$\newcommand\norm[1]{\lVert#1\rVert}\newcommand\abs[1]{\lvert#1\rvert}$Throughout my studying for some papers, in particular, the proof of localized Strichartz estimates, I encountered a use of the duality argument I could not fully understand. The outline of the problem is as follows:


Consider we have a dispersive nonlinear PDE with some not nice, not symmetrized, linear parts. Let's say that we have the unitary group $\{W(t)\}\_{t \in \mathbb{R}}$. Using some cut-off functions, we can define on dyadic numbers some projections for initial data and want to prove for $f \in L^2(\mathbb{T^2})$
$$
\norm{W(t)f}\_{L^2\_t([0, 2^{- (j+k)}]) L^\infty(\mathbb{T^2})} \le C\_\delta(2^k, 2^j) \norm f\_{L^2(\mathbb{T^2})}.
$$
It suffices to prove
$$
\norm{\chi\_{[0, 2^{- (j+k)}]}(\abs t) W(t)f}\_{L^2\_t(\mathbb{R})} L^\infty(X) \le C\_\delta(2^k, 2^j) \norm f\_{\ell^2(\mathbb{T}^2)}.
$$
The author said: by the duality; it suffices to prove that for any $g \in L^2\_t$
$$\newcommand{\Dt}{{\operatorname d}t}
\biggl\lVert\int\_{\mathbb{R}} g(t)\chi\_{[0, 2^{- (j+k)}]}(\abs t) W(t)\Dt\biggr\rVert\_{\ell^2(\mathbb{Z}^2))} \le C\_\delta(2^k, 2^j) \norm g\_{L^2\_t(\mathbb{R})}.
$$
By expanding the norm on the left-hand side we get
$$\newcommand{\DT}{{\operatorname d}t^\prime}
\left\lVert\int\_{\mathbb{R}} \int\_{\mathbb{R}} g(t) g(t')\chi\_{[0, 2^{- (j+k)}]}(\abs t) \chi\_{[0, 2^{- (j+k)}]}(|t'|) W(t) W(t')\Dt\,\DT \right\rVert\le C\_\delta(2^k, 2^j).
$$
In the last step, I believe the author used the $TT^\*$ argument, but I do not fully understand the idea of the steps above. Could you please explain the ideas behind these?



Duality argument
### Context


Let me first clarify the context of the notation, since there are minor typos in the question posed.


Throughout $f$ is an $L^2(\mathbb{T}^2)$ function with restricted frequency support; this implies that $f$ is in fact $C^\infty$. For convenience let $X$ denote the (finite dimensional) subspace of $L^2(\mathbb{T}^2)$ with that frequency restriction.


The operators $W(t)$ are Fourier multipliers. That is, there exists coefficients $\omega: \mathbb{R}\times \mathbb{Z}^2\to \mathbb{C}$ such that
$$ [W(t) f](x,y) = \sum\_{m,n\in \mathbb{Z}^2} e^{i (mx + ny)} \omega(t,m,n) \hat{f}(m,n) $$


Note that therefore $W(t)f$ has restricted frequency support, and hence $W(t)f\in C^\infty$ for every fixed $t$.


The goals is to study the norm of the mapping $X\ni f \mapsto W(\cdot)f \in L^2(I, L^\infty(\mathbb{T}^2))$


