Let $d\sigma$ denote the surface measure on $\mathbb{S}^2$. For each function $f\in L^2(\mathbb S^2)$, the Fourier transform $\widehat{fd\sigma}$ is defined as the integral $$ \int_{\mathbb S^2} f(\xi)e^{ix\cdot \xi}\, d\sigma(\xi), \qquad x\in \mathbb R^3,$$ and as Stein and Tomas proved, it satisfies the inequality $$\tag{1} \lVert \widehat{f d\sigma}\rVert_{L^4(\mathbb R^3)}\le C\lVert f\rVert_{L^2(\mathbb S^2)}.$$
Question. The function $u=\widehat{fd\sigma}$ satisfies the Helmholtz equation $\Delta u + u =0$ in $\mathbb R^3$. Is there a corresponding PDE interpretation of the estimate (1)?
A nice PDE interpretation is available for the Stein-Tomas estimate on the paraboloid $$\mathbb P^2=\{(\tau, \xi)\in \mathbb R\times \mathbb R^2\ :\ \tau=\lvert \xi\rvert^2\}.$$ Indeed, letting $$ d\mu:=\frac{\delta(\tau-\lvert \xi\rvert^2)}{(2\pi)^2}\, d\tau d\xi$$ we define a measure supported on $\mathbb P^2$, and the estimate analogous to (1) reads $$\tag{2} \lVert \widehat{f d\mu}\rVert_{L^4(\mathbb R^3)}\le C\lVert f\rVert_{L^2(\mathbb R^2)}.$$ If we denote $u(t, x)=\widehat{fd\mu}$, we see that it satisfies the initial value problem for the Schrödinger equation $$ \begin{cases} i\partial_t u = \Delta u, & t\in\mathbb R, x\in \mathbb R^2, \\ u|_{t=0}=\check{f}, \end{cases} $$ where $\check{f}$ denotes the inverse Fourier transform of $f$. By Plancherel's theorem, $\lVert f\rVert_{L^2}=\lVert \check{f}\rVert_{L^2}$, up to an irrelevant constant. Thus, (2) reads $$ \lVert u\rVert_{L^4(\mathbb{R}\times \mathbb{R}^2)}\le C\lVert f\rVert_{L^2(\mathbb R^2)}, $$ which is the celebrated Strichartz estimate.
I wonder if, similarly, the inequality (1) can be written as an estimate of the solution to the Helmholtz equation in terms of some kind of boundary values.
The answer is that, if $u$ solves $\Delta u + u=0$ on $\mathbb{R}^3$, writing $$ (x_1, x_2, x_3)=(\bar x, t), \qquad \bar x\in \mathbb R^2, t\in \mathbb R, $$ and $$ \bar \Delta = \partial_{x_1}^2+ \partial_{x_2}^2, $$ then the adjoint restriction inequality (1) above is equivalent to $$\tag{3} \lVert u\rVert_{L^4(\mathbb R^3)}^2\le 2C^2 \left(\int_{\lvert \bar \xi\rvert\le 1}\lvert \mathcal F(u|_{t=0})(\bar \xi)\rvert^2 (1-\lvert\bar\xi\rvert^2)^{1/2}\,d\bar\xi + \int_{\lvert\bar\xi\rvert\le 1} \lvert \mathcal F(u_t|_{t=0})(\bar \xi)\rvert^2 (1-\lvert\bar\xi\rvert^2)^{-1/2}\,d\bar\xi\right). $$ Unfortunately, this is not the prettiest thing around, but that's what we get. Here, and in the following, $\mathcal F$ refers to the Fourier transform in the variables $\bar x\to\bar \xi$, that is
$$\tag{4} \mathcal F(u|_{t=0})(\bar\xi)=\int_{\mathbb R^2} u(\bar x, 0)e^{-i\bar x\cdot \bar \xi}\, d\bar x,\quad \mathcal F(u_t|_{t=0})(\bar\xi)=\int_{\mathbb R^2} u_t(\bar x, 0)e^{-i\bar x\cdot \bar \xi}\, d\bar x. $$ Proof: each $f\in L^2(\mathbb S^2)$ can be written as $$f=f_+ + f_-,$$ where $f_+$ is supported in the upper hemisphere ($t>0$), while $f_-$ is supported in the lower hemisphere ($t<0$). We can regard $f_\pm$ as functions defined on the unit disk $\lvert \bar \xi \rvert\le 1$. Then
$$ u(\bar x, t)= \widehat{f d\sigma}= \int_{|\bar \xi|\le 1} e^{i \bar x \cdot \bar \xi + t\sqrt{1-\lvert \bar \xi \rvert^2}} f_+\, d\sigma + \int_{|\bar \xi|\le 1} e^{i \bar x \cdot \bar \xi - t\sqrt{1-\lvert \bar \xi \rvert^2}} f_-\, d\sigma. $$ Recall that the surface element on $\mathbb S^2$ reads $d\sigma=\pm(1-\lvert\bar\xi\rvert^2)^{-\frac12}d\bar \xi$. Thus, the above formula can be rewritten as $$\tag{5} u(\bar x, t)= e^{it\sqrt{1+\bar{\Delta}}}g_+ + e^{-it\sqrt{1+\bar{\Delta}}}g_-, $$ where $$\tag{6} \mathcal F g_{\pm}(\bar \xi) = f_{\pm}(\bar \xi)(1-\lvert\bar \xi \rvert^2)^{-\frac12} \mathbf 1_{\lvert\bar \xi\rvert\le 1}.$$ In particular, $$ \tag{7} u(\bar x, 0)=g_++g_-,\quad u_t(\bar x, 0)=i\sqrt(1+\bar{\Delta})(g_+-g_-).$$ To conclude, we observe that the Tomás-Stein estimate reads, in our notation, $$ \lVert u\rVert_{L^4(\mathbb R^3)}^2\le C^2\int_{\lvert\bar\xi\rvert\le 1} \lvert f_+(\bar \xi)\rvert^2 +\lvert f_-(\bar \xi)\rvert^2 \frac{d\bar \xi}{(1-\lvert\bar \xi\rvert^2)^{\frac12}},$$ and plugging (6) and (7) into it, we obtain (4), as we wanted. $\Box$