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How does one prove a Fourier transform is well defined?

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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.

Edited by asksciencequestions
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  • 1 month later...

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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  • 3 months later...

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.)

I hope that adds significantly to your question.

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