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Suppose $\tilde{u}(k,t)$ is the Fourier transform of a function $u(x,t)$.

Often, one makes the ansatz $$ \tilde{u}(k,t)=e^{-i\omega t}\tilde{u}(k),~~(*) $$ where $e^{-i\omega t}$ is an arbitrary expression for the time dependent part of $\tilde{u}(k,t)$.

I am trying to understand the idea behind this ansatz.

My conjecture is the following:

The Fourier operator $\mathcal{F}u\to\tilde{u}$ is often defined on $L^1(\mathbb{R}$), mapping into $L^1(\mathbb{R})$. So maybe the functions $e^{-i\omega t}$ are eigenfunctions of some operator $G$ on $L^1(\mathbb{R})$ and form an orthonormal basis of $L^1(\mathbb{R})$ with the consequence that the idea behind the ansatz is maybe that we can write each function in $L^1(\mathbb{R})$, e.g. the Fourier transform $\tilde{u}(k,t)=\mathcal{F}u(x,t)$ as some linear combination of the functions $e^{-i\omega t}$, in this case as exactly ONE $e^{-i\omega t}$?

Edit I should give more context!

Suppose we consider a function $u=u(x,t)$ defined on whole $\mathbb{R}$ and that, for each fixed $t$, can be expressed as Fourier integral: $$ u(x,t)=\int_{-\infty}^{+\infty}\, dk e^{ikx}\tilde{u}(k,x). $$

For the Fourier transform $\tilde{u}(k,x)$ we make the product-ansatz $$ \tilde{u}(k,t)=e^{-i\omega t}\tilde{u}(k) $$ as described above. That is, we try to write the solution as $$ u(x,t)=\frac{1}{2\pi}\int_{-\infty}^{+\infty}\, dk e^{i(kx-\omega t)}\tilde{u}(k). $$ The functions $e^{i(kx-\omega t)}$ are called Fourier modes.

(By the way: I thought that functions $e^{i\omega t}$ are called Fourier modes?!)

So now again my question: Why is it "reasonable" to make the product ansatz for the Fourier transform $\tilde{u}(k,t)$?

mathfemi
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    Two questions. First, for any $\omega$, is $e^{i\omega t }\in L^1(\mathbb{R})$? Second, what does "orthonormal" mean in the context of $L^1$? – Neal Aug 26 '16 at 19:13
  • Please correct me! Maybe domain and range of the Fourier transformation are different from what I claim. But this does not change my question whether the functions $e^{-i\omega t}$ are eigenfunctions of some operator (whereever it is defined on) and if the idea of the ansatz is that we can write the Fourier transform (as a function of the same functions space, whatever it is) as a linear combination of the eigenfunctions $e^{-\omega t}$ (which form an orthonormal basis)? – mathfemi Aug 26 '16 at 19:16
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    Don't have time to go in depth, so some pointers. First, think about $L^2$ as well as $L^1$. Second, we use the $e^{i\omega t}$ but they are not elements of $L^1$ or $L^2$; think of $\hat{f}(\omega)$ as "projection of $f$ onto $e^{i\omega t}$"; if the $e^{i\omega t}$s were elements of $L^1$ then their Fourier transforms would be Dirac deltas. (It is possible to extend the Fourier transformation to more general functionals, tempered distributions, which makes this intuition precise) – Neal Aug 26 '16 at 19:20
  • The ansatz $\tilde{u}(k,t)=\tilde{u}(k)e^{-i\omega t}$ is a planar wave ansatz, isn't it? So my question seems to reduce to the question why it is adequate to make this ansatz for the Fourier transform. – mathfemi Aug 26 '16 at 19:47
  • They form an orthonormal basis of $L^2([-1,1])$. However, there is no notion of "orthogonal" in $L^1(\Bbb R)$. – Ben Grossmann Aug 26 '16 at 21:06

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