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Suppose you started from $f(x)$, and did some hard work to get the Fourier transformation $F(p)$. Can you recover $f(x)$ from $F(p)$? Did you lose information when you throw away $f(x)$ and only keep $F(p)$?
If $f(x)$ is continuous and absolutely integrable, we can recover $f(x)$ from $F(p)$ by $$ f(x) = \frac{1}{2\pi} \int_{p \in \R} F(p) e^{ipx} dp $$ The proof of this theorem is beyond the scope of this class. You might be happy to just accept the formal 'rule' that $$ \frac{1}{2\pi} \int_\R e^{ipx - ipy} dp = \delta(x-y). $$ and that $$ f(x) = \int \delta(x-y) f(y) dy $$
We can try some example to see if it works.
If the function $f(x)$ is a periodic function, of period $L$, meaning $f(x) = f(x+L)$, then we cannot do Fourier transform (why?), but instead, we need to do Fourier series.
We are not going to use all $e^{ipx}$ for all $p$, but only those that satisfies $e^{ipx} = e^{ip(x+L)}$ have the same periodicity. Which mean $p$ needs to satisfy $$ p = (2\pi / L) n$$ for some integer $n$.
So, we define $$ e_n(x) = e^{i(2\pi/L) n x}. $$ We define the Fourer series coefficient as $$ c_n = \frac{1}{L} \int_{0}^L f(x) \overline{e_n(x)} dx $$
Given these coefficient $c_n$, can we recover $f(x)$? Yes, under some smoothness condition of $f(x)$, we have $$ f(x) = \sum_{n \in \Z} c_n e_n(x). $$