How can I show that the Fourier transform of an even integrable function $f\colon \mathbb{R}\to\mathbb{R}$ is even real-valued function? And the Fourier transform of an odd integrable function $f\colon \mathbb{R}\to\mathbb{R}$ is odd and purely imaginary function?
Fourier Analysis – Fourier Transform of Even and Odd Functions
analysisfourier analysisfourier transformfunctional-analysis
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The values of a frequency domain function represent how much of that frequency is "in" the function. For example, if you would take the fourier transform of a sine wave, you would get a delta function in the frequency domain: there's a lot of some specific frequency in that function.
Now, this is quite a simple way of saying it; we can deduct quite a bit more from the value, such as the "phase" of that frequency component. But it's also a bit more difficult, since for many signals the fourier transform is not composed of delta functions but a continuous function.
Don't worry too much about it, it takes a while getting used to the idea of fourier transforms. It's only now in my third year in Electrical Engineering that they really feel natural, I must say.
Actually, you have already solved the problem, but have not quite noticed it yet :)
Indeed, you have proven that $(x\mathcal{F}(f)(x))^2$ is integrable. But, by a Cauchy-Schwarz inequality,
$$ \int_{\mathbb{R}\backslash [-1,1]} |\mathcal{F}(f)(x)| \, dx = \int_{\mathbb{R}\backslash [-1,1]} |x\mathcal{F}(f)(x)| \times \frac{1}{|x|} \, dx $$
$$ \le \left( \int_{\mathbb{R}\backslash [-1,1]} |x\mathcal{F}(f)(x)|^2 \, dx \right)^{1/2} \times \left( \int_{\mathbb{R}\backslash [-1,1]} \frac{1}{x^2}\, dx \right)^{1/2} < +\infty.$$
So, it suffices to analyse the interval $[-1,1].$ But, as we assumed that $f \in L^1(\mathbb{R})$, we get that $\mathcal{F}(f)$ is continuous, and therefore bounded, in $[-1,1].$ This implies the required integrability of the Fourier transform.
On the other hand, we see that the integrability of $f$ is crucial. Indeed, one can show, by using elementary Sobolev space methods, that the condition that $g \in L^2(\mathbb{R})$ is equivalent to $f$ having a (distributional) derivative $f' \in L^2(\mathbb{R}).$ Of course, one needs to make sense of $f$ having a distributional derivative, but this happens if, for instance, $f \in \mathcal{S}'(\mathbb{R}).$
Proving that $g \in L^2(\mathbb{R})$ implies $f \in L^2(\mathbb{R})$ is standard, whereas the reverse implication follows from the fact that
$$ g(x) \le M(f')(x), \forall x \in \mathbb{R},$$
where $M$ stands for the usual Hardy-Littlewood maximal function on the real line. As $M:L^2 \to L^2$ boundedly, we are through.
To come up with a counterexample is not very difficult then: for any function $g$, continuous, of compact support and such that $\int g \ne 0,$ Let
$$G(t) = \int_{-\infty}^t g(x) \, dx.$$
This function satisfies the hypotheses of the problem, but clearly, as $\mathcal{F}(g)(x) \sim \int g$ for $x \sim 0,$ then, as $G' = g \Rightarrow \mathcal{F}(G)(x) \sim \frac{\int g}{x}$ for $x \sim 0,$ which does not integrate.
Of course, this all can be made formal, but this is intended as a sketch to just illustrate how important it is that $G \in L^1(\mathbb{R}).$
Best Answer
Let $f: \mathbb{R} \to \mathbb{R}$ be an integrable function and let $\hat{f}$ denote its Fourier transform, i.e. $$ \hat{f}(\xi)=\int_\mathbb{R}e^{ix\xi}f(x)dx. $$ We have $$ \overline{\hat{f}(\xi)}=\hat{f}(-\xi)=\int_\mathbb{R}e^{-ix\xi}f(x)dx=\int_\mathbb{R}e^{iy\xi}f(-y)dy. $$ If $f$ is even then $$ \overline{\hat{f}(\xi)}=\hat{f}(-\xi)=\int_\mathbb{R}e^{iy\xi}f(y)dy=\hat{f}(\xi), $$ i.e. $\hat{f}$ is an even real-valued function.
If $f$ is odd then $$ \overline{\hat{f}(\xi)}=\hat{f}(-\xi)=\int_\mathbb{R}-e^{iy\xi}f(y)dy=-\hat{f}(\xi), $$ i.e. $\hat{f}$ is an odd purely imaginary function.