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Here the equation reads sin2 ξ1 2 = 8g ξ2 2 + sin2 sin2 ξ1 2 sin2 or still ξ3 2 + sin2 ξ2 2 ξ1 2 + sin2 ⎛ 3 ξ3 2 sin2 + sin2 ξ2 2 sin2 ξ3 2 ⎞ 3 ⎜ ⎝1 − 4g j=1 sin2 k=1 ξk ⎟ 2 ⎠ sin 2 ξj 2 = 0. k=j Again, we may directly see that the only solution of this equation will be ξ = 0, provided that 1 − 4g sin2 ξk 2 ξ 2 + sin2 which will be fulfilled whenever g < > 0, 1 4(m−1) for all k, = 1, 2, 3, k = = 18 . The general case now easily follows. 9. 7) for the fundamental solution in frequency space, restricted to its basic period ([0, 2π[)m , holds for all ξ = 0, provided that the 1 metric scalar g is smaller then the critical value 4(m−1) .

I) The interior Hilbert transform H + : L2 (S m−1 ) → L2 (S m−1 ) is a self-adjoint bounded linear operator. (ii) The unique harmonic extensions to B + of f ∈ L2 (S m−1 ) and of H + [f ] are angular harmonic conjugates in B + , adding up to a left-monogenic function in B + . (iii) The interior Hilbert transform of a constant function on S m−1 is zero. In a similar way we are able to construct an exterior Hilbert transform H − on S based on the same concept of angular harmonic conjugate, however now in B − .

On account of Stokes’ theorem we have ˇ g=− (χB ∂) χB ∂[f ] = 0 since the last integral only involves points of B, where f is left discrete monogenic. 5. If B is a bounded set in Zm , then Proof. Take f = 1 in Cauchy’s theorem. χB ∂ˇ = 0. 46 F. Brackx, H. De Schepper, F. Sommen and L. 5, again the same two-dimensional set B as in Figure 1 is considered in the picture below. As opposed to what was mentioned in Section 3, at this moment, the exact position of the grid points does become important.

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Advanced calculus by Sagan H.


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