Dual local smoothing and retarded local smoothing for Schrodinger equation This exercise is from Tao's Nonlinear Dispersive Equations: Local and Global Analysis, Exercise 2.54.

Let $u$ be a solution to the inhomogeneous Schrodinger equation $i\partial_t u+\Delta u=F$, which is smooth in time and Schwarz in space. Establish the dual local smoothing estimate $$\sup_{t\in\mathbf{R}}\Vert u(t)\Vert_{\dot{H}_x^{1/2}(\mathbf{R}^3)}\lesssim_{\epsilon}\Vert u(0)\Vert_{\dot{H}_x^{1/2}(\mathbf{R}^3)}+\int_{\mathbf{R}}\int_{\mathbf{R}^3}\langle x\rangle ^{1+\epsilon}|F(t,x)|^2dxdt$$ for any $\epsilon>0.$ Also establish the retarded local smoothing estimate $$\int_{\mathbf{R}}\int_{\mathbf{R}^3}\langle x\rangle^{-1-\epsilon}|\nabla u(t,x)|^2+\langle x\rangle^{-3-\epsilon}|u(t,x)|^2dxdt\lesssim_{\epsilon}\Vert u(0,x)\Vert_{\dot{H}_x^{1/2}(\mathbf{R}^3)}^2+\int_{\mathbf{R}}\int_{\mathbf{R}^3}\langle x\rangle ^{1+\epsilon}|F(t,x)|^2dxdt.$$

My attempt: By Duhamel's formula, unitary property of Schrodinger propagator, and triangle inequality, one gets \begin{align}\Vert u(t)\Vert_{\dot{H}_x^{1/2}(\mathbf{R}^3)}&\leq \Vert e^{it\Delta/2}u(0)\Vert_{\dot{H}_x^{1/2}(\mathbf{R}^3)}+\Vert\int_0^te^{i(t-s)\Delta/2}F(s,x)ds\Vert_{\dot{H}_x^{1/2}(\mathbf{R}^3)}\\&\leq \Vert u(0)\Vert_{\dot{H}_x^{1/2}(\mathbf{R}^3)}+\int_{\mathbf{R}}\Vert F(s,x)\Vert_{\dot{H}_x^{1/2}(\mathbf{R}^3)}ds\end{align}
Therefore, it suffices to show that $\Vert F(s,x)\Vert_{\dot{H}_x^{1/2}(\mathbf{R}^3)}\lesssim_{\epsilon}\int_{\mathbf{R}^3}\langle x\rangle ^{1+\epsilon}|F(t,x)|^2dx$, and here is where I'm stuck.
How should I proceed here and for retarded estimates? The textbook provided a hint to use the homogeneous local smoothing estimate $$\int_{\mathbf{R}}\int_{\mathbf{R}^3}\langle x\rangle^{-1-\epsilon}|\nabla u(t,x)|^2+\langle x\rangle^{-3-\epsilon}|u(t,x)|^2dxdt\lesssim_{\epsilon}\Vert u(0,x)\Vert_{\dot{H}_x^{1/2}(\mathbf{R}^3)}^2,$$ but I have no idea how to do so.
Thanks in advance!
 A: I believe the inhomogeneous estimate should read
$\newcommand{\bbR}{\mathbb{R}} \newcommand{\jBra}[1]{\langle #1 \rangle}$
$$\sup_{t\in\mathbf{R}}\Vert u(t)\Vert^2_{\dot{H}_x^{1/2}(\mathbf{R}^3)}\lesssim_{\epsilon}\Vert u(0)\Vert^2_{\dot{H}_x^{1/2}(\mathbf{R}^3)}+\int_{\mathbf{R}}\int_{\mathbf{R}^3}\langle x\rangle ^{1+\epsilon}|F(t,x)|^2dxdt$$
since otherwise scaling arguments (replacing $F$ by $\lambda F$ and $u$ by $\lambda u$) suggest it cannot possibly hold.  
