-nj/2} 2^{-k/2} 
(\sum_{j_1} 2^{-|j_1-j|/2} \| \phi \|_{V_{j_1}})
(\sum_{j_2} 2^{-|j_2-k|/2} \| \psi \|_{V_{j_2}}).
$$
By \eqref{form}, the contribution to \eqref{h-est} is thus majorized
by
$$
\left(\sum_{j,k} (\sum_{j_1} 2^{-|j_1-j|/2} \| \phi \|_{V_{j_1}})^2
(\sum_{j_2} 2^{-|j_2-k|/2} \| \psi \|_{V_{j_2}})^2\right)^{1/2}.
$$
This factors as
$$
\left(\sum_j (\sum_{j_1} 2^{-|j_1-j|/2} \| \phi \|_{V_{j_1}})^2\right)^{1/2}
\left(\sum_k (\sum_{j_2} 2^{-|j_2-k|/2} \| \psi \|_{V_{j_2}})^2\right)^{1/2}.
$$
By Young's inequality for sums, and the fact that the kernel $2^{-|m|/2}$ is
summable in $m$, this is majorized by
$$
\left(\sum_{j_1} \| \phi \|_{V_{j_1}}^2\right)^{1/2}
\left(\sum_{j_2} \| \psi \|_{V_{j_2}}^2\right)^{1/2} = \| \phi\|_V \|\psi\|_V,
$$
which is acceptable by hypothesis.

\section{Proof of \eqref{v-est}}

To conclude the proof of the Proposition, we need to show that
\be{v-targ}
 (\sum_j \| Q^+_j( \phi \psi )\|_{V_j}^2)^{1/2} \lesssim 1.
\end{equation}
We split $Q^+_j(\phi \psi)$ as the sum of
\be{est}
Q^+_j(\phi) P^+_j(\psi) + P^+_j(\phi) Q^+_j(\psi)
\end{equation}
and
\be{error}
Q^+_j(\phi \psi) - Q^+_j(\phi) P^+_j(\psi) - P^+_j(\phi) Q^+_j(\psi).
\end{equation}

First consider \eqref{est}.  Since $\psi$, $\phi$ are in $L^\infty$ by hypothesis, and
$P_j$ is an approximation to the identity, we have
that $P^+_j(\psi)$, $P^+_j(\phi)$ are in $L^\infty$.  Since the
spaces $V_j$ are clearly closed under multiplication by $L^\infty$
functions, we thus have
$$ \|  Q^+_j(\phi) P^+_j(\psi) + P^+_j(\phi) Q^+_j(\psi) \|_{V_j}
\lesssim \| Q^+_j(\phi) \|_{V_j} + \| Q^+_j(\psi) \|_{V_j}.$$
Thus the contribution of \eqref{est} to \eqref{v-targ} is majorized by
$$ (\sum_j \| Q^+_j( \phi )\|_{V_j}^2)^{1/2} +
(\sum_j \| Q^+_j( \psi )\|_{V_j}^2)^{1/2} = \| \phi\|_V + \| \psi \|_V,$$
which is acceptable by hypothesis.  

It remains to control the error term \eqref{error}.
We will in fact prove that
\be{tricky}
\begin{split}
\|Q^+_j(\phi \psi) - Q^+_j(\phi) P^+_j(\psi) - & P^+_j(\phi) Q^+_j(\psi)\|_2
\lesssim 
2^{-nj/2} 2^{-j/2} \\
(\sum_{j_1} 2^{-|j_1-j|/2} \| Q^+_{j_1} \phi \|_{V_{j_1}})
(\sum_{j_2} 2^{-|j_2-j|/2} \| Q^+_{j_2} \psi \|_{V_{j_2}}).
\end{split}
\end{equation}

We will prove this estimate momentarily.  To apply \eqref{tricky},
we need the following

\begin{lemma}  If $\phi$ has frequency support in the region $|\xi| + |\tau| \sim 2^j$, then
$$ \| \phi \|_{V_j} \lesssim 2^{nj/2} 2^{j/2} \|\phi\|_2.$$
\end{lemma}

