I'm taking my definitions from Rudin Real & Complex Analysis. Let me use $F(f)=\hat{f}$ to represent the Fourier-Plancherel transform of a function $f\in L^2(\mathbb{R})$. If $f\in L^1(\mathbb{R})\cap L^2(\mathbb{R})$, then \begin{equation}\tag{1} \hat{f}(t)=\int_{-\infty}^\infty\!f(x)e^{-ixt}\,dm_1(x), \end{equation} the usual Fourier transform of a function in $L^1(\mathbb{R})$. (For those not familiar with Rudin's normalization conventions, $dm_n(t)=(2\pi)^{-n/2}dt$ in $\mathbb{R}^n$). $F$ is a linear extension of (1) from $L^1(\mathbb{R})\cap L^2(\mathbb{R})$ to an isometric Hilbert space isomorphism of $L^2(\mathbb{R})$ onto itself.
What I would like to know is, if I define \begin{equation}\tag{2} \check{f}(t)=\int_{-\infty}^\infty\!f(x)e^{ixt}\,dm_1(x)=\hat{f}(-t)=\bar{\hat{\bar{f}}}(t), \end{equation} for $f\in L^1(\mathbb{R})$ and extend by linearity from $L^1(\mathbb{R})\cap L^2(\mathbb{R})$ to $L^2(\mathbb{R})$ to get an isometric Hilbert space isomorphism $\mathscr{F}$, of $L^2(\mathbb{R})$ onto itself, in exactly the same way as $F$ is generated by $\hat{~}$, will it be true that $\mathscr{F}$ is the inverse of $F$? If so, is there a simple way to see that? If not, what goes wrong?
I've tried to prove it using approximations by Schwartz space functions (you can assume I'm also familiar with Rudin Functional Analysis Chapter 7), but haven't been able to work it out yet.
I just found the answer to my question in Rudin Real & Complex Analysis theorem 9.14. (Right after the Plancherel Theorem!!) It states (I'm replacing $f$ with $g$ in the statement):
If $g\in L^2$ and $\hat{g}\in L^1$, then $$g(x)=\int_{-\infty}^\infty\!\hat{g}(t)e^{ixt}\,dm_1(t)\quad\text{a.e.}$$
Since, in my question, $f\in L^1(\mathbb{R})\cap L^2(\mathbb{R})$, let $g=F^{-1}(f)\in L^2(\mathbb{R})$. Then $\hat{g}=f\in L^1(\mathbb{R})$, and so by the quoted Theorem, $\check{f}(x)=g(x)=(F^{-1}(f))(x)$ for almost all $x\in\mathbb{R}$; i.e., $\check{f}=F^{-1}(f)$ as a member of $L^2(\mathbb{R})$. (After switching $x$ and $t$, this is what I originally asked to prove.)