I have the following exercise:
Let $f \in C(\mathbb{R})\cap \cal{L}^1$. Show that:
$$\hat{f}(\xi) \in \mathbb{R} \;\;\forall \xi \in \mathbb{R} \iff f \text{ is even.} $$
I did the ($\Leftarrow$) part just using basic properties of Fourier and change of variable. But my problem is at the ($\Rightarrow$) part.
My attempt:
$$\overline{\hat{f}(\xi)}=\hat{f}(\xi)$$
$$\overline{\frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} f(x)e^{-ix\xi} \, dx} = \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} f(x)e^{-ix\xi} \, dx$$
$$\frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} f(x)e^{ix\xi} \, dx =\frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} f(x)e^{-ix\xi} \, dx$$
$$\int_{\mathbb{R}} f(x)e^{ix\xi}\,dx=\int_{\mathbb{R}} f(x)e^{-ix\xi}\,dx$$
$$\int_{\mathbb{R}} f(x)e^{ix\xi}\,dx=-\int_{\mathbb{R}} f(-y)e^{iy\xi}\,dy$$
And I'm failing to conclude $f(x)=f(-x) \forall x \in \mathbb{R}$ from this...
Any help would be appreciated.
Essentially you have $f$ and $g$, where $g(x)=f(-x)$, and $$\int_{\mathbb{R}} f(x)e^{ix\xi}dx=\int_{\mathbb{R}}g(x)e^{ix\xi}dx.$$
By the Fourier inversion theorem we have $$\int_{\mathbb{R}}\left(\int_{\mathbb{R}} f(x)e^{ix\xi}dx\right)e^{-i\xi y}dy=\sqrt{2\pi} f(y)$$ and $$\int_{\mathbb{R}}\left(\int_{\mathbb{R}} g(x)e^{ix\xi}dx\right)e^{-i\xi y}dy=\sqrt{2\pi} g(y).$$
Therefore, $f=g$.