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In normal summations, like 2+3=5, the information about the original numbers is lost. But in infinite summations like integral transforms, no information is lost and the function can still be recovered. What about integral transforms makes this possible? Give a less formal explanation (I mean, just avoid complex terminology. Use high school familiar terms)

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    $\begingroup$ To use your analogy, yes, $2+3 = 5$ alone causes a loss of information, but if you add a second equation, say $2-3 =-1$, then you've recovered the lost information. The equation $\begin{pmatrix}1 & 1 \\ 1 & -1\end{pmatrix}\begin{pmatrix}2 \\ 3\end{pmatrix} = \begin{pmatrix} 5 \\ -1\end{pmatrix}$ is of the form $Ax = b$, where $A$ is invertible. Similarly, on certain domains of functions, the Fourier and other transforms are invertible operations. For example, the Fourier transform is an invertible mapping from $L^2$ to itself. $\endgroup$
    – user169852
    Sep 5, 2019 at 5:00
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    $\begingroup$ @Bungo That's really good. I want to know how they are invertible and other made-up transforms (like $\int_0^1 f(x)^bdx$) are not. Also, summations can be invertible because they form systems of linear equations, like in your example. But how can integrals be invertible? Won't there be a system of infinite equations or something? $\endgroup$
    – Ryder Rude
    Sep 5, 2019 at 5:04
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    $\begingroup$ I think this is an interesting question, with one major issue: it's too broad. Could you specify one particular transformation that you'd like explained? If the explanation doesn't apply to other transformations, then ask another question. $\endgroup$ Sep 5, 2019 at 5:05
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    $\begingroup$ It actually takes quite a lot of work to show that the Fourier transform is invertible on certain spaces. If I recall correctly, you start by showing that it's an invertible mapping from the Schwartz space to itself, and then use the fact that the Schwartz functions are dense in $L^2$ to prove that there is a unique continuous extension of the Fourier transform to $L^2$. Then you show that this extension agrees with the usual Fourier transform defined via integration for functions that are also in $L^1$. $\endgroup$
    – user169852
    Sep 5, 2019 at 5:08
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    $\begingroup$ @Bungo there isn't anything common in invertible transforms which makes them invertible? Invertibility has to be proven for each transform separately? $\endgroup$
    – Ryder Rude
    Sep 5, 2019 at 5:11

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Summations (or integrals) in discrete and continuous Fourier transforms merely takes points in one infinite dimensional space (e.g., continuous functions of a real variable) and express them as a different continuous function. No information is lost (in typical cases) because the basis set spans the space.

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