By applying Duhamel's formula and the triangle inequality, it suffices obtain the bound
$$\left\Vert\int_0^te^{i(t-s)\Delta/2}F(s,x)\;ds\right\Vert_{\dot{H}_x^{1/2}(\mathbf{R}^3)}^2 \lesssim_\epsilon \int_\bbR\int_{\bbR^3} \jBra{x}^{1+\epsilon} |F(s,x)|^2 \;dxdt = \lVert \jBra{x}^{\frac{1}{2}(1+\epsilon)} F\rVert_{L^2_{x,t}}^2$$
Now, by duality and the fact that $e^{it\Delta}$ is an isometry on $\dot H^{1/2}$:
$$\begin{align*}\left\Vert\int_0^te^{i(t-s)\Delta/2}F(s,x)\;ds\right\Vert_{\dot{H}_x^{1/2}(\mathbf{R}^3)} &= \sup_{\lVert v \rVert_{\dot H^{1/2}} = 1} \int_0^t\langle e^{i(t-s)\Delta/2}F(s,x), e^{it\Delta/2} v \rangle_{\dot H^{1/2}}\;ds
\\&=\sup_{\lVert v \rVert_{\dot H^{1/2}} = 1}\int_0^t\langle F(s,x), v(s,x) \rangle_{\dot H^{1/2}}\;ds\\
&\leq\sup_{\lVert v \rVert_{\dot H^{1/2}} = 1}\int_\bbR \langle \jBra{x}^{\frac{1}{2}(1+\epsilon}F, \jBra{x}^{\frac{-1}{2}(1+\epsilon} v(s,x) \rangle_{\dot H^{1/2}}\;ds\end{align*}$$
where $v(s,x) = e^{is\Delta/2}v(x)$ is the solution to the homogeneous Schrodinger equation with initial data $v$.  Now, the $\dot H^{1/2}$ pairing can be written as
$$\langle u, v \rangle_{\dot H^{1/2}} = \int_{\bbR^3} |\xi| \hat u(\xi \hat v(\xi)\;dx \leq \lVert \hat u \rVert_{L^2_\xi}\lVert |\xi|\hat v \rVert_{L^2_\xi} = \lVert u \rVert_{L^2_x} \lVert \nabla v \rVert_{L^2}$$
so, by Cauchy-Schwartz, we find that
$$\begin{align*}
\left\Vert\int_0^te^{i(t-s)\Delta/2}F(s,x)\;ds\right\Vert_{\dot{H}_x^{1/2}(\mathbf{R}^3)} &\leq \sup_{\lVert v \rVert_{\dot H^{1/2}} = 1}\int_\bbR \lVert \jBra{x}^{\frac{1}{2}(1+\epsilon)}F(s,x)\rVert_{L^2_x} \lVert \nabla\left(\jBra{x}^{\frac{-1}{2}(1+\epsilon)} v(s,x)\right)\rVert_{L^2_x} \;ds
\\&\leq \sup_{\lVert v \rVert_{\dot H^{1/2}} = 1}\left(\int_\bbR \int_{\bbR^3} \jBra{x}^{1+\epsilon} |F|^2\;dxds \int_\bbR \int_{\bbR^3} \jBra{x}^{-1-\epsilon}|\nabla v|^2 + \jBra{x}^{-3-\epsilon}|v|^2\;dxds \right)^{1/2}
\end{align*}$$
But, the last term we can estimate using the homogeneous estimate to obtain
$$\left\Vert\int_0^te^{i(t-s)\Delta/2}F(s,x)\;ds\right\Vert_{\dot{H}_x^{1/2}(\mathbf{R}^3)} \lesssim \left(\int_\bbR \int_{\bbR^3} \jBra{x}^{1+\epsilon} F(s,x)\;dxds\right)^{1/2}$$
from which the inhomogeneous estimate follows.  This method of argument (where we use the homogeneous estimate to prove the inhomogeneous estimate) is known as the $TT^*$ argument.
