Recently I am reading Stein's Complex Analysis but I haven't read the book Fourier Analysis before. So I do not have any knowledge about Fourier Transformation.
In Chapter 4.3 (p.121), it tells that the Fourier inversion formula
$$f(x)=\int^\infty_{-\infty}\hat f(\xi)e^{2\pi ix\xi}d\xi\hspace{.5in}\text{if}\hspace{.5in}\hat f(\xi)=\int^\infty_{-\infty} f(x)e^{-2\pi ix\xi}dx$$
holds under the conditions $|f(x)|\le A/(1+x^2)$ and $|\hat f(\xi)|\le A'/(1+\xi^2)$.
I know why the formula holds when $f$ is analytic, which is proved in the previous section using contour integral. However, when $f$ is not analytic (even not continuous), can anyone give me a fundamental proof without using much fact about Fourier analysis?
A simple example of a family of functions with the properties you cite is $$ G_{y,\alpha}(x)=e^{-\alpha(x-y)^{2}},\;\;\;\alpha > 0,y\in\mathbb{R}. $$ This function is analytic in $x$, and presumably that means you know that $(G_{y,\alpha}^{\wedge})^{\vee}=G_{y,\alpha}$ for all $\alpha > 0$, $y\in\mathbb{R}$. (Here I have used the symmetric notation of $\wedge$ for Fourier transform and $\vee$ for inverse Fourier transform, which I greatly prefer because it indicates order of application of these operators, and is symmetric looking. So, any of the notation police reading this who are tempted to change it, please don't.)
Suppose $f,g\in L^{1}(\mathbb{R})$. Then $f^{\wedge},g^{\wedge}\in L^{\infty}(\mathbb{R})$, and a simple interchange of order of integration gives the following: $$ \int_{-\infty}^{\infty}f(s)g^{\wedge}(s)ds = \int_{-\infty}^{\infty}f^{\wedge}(s)g(s)ds \\ \int_{-\infty}^{\infty}f(s)g^{\vee}(s)ds = \int_{-\infty}^{\infty}f(s)g^{\vee}(s)ds $$ Then your function $f$ and $G_{y,\alpha}$ have the properties to allow one to apply the above repeatedly: $$ \begin{align} \int_{-\infty}^{\infty}(f^{\wedge})^{\vee}(x)G_{y,\alpha}(x)dx & =\int_{-\infty}^{\infty}f^{\wedge}(s)G_{y,\alpha}^{\vee}(s)ds \\ & =\int_{-\infty}^{\infty}f(x)(G_{y,\alpha}^{\vee})^{\wedge}(x)dx \\ & =\int_{-\infty}^{\infty}f(x)G_{y,\alpha}(x)dx. \end{align} $$ The Gaussians $G_{y,\alpha}$ are a rich enough set of functions that one has $$ (f^{\wedge})^{\vee} = f \;\;\; a.e. $$ Almost everywhere is the best you can do unless you know that $f$ is continuous, in which case $(f^{\wedge})^{\vee}=f$ everywhere because both are continuous and equal a.e..