Relation between Stein-Tomas adjoint restriction estimate and the Helmholtz equation 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.

 A: 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. $\Box$
