Riemann Hypothesis Proof

Proof of Riemann Hypothesis

We consider H(x, y) = H(z, ¯z) = e−ζ(z)2e−ζ(¯z)2
where ζ is the Riemann zeta function. We calculate the Laplacian of H(x, y) using
∆H(x, y) = 4∂¯∂H (z, ¯z) = 4(−2ζ(z)ζ0(z))e−ζ(z)2(−2ζ(¯z)ζ0(¯z))e−ζ( ¯z)2= 16(ζ(z)ζ(¯z))(ζ0(z)ζ0(¯z))H(x, y)≥0

So if we consider the domain enclosed by the lines a1, a2, a3 where
a1={0≤x≤1/2, y = 0},
a2={x= 1/2,0≤y},
a3={x= 0,0≤y}

Rieman hypothesis proof

The max of H are on the boundaries ,since we have shown that H is subharmonic, but we know that the maxima of H is 1 and that the zeta function has no zeros ona1 and a3so the maximas must lie on a2.So we have shown that the RiemannHypothesis is true and all the zeros of the zeta function lie on Re(z)=1/2. The fuction H(z, ¯z) would change if we enter Re(z)>1/2 because of the function alequation ζ(z)=2zπz−1sin(πz2)Γ(1 −z)ζ(1 −z) ,so we restricted reasonably Hto this boundary and Re(z) = 1/2 is a sort of a symmetry axis.

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