rangle^{2b}|f_{n,m}|^2\bigg)^{1/2}
$$
is finite, where the infinum is taken over all representations \eqref{(2.8)}.  We also refer the reader
to the works \cite{B-GAFA1}-\cite{B-GAFA2}, where these spaces were first introduced.

Moreover, when $f:I\times B\rightarrow\mathbb{C}$ has a representation \eqref{(2.8)}, we shall define the
function $T_{s,b}f$ via
\be\label{(2.9)}
(T_{s,b}f)(x,t)=\sum_{n,m}\langle n\rangle^{s}\langle n^2-m\rangle^{b}f_{n,m}e_n(x)e(mt).
\ee

Our analysis requires to consider short time intervals $[0, T]$, where $T$ will depend on the truncation
parameters.
In order to establish contractive estimates for the nonlinear term, we need a variant of the 
$\Vert \cdot \Vert_{0, \frac 12}$-norm adapted to the time interval.
We denote this norm by $\ltriple \cdot \ltriple_{0, \frac 12; T}$, and its
unit ball is generated by functions of the form
\be\label {(2.10)}
\sum_{n, m} \frac {a_{n, m}}{(|n^2-m|+\frac 1T)^{\frac 12}} \, e_n(x) e(mt) +\sum_{\substack {n, m\\
|n^2-m|>\frac 1T}} \frac {a_n}{|n^2-m|}\, e_n(x) e(mt)
\ee
with
$$
\sum_{n, m}|a_{n, m}|^2 \leq 1\quad\textrm{and}\quad\sum_n |a_n|^2\leq 1.
$$

Obviously, $\Vert \cdot \Vert_{0, b}\lesssim \ltriple\,\cdot\,\rtriple$ 
for $b<\frac 12$.  One can similarly introduce norms  
$\ltriple\,\cdot\,\rtriple_{s, \frac 12; T}$ for 
$s>0$, but we will not need them for our purposes.

The next few lemmas put into evidence some basic properties of the norm $\ltriple \cdot \rtriple$.

\begin{lemma}\label{Lemma1}
Let $\ltriple f\ltriple\leq 1$. Then
\be\label{(2.11)}
\frac 1T \int_0^T \Vert f(t) \Vert^2_{L^2_x} dt< O(1).
\ee
\end{lemma}

\begin{proof}
We first write $f$ as in \eqref{(2.10)}.  Then
\begin{align*}
\Vert f(t)\Vert^2_{{L^2_x}} &=\sum_n\bigg|\sum_m \frac {f_{n, m}}{(|n^2-m|+\frac 1T)^{\frac{1}{2}}}
\, e(mt)\bigg|^2+\sum_n \bigg|\sum_{|n^2 -m|>\frac 1T} \ \frac {e(mt)}{n^2-m}\bigg|^2|f_n|^2\\
&= (I)+ (II).
\end{align*}

Taking $0\leq \vp \leq 2$ such that $\vp\geq 1$ on $[0, 1]$ and supp\,$\hat\vp \subset [-1,1]$, we have
$$
\begin{aligned}
\int_0^T (I) dt\leq \int (I) \vp\Big(\frac tT\Big)dt&\leq T\sum_n \, \sum_{\substack{m, m',\\|m-m'|\leq \frac 1T}} \frac{|f_{n,m}|\,|f_{n,m'}|}{|n^2-m|+\frac{1}{T}}\\
&\leq T^2\sum_{|k|\leq \frac 1T} \ \sum_{n, m} |f_{n, m}| \, |f_{n, m+k}|\\
&\lesssim T\Vert f\Vert^2_{L_{t,x}^2}\\
&\lesssim T
\end{aligned}
$$
and similarly
$$
\begin{aligned}
\int_0^T (II)dt&\lesssim T\sum_n \sum_{\substack {m,m',\,|m-m'|\lesssim \frac 1T\\ 
|n^2-m|>\frac 1T,\,|n^2-m'|>\frac 1T}}\ \frac {|f_n|^2}{|n^2-m| \, |n^2-m'|}\\
&\lesssim \sum_{\substack{n, m\\ |n^2-m|>\frac 1T}} \ \frac {|f_n|^2}{|n^2-m|^2} \\
&\lesssim T.
\end{aligned}
$$
The combination of these two bounds suffices to prove the claim.
\end{proof}

The next statement expresses an important duality property with respect to
the Duhamel formula \eqref{(1.2)}.
\begin{lemma}\label{Lemma2}
Assume $\displaystyle f(x, t) =\sum_{{n\in\mathbb Z_+, m\in\mathbb Z}} f_{n, m} e_n(x) e(mt)$.
Then
$$
\Bigltriple \int^t_0 e^{i(t-\tau)\Delta} f(\tau) d\tau\Bigrtriple
\lesssim \max_{\ltriple g\ltriple_{0, \frac 12; T\leq 1}} \Big|\sum_{n,m}
f_{n, m} g_{n, m}\Big|,\eqno{(2.14)}
$$
where $\displaystyle g(x, t) =\sum_{\substack {n\in\mathbb{Z_+}, m\in \mathbb{Z}}} g_{n,m}e_n(x)e(mt)$.
\end{lemma}

\begin{proof}
Write
$$ \int^t_0 e^{i(t-\tau)\Delta} f(\tau)d\tau =\sum_{n,m} f_{n, m}\, e_n \frac {e(mt) - e(n^2t)}{m-n^2}
$$
and decompose this as
$$
\hspace{0.1in}\sum_{|m-n^2|>\frac 1T} \frac {f_{n, m}}{m-n^2} \, e_n \, e(mt)\eqno{(2.15)}\\
$$
$$
-\sum_{|m-n^2|> \frac 1T} \, \frac {f_{n, m}}{m-n^2} \, e_n \, e(n^2t)\eqno{(2.16)}\\
$$
$$
\hspace{0.3in} +\sum_{|m-n^2|\leq\frac 1T} f_{n, m} e_n\, \frac {e(mt)-e(n^2t)}{m-n^2}.\eqno{(2.17)}
$$
Hence we may write (2.15) as
$$
\sum_{|m-n^2|> \frac 1T} \frac{ b_{n, m}}{|m-n^2|^{\frac 12}} e_n \, e (mt)
$$
with
$$
b_{n, m} =\frac {\pm f_{n, m}}{|m-n^2|^{\frac 12}},
$$
which satisfies
\begin{align*}
\Big(\sum_{n,m}|b_{n, m}|^2\Big)^{\frac 12} &= \Big(\sum_{|m-n^2|>\frac 1T} \ \frac
{|f_{n, m}|^2}{|m-n^2|}\Big)^{\frac 12}
= \max\bigg|\sum_{|m-n^2|>{\frac 1T}} f_{n, m} \frac {a_{n, m}}{|m-n^2|^{\frac
12}}\bigg|
\end{align*}
where the maximum is over sequences $(a_{n,m}$) with
$$
\Big(\sum_{n,m} |a_{n, m}|^2\Big)^{1/2}\leq 1,
$$
which takes care of the contribution of (2.15) to the left-