06/18/2022, 10:40 PM

(08/23/2012, 04:26 PM)tommy1729 Wrote: [ -> ]here i give a ( nonunique short ) proof of TPID 4.

remember that entire taylor series are coo everywhere. ( infinitely differentiable for all finite complex )

let f(z,1) be an entire periodic function with f(0,1)=f(1,1)=1 and period 1.

and f(z,1) is not identically 1 for all z.

we will prove that for complex b with arg(b) <> 0 , the only solution to the equations is f(z,1) * b^z and hence the proof follows.

let k and n be positive integers.

f(0) = 1

f(z+k) = b^k f(z)

f = entire

then

take the derivative of the equation f(z+k) = b^k f(z) on both sides

f ' (z+k) = b^k f ' (z)

again

f '' (z+k) = b^k f '' (z)

and in general

f^(n) (z+k) = b^k f^(n) (z)

hence because of taylors theorem we must conclude

f(z) = f(0) * f(z,1) * b^z in the neighbourhood of 0.

but since f is entire it must be true everywhere and f(0) = 1 hence

f(z) = f(z,1) b^z

for all z.

if arg(b) <> 0 then the period of b^z does not have Re <> 0 and hence b^z is unbounded on the strip.

if f(z) needs to be bounded and b^z is not bounded , this implies that f(z,1) needs to be bounded.

but this is impossible since f(z,1) has a real period and is entire , it must be unbounded on the strip.

( remember f(z,1) =/= 1 everywhere by definition )

the product of two functions unbounded in the same region must be unbounded in that region.

...

...unless they cancel each other.

but notice the general solution is (also) b^(z + theta(z)) = b^z f(z,1)

now since theta is entire and 1-periodic , this implies that b^theta(z) = f(z,1) grows double exp.

and therefore b^z f(z,1) grows double exponential ( double exponential dominates exponential ).

So going up or down the strip we must get bigger and bigger values ; unbounded.

QED

tommy1729