I'm reading a proof of the Strichartz inequalities. It shows that $$ \| \int_\mathbb{R} e^{-is\Delta}F(s) \, ds \|_{L^2_x} \lesssim \|F\|_{L^{q'}_t L^{r'}_x}, $$ and then says that by duality, $$ \|e^{it\Delta}u_0\|_{L^{q}_t L^{r}_x} \lesssim \|u_0\|_{L^2_x}. $$

This is some standard argument using $(L^2)^* = L^2$ and $(L^qL^r)^* = L^{q'}L^{r'}$, but I'm just not seeing it...

I know that the first inequality implies that for any $g\in L^2_x$

$$ \left| \int_{\mathbb{R}^d} \int_\mathbb{R} g(x) e^{is\Delta}F(s) ds dx \right| = \left| \int_{\mathbb{R}^d} \int_\mathbb{R} \int_{\mathbb{R}^d} g(x) e^{i|\xi|^2s + ix\cdot \xi}\hat{F}(\xi,s) \,d\xi \, ds \, dx \right|\lesssim \|F\|_{L^{q'}_t L^{r'}_x} \|g\|_{L^2_x}, $$ and that $$ \|e^{it\Delta}u_0\|_{L^{q}_t L^{r}_x} = \sup_{G \in L^{q'}_tL^{r'}_x\backslash\{0\}} \left| \int_{\mathbb{R}^d} \int_\mathbb{R} G(x,t) e^{-it\Delta}u_0 \, dt \, dx \right| / \|G\|_{L^{q'}_tL^{r'}_x} $$ $$=\sup_{G \in L^{q'}_tL^{r'}_x\backslash\{0\}} \left| \int_{\mathbb{R}^d} \int_\mathbb{R} \int_{\mathbb{R}^d} G(x,t) e^{-i|\xi|^2t+ix\cdot\xi}\hat{u}_0 \, d\xi \, dt \, dx \right| / \|G\|_{L^{q'}_tL^{r'}_x}$$ So what I want amounts to exchanging $\hat{F}$ for $G$ and $g$ for $\hat{u}_0$. I think I need to choose a clever $G$ and then use Holder, but I don't see how. Any advice would be appreciated.


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