Duality argument $\newcommand\norm[1]{\lVert#1\rVert}\newcommand\abs[1]{\lvert#1\rvert}$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?
 A: 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

*

*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 $\|f\|_{L^2}$

*Next, we use the duality method, where, for a fixed path $(x(t),y(t))$, we write
$$ \| [W(\cdot)f](x(\cdot),y(\cdot)) \|_{L^2(I)} = \sup_{g\in L^2(I)\setminus \{0\}} \frac{ \int_I g(t) [W(t)f](x(t),y(t)) ~dt }{\|g\|_{L^2(I)}} $$

*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
$$ \| [W(\cdot)f](x(\cdot),y(\cdot)) \|_{L^2(I)} \leq \sup_{g\in L^2(I)\setminus \{0\}} \frac{ \mathscr{V}_{g,x,y}(f)}{\|g\|_{L^2(I)}} \leq \sup_{g\in L^2(I)\setminus \{0\}} \frac{\|\mathscr{V}_{g,x,y}\|_{L^2(\mathbb{T}^2)} \|f\|_{L^2(\mathbb{T}^2)}}{\|g\|_{L^2(I)}} $$

*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
$$ \| \mathscr{V}_{g,x,y}\|_{L^2(\mathbb{T}^2)} = \left\| \int_I g(t) e^{imx(t) + iny(t)} \omega(t,m,n) ~dt \right\|_{\ell^2(\mathbb{Z}^2)} \tag{N}$$

*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).
