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I'm wondering what the general process is for this. In general being well defined means there is only exactly one output for each input or that if x=y, f(x) = f(y) which looks like the reverse of proving injectivity, I don't know if that is a coincidence or not.

Is a Fourier transform of a real function is still always real? I suppose the idea is that the imaginary component decays to 0 as you take the integral from -infinity to infinity so that it evaluates to a single finite real number, or actually, does the output of a the Fourier transform of a real valued function need to be real? Why do I generally see absolute value arguments in proving well-defined properties if complex functions have even more possible they can take? If you take an absolute value that only tells you anything about the magnitude of uncountably infinite numbers.

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The Fourier transform of an even real function is real. More generally, it is real for the function $$f$$ if $$f(-x) = f(x)$$ holds for almost every $$x \in \mathbb{R}.$$

The Fourier transform of a real function is otherwise not itself a real-valued function.

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Taeto has answered very proficiently your longer-worded question, which has a different scope than the title really. Namely: "How does one prove that a Fourier transform is well defined?"

I don't remember the details, nor can I find them on Wikipedia, but a Fourier transform is well-defined when your function is piecewise-continuous. That means it better not have an uncountable number of discontinuities. But the definition is very solid, in the sense that you can even define it for some non-integrable functions or even temperate distributions (strange objects, like the Dirac delta function, that your garden-variety functions can be integrated against.)

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