P_{2j}(0)} \sim \frac{1}{\sqrt{2j+1}}$ if  $j \to \infty$.

The first two equations above are equivalent to the system
\begin{align*}
 \left \{
 \begin{array}{lll}
 N{\mathcal S}{\bf y}^{(1)}_{N \ell}+{\mathcal S}{\bf y}^{(2)}_{N \ell}
  &=
  -\frac{{\bf y}^{(1)}_{N \ell}}{P_{N}(0)}-\frac{{\bf y}^{(2)}_{N \ell}}{N P_{N}(0)}
 \\
  -(N+1){\mathcal S}{\bf y}^{(1)}_{N \ell}+{\mathcal S}{\bf y}^{(2)}_{N \ell}
 &=
  -\frac{{\bf y}^{(1)}_{N \ell}}{P_{N}(0)}+ \frac{{\bf y}^{(2)}_{N  \ell}}{(N+1)P_{N}(0)}\,.
  \\
 \end{array}
\right.
\end{align*}

 Solving this system, we obtain the desired
\begin{align*}
  {\mathcal S}{\bf y}^{(1)}_{N \ell}=
         \frac{-{\bf y}^{(2)}_{N \ell}}{P_{N}(0)N(N+1)},
\ \
 {\mathcal S}{\bf y}^{(2)}_{N \ell}=
    \frac{-1}{P_{N}(0)}
   \Big  ( {\bf y}^{(1)}_{N \ell}
       + \frac{{\bf y}^{(2)}_{N \ell}}{N(N+1)} \Big  ) .
\end{align*}
\subsection{Proof Theorem \ref{th:main1}}
We recall some of the basic properties that are implied in our proof.
The Funk--Minkowski transform is even, $\{ {\mathcal F}f\}(-{\bm
\xi})=\{ {\mathcal F}f\}({\bm \xi})$, and ${\mathcal F}
f_{odd} =0$,
 but spherical transform  ${\mathcal S}$ is odd,
   $\{{\mathcal S} f\}(-{\bm \xi})=- \{ {\mathcal S} f\}({\bm
   \xi})$,
and ${\mathcal S} f_{even} =0$. It is obviously that the surface
gradient  $\nabla$, scalar (dot) product ${\bm \eta} \, \centerdot
$ and vector (cross) product ${\bm \eta} \times $ change the
parity.
 We also recall the  parity rules for scalar and vector spherical harmonics :
  $  Y_{N\ell}(-{\bm \xi})=(-1)^NY_{N\ell}({\bm \xi})$,
  ${\bf y}^{(1)}_{N\ell}(-{\bm \xi})= (-1)^{N+1}{\bf y}^{(1)}_{N\ell}({\bm \xi})$,
  ${\bf y}^{(2)}_{N\ell}(-{\bm \xi})= (-1)^{N+1}{\bf y}^{(2)}_{N\ell}({\bm \xi})$,
  ${\bf y}^{(3)}_{N\ell}(-{\bm \xi})= (-1)^{N}{\bf y}^{(3)}_{N\ell}({\bm \xi})$.



