On the Fourier inversion formula

Question

For a given function $f\in L^1(\mathbb{R})$, suppose that the
$$\check{f}(x)=\int_\mathbb{R} \hat{f}(\zeta)e^{2\pi i\zeta x}d\zeta$$ almost every where converges in $\mathbb{R}$. Then, can we say that $f=\check{f}$ almost everywhere? If the answer is NO, is it possible that the Lebesgue measure of $\{x: f(x)\neq\check{f}(x)\}$ to be infinite?

Answer 1

Note: I am not sure if I understand the word "converges" correctly.

This is completely analogous to the similar question regarding convergence of Fourier series, which is classical.

Let $$g(x,r) = \int_{-r}^r \hat f(\zeta) e^{2\pi i \zeta x} d\zeta$$ by "partial sums" of the inverse Fourier transform, and denote by $$h(x, r) = \int_0^1 g(x, r t) dt = \int_{-r}^r \hat f(\zeta) e^{2\pi i \zeta x} (1 - \tfrac{|\zeta|}{r}) d\zeta $$ the Cesàro averages of $g$.

By Plancherel's theorem, $g(\cdot, r)$ is the convolution of $f$ with the function $\phi_r(x) = 2 r \operatorname{sinc}(\pi r x)$ (which plays the same role as the Dirichlet kernel in the theory of Fourier series). In a similar way, $h(\cdot, r)$ is the convolution of $f$ with an $\psi_r(x) = r (\operatorname{sinc}(\pi r x))^2$ (which serves as the continuous counterpart of the Fejér kernel).

Since $\psi_r(x)$ is an approximate identity as $r \to \infty$ (that is: $\psi_r(x) = r \psi_1(r x)$, $\psi_r(x) \ge 0$ and $\int_{-\infty}^\infty \psi_r(x) dx = 1$), and additionally $\psi_1$ is bounded by a "radially decreasing" and integrable function: $\psi_1(x) \leqslant \min\{1, 1 / (\pi x)^2\}$. This implies that the functions $f * \psi_r$ converge to $f$ as $r \to \infty$ almost everywhere (and also in $L^1$); see, for example, Corollary 2.43 in Advanced Real Analysis by David McCormick and José Luis Rodrigo, available here. Hence, $h(x, r) \to f(x)$ almost everywhere as $r \to \infty$ (this is stated just below the proof of Corollary 2.43 in the book linked above).

For a fixed $x$, if $g(x, r)$ has a limit as $r \to \infty$, then the limit is necessarily equal to the limit of Cesàro means $h(x, r)$. Thus, if $g(x, r)$ converges for almost all $x$ as $r \to \infty$, then the limit is equal to $f(x)$ almost everywhere.

Answer 2

I may be wrong, but the following reformulation of your question seems the widest possible one.

Question: let $u\in L^1$, so that $\hat u\in L^\infty$ is a tempered distribution and hence $v=F^{-1}\hat u$ is a well defined tempered distribution. Suppose $v$ is a function, meaning that it coincides with a locally integrable function $w$ in distribution sense. Is it true then that $w=u$ a.e.?

Thus the assumption is that for any test function $\phi$ we have $v(\phi)=\int w \phi$, which implies, by definition of the Fourier transform of a distribution, $\int\hat u\cdot F^{-1}\phi=\int w\phi$ i.e. $\int u FF^{-1}\phi=\int w\phi$ for all test functions $\phi$. It obviously follows $u=w$ a.e.