Well, the first exercise basically consisted only in evaluating this integral, so let's
rewrite kxy integrating over the surface measure, with respect to the surface measure over the
house space boundary of Rn plus as the integral of Rn minus 1 of 1 over Xn squared plus norm
of Xn hat minus Z squared to the power n over 2 into Z, where X is equal to Xn hat Xn.
And here there is a constant times Xn. By doing the simple change of variables, Z equal
to Xn hat plus Xn W, notice that Xn is always strictly positive, so this is perfectly fine,
you get a factor Xn, sorry, to the power n here, the denominator, a factor Xn to the
power n minus 1 from the differential, and there is an Xn here, they all simplify. So
what remains is a constant, integral of Rn minus 1, 1 over 1 plus norm of Y squared to
the power n over 2 into Y, where at this point it's enough to use the spherical change of
coordinates, because this function is radial. We are in Rn minus 1, and therefore the constant
which is the surface measure of the sphere is n minus 1 omega n minus 1, integral from
0 to plus infinity, rho to the n minus 2, 1 plus rho squared to the n over 2, because
we are in the measure n minus 1. Now let's change variables like rho is equal to square
root of t, it's just a way to get rid of this rho squared here, there are of course other
possible. Now we want to bring the integral from 0 to plus infinity to 0, 1, I mean, there
are many ways to evaluate this integral, one way that I thought could be possible would
be to reduce this integral to the Euler beta function, so the change of variables is t
equal to 1 over 1 minus 1, which give us this integral over here, it's just calculations,
so this is the beta function of 1 half n minus 1 over 2, then you use the formula which relates
Euler's beta and gamma functions, the formula which gives you the definition of omega n,
the fact that gamma of n over 2 plus 1 is n over 2 gamma of n over 2 and the same for
gamma of n minus 1 over 2 plus 1, and finally the fact that gamma of 1 half is square root
of pi, and we are done. Then notice that kxy is positive for part b, and so basically
part a is telling you that the integral, sorry, the del 1 norm of k is equal to 1 over the
boundary of Rn plus, this will be already enough, but if you want to write it explicitly
of course the absolute value of the integral of the boundary of Rn plus of kxy g of y in
the s of y is controlled by the integral of Rn plus of kxy norm of g of y in the s of
y, bring out the l infinity norm of g over the boundary of Rn plus, it remains only this
integral but this is 1, so this means this function is well defined for every x in Rn
and its absolute value is controlled by the l infinity norm of g, and so we are done.