When it is given that a signal $x(t)$ has a real-valued Fourier transformation $X(f)$ then is the signal necessarily real-valued?
I have thought the following:
We have that the real and imaginary parts of the Fourier transform of a signal $x(t)$ are the Fourier transforms of the signal's even and odd parts, respectively:
$X_R(ω)=\frac{1}{2}[X(ω)+X^{\star}(ω)]⟺\frac{1}{2}[x(t)+x^{\star}(−t)]=x_e(t) \\ X_I(ω)=\frac{1}{2i} [X(ω)−X^{\star}(ω)]⟺ \frac{1}{2i}[x(t)−x^{\star}(−t)]=−i⋅x_o(t)$
where $X_R(ω)$ and $X_I(ω)$ are the real and imaginary parts of $X(ω)$, and $x_e(t)$ and $x_o(t)$ are the even and odd parts of $x(t)$, respectively.
So the odd part of $x$ is $0$ and the even one is real-valued, and so the signal $x(t)$ is real-valued.
Is everything correct?