### Method


1. Noting that $W(t)f$ is continuous, there exists a function $I\ni t \mapsto (x(t),y(t))\in \mathbb{T}^2$ such that $\|W(t) f\|\_{L^\infty(\mathbb{T}^2)} = |[W(t)f](x(t),y(t))|$. Hence it suffices to prove that, for all (measurable) paths $(x(t),y(t))$, the scalar function $[W(t) f](x(t),y(t))$ belongs to $L^2(I)$, with norm uniformly bounded by $\|f\|\_{L^2}$
2. Next, we use the duality method, where, for a fixed path $(x(t),y(t))$, we write
$$ \| [W(\cdot)f](x(\cdot),y(\cdot)) \|\_{L^2(I)} = \sup\_{g\in L^2(I)\setminus \{0\}} \frac{ \int\_I g(t) [W(t)f](x(t),y(t)) ~dt }{\|g\|\_{L^2(I)}} $$
3. For a fixed function $g$ and path $(x(t),y(t))$, observe the mapping
$$ X\ni f \mapsto \int\_I g(t) [W(t)f](x(t),y(t)) ~dt $$
is a linear functional (let's write it as $\mathscr{V}\_{g,x,y}$). So using Riesz representation we have that $\mathscr{V}\_{g,x,y}$ can be regarded as an element of $L^2(\mathbb{T}^2)$ (or actually $X^\*$). So we have that
$$ \| [W(\cdot)f](x(\cdot),y(\cdot)) \|\_{L^2(I)} \leq \sup\_{g\in L^2(I)\setminus \{0\}} \frac{ \mathscr{V}\_{g,x,y}(f)}{\|g\|\_{L^2(I)}} \leq \sup\_{g\in L^2(I)\setminus \{0\}} \frac{\|\mathscr{V}\_{g,x,y}\|\_{L^2(\mathbb{T}^2)} \|f\|\_{L^2(\mathbb{T}^2)}}{\|g\|\_{L^2(I)}} $$
4. Hence for proving what you want (that "the scalar function $[W(t) f](x(t),y(t))$ belongs to $L^2(I)$, with norm uniformly bounded by $\|f\|\_{L^2}$", as indicated in step 1), it is enough to find a bound of $\|\mathscr{V}\_{g,x,y}\|\_{L^2(\mathbb{T}^2)}$ by $\|g\|\_{L^2(I)}$ that is uniform over the choice of paths $(x(t),y(t))$. To do this, we use the Fourier representation and Plancherel. Note that
$$ \mathscr{V}\_{g,x,y}(f) = \int\_I g(t) \sum\_{m,n\in\mathbb{Z}^2} e^{imx(t)+iny(t)} \omega(t,m,n) \hat{f}(m,n) ~ dt $$
we find
$$ \| \mathscr{V}\_{g,x,y}\|\_{L^2(\mathbb{T}^2)} = \left\| \int\_I g(t) e^{imx(t) + iny(t)} \omega(t,m,n) ~dt \right\|\_{\ell^2(\mathbb{Z}^2)} \tag{N}$$
5. The final step, *which has nothing to do with $TT^\*$*, just expands this $\ell^2$ norm as
$$ \sum\_{m,n\in \mathbb{Z}^2} \iint\_{I\times I} g(t) \bar{g}(t') e^{im[x(t)- x(t')] + i n[y(t)-y(t')]} \omega(t,m,n)\bar{\omega}(t',m,n) ~dt~dt' \tag{P}$$
You need then to prove that this is uniformly bounded by $\|g\|\_{L^2(I)}^2$. If you now define the kernel (the sum is well-defined since the frequency restriction on $f$ means that $\omega$ is only supported on finitely many $(m,n)$ pairs)
$$ k(t,t') = \sum\_{m,n\in \mathbb{Z}^2} e^{im[x(t)- x(t')] + i n[y(t)-y(t')]} \omega(t,m,n)\bar{\omega}(t',m,n) $$
it finally is enough to prove that this kernel belongs to $L^2(I\times I)$, with its norm the constant you seek in step 1.




---


The OP asked for more details on Step 5. We will start from equation (N). Note that we are looking for the $\ell^2(\mathbb{Z}^2)$ norm of the mapping
$$ (m,n) \mapsto \int\_I g(t) e^{imx(t) + iny(t)} \omega(t,m,n) ~dt =: \mathfrak{v}(m,n) $$
Here we also introduce the notation $\mathfrak{v}(m,n)$ for convenience to represent the integral (for fixed $(m,n)$).


Its $\ell^2$ norm is simply
$$ \|\mathfrak{v}\|\_{\ell^2}^2 = \langle \mathfrak{v}, \overline{\mathfrak{v}} \rangle = \sum\_{m,n} \mathfrak{v}(m,n) \overline{\mathfrak{v}(m,n)} $$
If you plug in the definition of $\mathfrak{v}$ in terms of the integral (separately for each instance, so one integration for $\mathfrak{v}$ and one integration for its complex conjugate $\overline{\mathfrak{v}}$), you get exactly equation (P).


