alternate derivation of C_44
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Phil's working note dated 4.16.13, later incorporated at the end of Appendix C of his spectral theory book and marked as no longer needed. It computes the Hilbert transform of a Fourier transform, (f^)h, starting from (C.26) and (C.27) as a convolution. It diagonalizes in Fourier space using the transform of the kernel, -i k sgn(ω), and Fact 3 to get (f^)h(ω) = i[sgn(s) f(s)]^(ω).
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Alternate derivation of (C.44) PhL 4.16.13
This is incorporated at the end of Appendix C. The task is to compute (f^)h as opposed to (fh)^.
This, this file is really no longer needed.
I want an expression for this object: (f^)h since I know it in the reverse order.
Start with (C.26) applied to f^
(f^)h(t) ≡ (1/π) dω
Rewrite as
(f^)h(t) = dt' f^(t') = !Syntax Error, I dt' pf() f^(t') (C.27)
or
(f^)h = pf( ) * f^
The diagonalized version is then
a^(ω) = k-1 b^(ω) c^(ω)
or
((f^)h)^ = k-1 pf( )^(ω) f^^(ω)
But use
( )^(ω) = -i k sgn(ω)
to get
((f^)h)^ = k-1 [-i k sgn(ω)] f^^(ω) = -i sgn(ω) f^^(ω)
But we know that
f^^(ω) = 2πk2 f(-ω) from Fact 3
We then have
((f^)h)^(ω) = -i sgn(ω) 2πk2 f(-ω) = -2πk2 i sgn(ω) f(-ω) = 2πk2 i sgn(-ω) f(-ω)
Now take the FT on both sides
((f^)h)^^(ω) = 2πk2i [sgn(-s) f(-s)]^(ω)
Use Fact 3 on the LHS to get
((f^)h)^^(ω) = 2πk2((f^)h)(-ω)
Then we have
2πk2((f^)h)(-ω) = 2πk2i [sgn(-s) f(-s)]^(ω)
((f^)h)(-ω) = i[sgn(-s) f(-s)]^(ω)
((f^)h)(ω) = i [sgn(-s) f(-s)]^(-ω) = i[sgn(s) f(s)]^(ω)