I'm trying to prove the following:
Let the extension of the fourier transformation from $L^1(\mathbb R)$ to $L^2(\mathbb R) $ be denoted by $T$. Let$f \in L^1(\mathbb R) \cap L^2(\mathbb R)$ . Then for $f_r(x) :=f(-x)$,
$$TT(f_r)=f$$ holds.
Here's my attempt:
I know $T(f_r)=F(f_r) =\bar F (f) \in L^2(\mathbb R)$.
$ L^1(\mathbb R) \cap L^2(\mathbb R)$ dense in $L^2(\mathbb R)$, so $$\exists f_n \in L^1(\mathbb R) \cap L^2(\mathbb R) \ , \ ||f_n - \bar F(f)||_2\rightarrow 0 $$ Enough to show $$||F(f_n)-f||_2 \rightarrow 0$$ which by definition of $T$ implies $TT(f_r)=T(\bar F(f))=f$ , hence completing the proof.
I guess I need to use the isometric property of $T$, but it does not work well. Any hint is appreciated!
p.s. I'm trying to prove the above lemma to prove the extension of inverse Fourier transform.
The Fourier transform $Ff = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)e^{-ist}dt$ extends to an isometry on $L^2$. The same is true of the inverse Fourier transform $Gf = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(s)e^{ist}ds$. Because $FGf=f=GFf$ for a dense set of functions $f$ in $L^2$, then $FGf=f=GFf$ holds for all $f\in L^2$, assuming $F$, $G$ refer to the extension of these transforms to $L^2$. Therefore, $$ Ff_r = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(-t)e^{-ist}dt \\ = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)e^{ist}dt=Gf. $$ Therefore, $$ FFf_r = FGf=f. $$