 Now we can proceed to our
formula (\ref{SEMR:representation}) and without loss of
generality, we assume that
 $f({\bm \theta}) \in H^{1}({\mathbb S}^2)/{\mathbb R}$, then we have
\begin{align*}
 f({\bm \theta})&=
        \frac{1}{4\pi} \int_{{\mathbb S}^2}
   \frac{
    ({\bm \eta} + {\bm \theta}) \, \centerdot
    \Big \{
    \Big [ {\mathcal F}, \nabla   \Big ] f  \Big   \}   (\bm \eta)
      }
   {{\bm \eta} \centerdot{\bm \theta}}\,
   {\rm d}{\bm \eta}
\\ &=
        \frac{1}{4\pi} \int_{{\mathbb S}^2}
   \frac{
    {\bm \eta}     \centerdot \Big \{  \Big [ {\mathcal F}, \nabla   \Big ]
 f \Big \}
    (\bm   \eta)+
   {\bm \theta} \, \centerdot  \Big \{ \Big [ {\mathcal F}, \nabla   \Big ]
   f \Big \}(\bm \eta)
      }
   {{\bm \eta} \centerdot{\bm \theta}}\,
   {\rm d}{\bm \eta}
   \\
  &=  \frac{1}{4\pi} \int_{{\mathbb S}^2}
   \frac{
    {\bm \eta}   \centerdot  {\mathcal F}_{\bm   \eta} \nabla  f -
\overbrace{ {\bm \eta}   \centerdot  \nabla{\mathcal F}_{\bm   \eta} f }^{=0}
    -
   {\bm \theta}  \centerdot \nabla{\mathcal F}_{\bm   \eta} f
     }
   {{\bm \eta} \centerdot{\bm \theta}}\,
   {\rm d}{\bm \eta}+
       \frac{ {\bm \theta}  \centerdot}{4\pi} \int_{{\mathbb S}^2}
   \frac{
   \overbrace{  {\mathcal F}_{\bm   \eta} \nabla  f}^{eveen}
      }
   {{\bm \eta} \centerdot{\bm \theta}}\,
   {\rm d}{\bm \eta}
      \\
 &=       \frac{1}{4\pi} \int_{{\mathbb S}^2}
   \frac{
    {\bm \eta}   \centerdot {\mathcal F}_{\bm   \eta} \nabla  f-
   {\bm \theta} \centerdot  \nabla {\mathcal F}_{\bm   \eta} f
      }
   {{\bm \eta} \centerdot{\bm \theta}}\,
   {\rm d}{\bm \eta}
   = {\mathcal S}_{\bm \theta} {\bm \eta}   \centerdot {\mathcal F}_{\bm \eta} \nabla  f -
    {\mathcal S}_{\bm \theta}{\bm \theta}  \centerdot \nabla {\mathcal F}_{\bm \eta} f.
   \end{align*}

   It is clear that
   ${\rm ker} ({\mathcal S}_{\bm \theta} {\bm \eta}
    \centerdot {\mathcal F}_{\bm \eta} \nabla)=H^{s}_{even}({\mathbb S }^2)$
   and
     ${\rm ker} ({{\bm \theta}  \centerdot}\,
   {\mathcal S}_{\bm \theta} \nabla{\mathcal F}_{\bm \eta})
 =H^{s}_{odd}({\mathbb S }^2)$.
 From the  Lemmas  \ref{l:F},  \ref{l:S} we have
  ${\mathcal S} {\bm \eta}  \centerdot
   {\mathcal F}_{\bm \eta}  \nabla :H^{s}({\mathbb S }^2)\to H^{s}({\mathbb S }^2)$
   if $s\geq 1$, that looks on the diagram
\[
   \begin{CD} H^{s}({\mathbb S }^2)  @>  \nabla     >>
{\bf H}^{s-1}_{tan}({\mathbb S }^2)
                         @>  {\bm \eta }  \centerdot {\mathcal F}_{\bm \eta}          >>
           H^{s-1/2}({\mathbb S }^2)  @>  {\mathcal S}       >>
             H^{s}({\mathbb S }^2).
   \\
   \end{CD}
\]
Similarly, $ {\bm \theta} \centerdot
  {\mathcal S}_{\bm \theta} \nabla {\mathcal F}:
  H^{s}({\mathbb S }^2)  \to   H^{s}({\mathbb S }^2)$ if $s\geq 1/2$,
  which is also confirmed by the diagram
  \[
  \begin{CD} H^{s}({\mathbb S }^2)  @>  {\mathcal F}      >>
  H^{s+1/2}({\mathbb S }^2)          @>   \nabla        >>
 {\bf H}^{s-1/2}_{tan}({\mathbb S }^2)
           @>  {\bm \theta}  \centerdot {\mathcal S}_{\bm \theta}       >>
     H^{s}({\mathbb S }^2).
   \\
   \end{CD}
 \]
 For further calculations we take specifically $f={Y}_{N\ell}$, then
 from (\ref{SEMR:Fy2}) and (\ref{SEMR:Fy3}) we have
 \begin{align*} \text{for} \ N=2j &:
 {\mathcal F}_{\bm \eta} \nabla {Y}_{N\ell}= 0,
 \ \
  \nabla {\mathcal F}_{\bm \eta} {Y}_{N\ell}
 =  P_{N}(0) {\bf y}^{(2)}_{N\ell} ({\bm \eta}),
  \\ \text{for} \ N=