Confused about Fourier's Inversion Theorem with another Proposition Consider the space $L^{1}(\mathbb{R})$.
The Inversion theorem states :

If $f$ and $\widehat{f}$ are in $L^{1}(\mathbb{R})$, then $\overline{\mathscr{F}}\widehat{f}(t)=f(t)$ at all points where $f$ is continuous.

There is another proposition that is similar which states :

If $f$ is continuous and integrable and if $\widehat{f}\in L^{1}(\mathbb{R})$, then for all $x\in\mathbb{R}$ :
$$
\mathscr{F}\widehat{f}(x)=f(-x)
$$

I am lost in trying to differentiate between these two results. Is the proposition (second statement) saying that the inverse of Fourier is now just $f(-x)$ given that $f$ is also integrable? because I have seen examples where this proposition is being used to compute inverse Fourier of some functions.
 A: The notation of the first theorem is slightly nonconventional. The $\overline{\mathscr{F}}$ is meant to be the inverse Fourier transform, given by the formula
$$
\overline{\mathscr{F}}g(x) = \int_{\mathbb{R}} e^{2\pi i x \cdot \xi} g(\xi)\,d\xi.
$$
This is what you get if you "complex conjugate" the formula for the Fourier transform
$$
\mathscr{F} f(\xi) = \int_{\mathbb{R}} e^{-2\pi i x \cdot \xi} f(x)\,dx.
$$
This is not true complex conjugation because $g$ could be complex valued, but the bar didn't get applied to $g$.
The first theorem thus tells you that $\overline{\mathscr{F}}$ actually inverts the Fourier transform and gives you the function back at points of continuity of the function.
The second theorem is not about applying $\mathscr{F}$ then $\overline{\mathscr{F}}$, but rather about applying $\mathscr{F}$ twice. Essentially, applying the Fourier transform twice just reflects the function. And hence applying it four times gets you back to the original function. In other words, under suitable continuity and integrability assumptions: $\mathscr{F}^{-1} = \mathscr{F}^3 = \overline{\mathscr{F}}$.
Thus, to answer your question, NO $\mathscr{F} \hat f(x) = f(-x)$ is NOT saying that if $f$ is integrable then the inverse of the Fourier transform is just $f(-x)$. It is saying that the square of the Fourier transform is the reflection map, the map that sends $f(x) $to $f(-x)$.