\begin{proof}  By partitioning $\phi$ and multiplying by a harmless phase
function to translate the frequency support, we may assume that
$\phi$ has Fourier transform on the set $\{ (\tau, \xi): \tau = |\xi| + O(2^j),
\xi \in A_j \}$, where $A_j$ is an annulus on which the symbol of
$Q^+_j$ is bounded from below.  We then apply the formula
$$ \hat{\phi}(\tau,\xi) = \int_{|\lambda| \lesssim 2^j}
\hat{\phi}(|\xi| + \lambda, \xi) \delta(\tau - \lambda - |\xi|)\ d\lambda.$$
The inverse Fourier transform of $\hat{\phi}(|\xi| + \lambda, \xi) \delta(\tau - \lambda - |\xi|)$ is equal to a phase function multiplied by
a solution to the free wave equation, with $\dot H^{n/2}$ norm comparable to
$$ 2^{nj/2} (\int_{A_j} |\hat{\phi}(|\xi| + \lambda, \xi)|^2\ d\xi)^{1/2}.$$
Thus by the integral form of the triangle inequality, we have
$$ \| \phi \|_{V_j} \lesssim 2^{nj/2} \int_{|\lambda| \lesssim 2^j}
(\int_{A_j} |\hat{\phi}(|\xi| + \lambda, \xi)|^2\ d\xi)^{1/2}\ d\lambda.$$
By Cauchy-Schwarz, the right-hand side is majorized by
$$  2^{nj/2} 2^{j/2} (\int_{|\lambda| \lesssim 2^j}
\int_{A_j} |\hat{\phi}(|\xi| + \lambda, \xi)|^2\ d\xi\ d\lambda)^{1/2},$$
and the claim follows from a change of variables and Plancherel's theorem.
\end{proof}

From this lemma, we see that the contribution of \eqref{error} to
\eqref{v-targ} is majorized by
$$
(\sum_j 
(\sum_{j_1} 2^{-|j_1-j|/2} \| Q^+_{j_1} \phi \|_{V_{j_1}})^2
(\sum_{j_2} 2^{-|j_2-j|/2} \| Q^+_{j_2} \psi \|_{V_{j_2}})^2)^{1/2}.$$
By Cauchy-Schwarz, this is majorized by
$$
(\sum_j 
(\sum_{j_1} 2^{-|j_1-j|/2} \| Q^+_{j_1} \phi \|_{V_{j_1}})^4)^{1/4}
(\sum_j 
(\sum_{j_2} 2^{-|j_2-j|/2} \| Q^+_{j_2} \psi \|_{V_{j_2}})^4)^{1/4}.$$
By Young's inequality for sums and the fact that $2^{-|m|/2}$ is 
in $l^{4/3}$, this is majorized by
$$ 
(\sum_{j_1}  \| Q^+_{j_1} \phi \|_{V_{j_1}})^2)^{1/2}
(\sum_{j_2}  \| Q^+_{j_2} \psi \|_{V_{j_2}})^2)^{1/2} = \|\phi\|_V \|\psi\|_V,$$
which is acceptable by hypothesis.

It remains to prove \eqref{tricky}.  We decompose this expression dyadically
as
\bas
Q^+_j(\phi \psi) - &Q^+_j(\phi) P^+_j(\psi) - P^+_j(\phi) Q^+_j(\psi)
=\\
&\sum_{j_1} \sum_{j_2} 
Q^+_j(Q^+_{j_1}\phi Q^+_{j_2}\psi) - Q^+_j(Q^+_{j_1} \phi) 
P^+_j(Q^+_{j_2} \psi) - P^+_j(Q^+_{j_1} \phi) Q^+_j(Q^+_{j_2} \psi).
\end{align*}
It suffices to prove \eqref{tricky} for a single pair of indices
$j_1$, $j_2$, with an exponential decay in $|j_1-j|$, $|j_2-j|$.  More
precisely, it suffices to show that
\be{dyadic}
\begin{split}
\|
Q^+_j(Q^+_{j_1}\phi Q^+_{j_2}\psi) &- Q^+_j(Q^+_{j_1} \phi) 
P^+_j(Q^+_{j_2} \psi) - P^+_j(Q^+_{j_1} \phi) Q^+_j(Q^+_{j_2} \psi)
\|_2 \lesssim \\
&2^{-|j_1-j|/2} 2^{-|j_2-j|/2}
2^{-nj/2} 2^{-j/2} 
\| Q^+_{j_1} \phi \|_{V_{j_1}}
\| Q^+_{j_2} \psi \|_{V_{j_2}},
\end{split}
\end{equation}
since \eqref{tricky} follows by summing in $j_1, j_2$, then
using the triangle inequality and Cauchy-Schwarz.

We may apply the
rescaling $\tilde \phi(t,x) = \phi(2^{-j}t, 2^{-j}x)$, 
$\tilde \psi(t,x) = \psi(2^{-j}t, 2^{-j}x)$ and assume that $j=0$.  By
symmetry we may take $j_1 \geq j_2$.

We first consider the case when $j_1 \geq j_2 > -C$ for some large $C$.
We then estimate \eqref{dyadic} crudely as
$$ \| Q^+_0(Q^+_{j_1}\phi Q^+_{j_2}\psi) \|_2 +
\| Q^+_0(Q^+_{j_1} \phi) P^+_0(Q^+_{j_2} \psi)\|_2 +
\| P^+_0(Q^+_{j_1} \phi) Q^+_0(Q^+_{j_2} \psi)\|_2.$$
Discarding the $Q^+_0$ projection from the first term, we 
apply Proposition \ref{useful}
to estimate this by
$$ 2^{-n j_1/2} 2^{-j_2/2} \| Q^+_{j_1}\phi \|_{V_{j_1}} 
\| Q^+_{j_2}\psi \|_{V_{j_2}},$$
which implies \eqref{dyadic} for this case.

