Let $f\in S_{\infty}$ be a Schwartz function and let us define a linear functional,for any $\varphi\in S_{\infty}$, $S_{\infty}\to\mathbb{C}$, $\varphi\mapsto (f,\varphi)$ by$$(f,\varphi):=\int_{\mathbb{R}}f\varphi d\mu$$where $\mu$ is the usual linear Lebesgue measure. I read in Kolmogorov-Fomin's Elements of the Theory of Functions and Functional Analysis (p. 453) that, if we call the Fourier trasnform operator $F:f\mapsto\int_{\mathbb{R}}f(x)e^{-i\lambda x}d\mu_x$, then$$2\pi(f,\varphi)=(F[f],F[\varphi])$$accordingly with Plancherel theorem. I know a corollary of Plancherel theorem (p. 442 here) according to which $$\forall f_1,f_2\in L_2(-\infty,\infty)\quad\int_{\mathbb{R}}f_1 \bar{f_2}d\mu=\frac{1}{2\pi}\int_{\mathbb{R}}F[f_1]\overline{F[f_2]}d\mu$$therefore I would say that $2\pi(f,\bar{\varphi})=(F[f],\overline{F[\varphi]})$, but I am not sure how $2\pi(f,\varphi)=(F[f],F[\varphi])$ can hold... Is it valid for real valued functions only? Even if $\varphi=\bar{\varphi}$, I do not think that $F[\varphi]$ is real valued if $\varphi$ is, in general. Could anybody give a proof of $2\pi(f,\varphi)=(F[f],F[\varphi])$? (I am assuming that there is not a typographical error in the book, but it is not impossible that there is one). I heartily thank you!
After stating that, the text says that $2\pi(f,\varphi)=(F[f],F[\varphi])$ holds not only for $f\in S_\infty$, but for any $f\in L_1(-\infty,\infty)$, while I did not even know that, such a generic function, $F[f]\cdot F[\varphi]$ is summable...