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I am relatively new to Fourier transforms, so I apologize for the rather basic nature of my question, but, despite much googling, I was not able to find a clear answer on-line, so I must be stuck with some misconception.

I do understand that the Fourier transform of a real valued function f: R => R is, in general, a complex valued function, which encodes both the amplitude and phase of the harmonics that make up the function itself. What I am having trouble with is understanding how the inverse transform of the Fourier transform of a real valued function f (which is complex valued) can return the original real valued function f. As far as I can tell, the inverse Fourier transform will, in general, return a complex valued function, not a real valued one. I am obviously missing something. Could anybody point me to the right direction and/or suggest some resources that would help me clarify this point?

Thank you so much for your help.

Myron
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If $f(t)$ is a real-valued function, then its Fourier transform $F(\omega)$ may be a complex function, but $F(\omega)$ has the property that $F(-\omega)$ is the complex conjugate of $F(+\omega)$. The inverse Fourier transform of any function having this property is real.

A related fact is that any real function that is symmetric about the origin, so that $f(t)=f(-t)$, has a Fourier transform $F(\omega)$ that is a real function.

macrofish
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As an analogy, multiplying a complex number by (e.g.) $1+i$ will, generally, give you a complex number. But, $(1+i)(1-i)$ is a real number.

Similarly, an inverse Fourier transform will generally transform a complex function to another complex function. But, it might happen that a particular complex function, after an inverse Fourier transform, will happen to have all of its output contained in the real line.

Real-valued functions are a subset of complex-valued functions, since the real numbers are a subset of the complex numbers. Sure, almost all complex-valued functions will not give real-valued functions after an inverse Fourier transform. But there will be some complex-valued functions that inverse Fourier transform into a real-valued function. The Fourier transforms of real-valued functions will definitely have this property, because the inverse Fourier transform is designed to invert the original Fourier transform.

Sam Jaques
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