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Can we solve this strange functional equation? $$ f(x+i\epsilon)-f(x-i\epsilon) = g(x) $$

I believe that the solution is the Hilbert (finite part) transform of the function g(x) however I do not know it exactly.

I had thought taking in both sides the Fourier transform in tihs case i believe that $$2 i F(p)\sin(p\epsilon)=G(p)$$ so from this algebraic equation we could evaluate $f(x)$.

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IS $\epsilon$ a fixed constant? –  Henning Makholm Oct 26 '11 at 12:06
    
'epsilon tends to .. ups i forgot :( sorry $ \epsilon \to 0 $ –  Jose Garcia Oct 26 '11 at 12:19
    
Wouldn't the left-hand side of your equation tend to zero too, then -- or are you looking for an everywhere discontinuous $f$? –  Henning Makholm Oct 26 '11 at 12:22
    
Is $x$ required to be real? –  Craig Oct 26 '11 at 12:32
    
If $f$ is analytic in a strip along the real axis then $f(x+i\epsilon)-f(x-i\epsilon\doteq 2i\epsilon f'(x)$. –  Christian Blatter Oct 26 '11 at 12:57
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3 Answers

up vote 4 down vote accepted

Since $f$ cannot be continuous along the real line, the $f(x+i\epsilon)$ and $f(x-i\epsilon)$ probably should be not-necessarily-related functions. Thus, if $g(x)=\lim_{\epsilon\rightarrow 0^+} f(x+i\epsilon)+F(x-i\epsilon)$ with $f$ holomorphic on the upper half-plane and $F$ holomorphic on the lower, we are expressing $g$ as a hyperfunction, by definition. (This connects to the Riemann-Hilbert business, also.)

Or, for real-valued $g$, we might require that $f$ be the real part of a holomorphic function, and then the Hilbert transform of $g$ (under various hypotheses) would be the imaginary part. This is obviously related to the previous, but the goals may be different.

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If the function is continuous, g(x)=0.

Otherwise substitute in $x=i\epsilon$ then $x=-i\epsilon$:

$f(2i\epsilon)-f(0)=g(i\epsilon)$

$f(0)-f(-2i\epsilon)=g(-i\epsilon)$

$g(-i\epsilon)=-g(i\epsilon)$

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$f(x+i\epsilon)-f(x-i\epsilon)=g(x)$

$x\to x+i\epsilon$ :

$f(x+2i\epsilon)-f(x)=g(x+i\epsilon)$

$x\to 2i\epsilon x$ :

$f(2i\epsilon x+2i\epsilon)-f(2i\epsilon x)=g(2i\epsilon x+i\epsilon)$

$f(2i\epsilon(x+1))-f(2i\epsilon x)=g(i\epsilon(2x+1))$

$f(2i\epsilon x)=\sum\limits_xg(i\epsilon(2x+1))+\Theta_1(x)$, where $\Theta_1(x)$ is an arbitrary periodic function with unit period

$f(x)=\left(\sum\limits_xg(i\epsilon(2x+1))\right)_{x\to -\frac{ix}{2\epsilon}}+\Theta(x)$, where $\Theta(x)$ is an arbitrary periodic function with period $2i\epsilon$

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