We can therefore assume that $j_2 < -C$; this implies
$j_1 = O(1)$ since \eqref{dyadic}
vanishes otherwise.  The left-hand side of \eqref{dyadic} then
simplifies to
\be{cancel}
\|
Q^+_0(Q^+_{j_1}\phi Q^+_{j_2}\psi) - Q^+_0(Q^+_{j_1} \phi)
Q^+_{j_2} \psi \|_2.
\end{equation}
The idea is to exploit the cancellation between the two terms on
the multiplier level.
The expression inside the norm can be written as
\be{complex}
[Q^+_0(Q^+_{j_1}\phi Q^+_{j_2}\psi) - Q^+_0(Q^+_{j_1} \phi)
Q^+_{j_2} \psi](\tilde x) = 
\int_{|\tilde \xi| \sim 2^{j_1}} \int_{|\tilde \eta| \sim 2^{j_2}}
e^{2 \pi i \tilde x \cdot (\tilde \xi + \tilde \eta)} 
m(\tilde \xi,\tilde \eta,) \hat{\phi}(\tilde \xi) \hat{\psi}(\tilde \eta)
\ d\tilde \xi d\tilde \eta
\end{equation}
where to avoid notational complexity we have
used spacetime variables $\tilde \xi, \tilde \eta$, $\tilde x$ in
the Euclidean space $\R^{n+1}$, as opposed to the more usual
separate variables for space and time; $m$ is the multiplier
$$
m(\tilde \xi,\tilde \eta) = (\psi(\tilde \xi + \tilde \eta) - \psi(\tilde \xi))
\psi(2^{-j_1}\tilde \xi) \psi(2^{-j_2}\tilde \eta),$$
and $\psi$ is the symbol of the Littlewood-Paley projection $Q^+_0$.

Since $\tilde \eta$ is constrained to be about $2^{j_2}$, the expression
$\psi(\tilde \xi + \tilde \eta) - \psi(\tilde \xi)$ is $O(2^{j_2})$.
In fact, we have the symbol bounds
$$ \|\partial_{\tilde \xi}^\alpha \partial_{\tilde \eta}^\beta m\|_\infty
\leq C_{\alpha,\beta} 2^{j_2} 2^{-j_1 |\alpha|} 2^{-j_2 |\beta|}$$
for all $\alpha, \beta$.  We can therefore
decompose $m$ as a Fourier series in the box $\tilde \xi \in [- C2^{j_1},
C 2^{j_1}]^{n+1}$, $\tilde \eta \in [- C2^{j_2}, C 2^{j_2}]^{n+1}$ as
$$ m(\tilde \xi,\tilde \eta) = 2^{j_2} \sum_{q \in \Z^{n+1}} 
\sum_{r \in \Z^{n+1}}
c_{q,r} e^{2 \pi i C^{-1} 2^{-j_1} q \cdot \tilde \xi} 
e^{2 \pi i C^{-1} 2^{-j_2} r \cdot \tilde \eta}
\psi(2^{-j_1}\tilde \xi) \psi(2^{-j_2}\tilde \eta)$$
where the Fourier coefficients $c_{q,r}$ are rapidly decreasing.  
Inserting this into \eqref{complex} and simplifying using the Fourier
inversion formula, we can rewrite \eqref{complex} as
$$ 
Q^+_0(Q^+_{j_1}\phi Q^+_{j_2}\psi) - Q^+_0(Q^+_{j_1} \phi)
Q^+_{j_2} \psi =
2^{j_2} \sum_{q} \sum_{r}
c_{q,r} (\tau_{C^{-1} 2^{-j_1} q} Q^+_{j_1} \phi)
(\tau_{C^{-1} 2^{-j_2} r} Q^+_{j_2} \psi),$$
where $\tau_y$ denotes a translation in physical spacetime by $y$.

We now apply the triangle inequality and
Proposition \ref{useful} to estimate \eqref{cancel} by
$$ 2^{j_2} \sum_{q} \sum_{r}
|c_{q,r}| 2^{-nj_1/2} 2^{-j_2/2}
\| \tau_{C^{-1} 2^{-j_1} q} Q^+_{j_1} \phi \|_{V_{j_1}}
\| \tau_{C^{-1} 2^{-j_2} r} Q^+_{j_2} \psi \|_{V_{j_2}}.$$
The norms $V_j$ are easily seen to be translation invariant,
so we can rewrite this as
$$ 2^{-n j_1/2} 2^{j_2/2} (\sum_{q} \sum_{r}
|c_{q,r}|)
\| Q^+_{j_1} \phi \|_{V_{j_1}}
\| Q^+_{j_2} \psi \|_{V_{j_2}}.$$
Since $j_1 = O(1)$, $j_2 < -C$, and $c_{q,r}$ is rapidly decreasing,
we thus obtain \eqref{dyadic} as desired.

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\end{document}
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