TITLE: "eigenfunction" of a transformation
QUESTION [5 upvotes]: Fourier transform of a gaussian is another gaussian. Fourier/Laplace transforms of $\frac{1}{\sqrt t}$ is something like $\frac{1}{\sqrt \omega}$.
I realize that we can't call these eigenfunctions since the variable is also changed, but is there a special name, are there other interesting stuff to know about these?

REPLY [6 votes]: I'll just add some detail to to Cameron William's answer. Cameron's answer is correct.
The term eigenvector is usually reserved for a linear a operator $A : X\rightarrow X$ on some particular linear space $X$; $x$ is an eigenvector of $A$ if $Ax=\lambda x$ for some scalar $\lambda$. If the linear space consists of functions, then eigenfunction is used interchangeably with eigenvector.
The Fourier transform $\mathscr{F}$ maps $X=L^{2}(\mathbb{R})$ to itself. So it makes sense to talk about eigenfucntions of $\mathscr{F}$ on $L^{2}(\mathbb{R})$. $\mathscr{F}$ is unitary, meaning that (a) $\|\mathscr{F}f\|=\|f\|$ for all $f\in L^{2}$, and (b) $\mathscr{F}$ is surjective. It turns out that $\mathscr{F}$ has an orthonormal basis of eigenfunctions, and the eigenvalues are $1,i,-1,-i$. The normalized eigenfunction are the Hermite functions
$$
         h_{n}(x) = \frac{(-1)^{n}e^{x^{2}/2}}{(2\pi)^{1/4}2^{n/2-1/4}\sqrt{n!}}\frac{d^{n}}{dx^{n}}e^{-x^{2}},\;\;\; n=0,1,2,3,\cdots .
$$
The Hermite functions satisfy
$$
              \mathscr{F} h_{n} = i^{n}h_{n},\;\;\|h_{n}\|_{L^{2}}=1,\;\; n=0,1,2,3,\cdots .
$$
These functions also satisfy the Hermite differential equation
$$
                  -h''+x^{2}h = 0.
$$
This equation is invariant under the Fourier transform.
The Laplace transform of $1/\sqrt{t}$ is
\begin{align}
    \int_{0}^{\infty}\frac{1}{\sqrt{t}}e^{-st}dt 
        & = \frac{1}{\sqrt{s}} \int_{0}^{\infty}\frac{1}{\sqrt{st}}e^{-st}d(ts) \\
        & = \frac{1}{\sqrt{s}}\int_{0}^{\infty}\frac{1}{\sqrt{u}}e^{-u}du \\
        & = \Gamma(1/2)\frac{1}{\sqrt{s}} = \sqrt{\pi}\frac{1}{\sqrt{s}}
\end{align}
In this context, it is unusual to think of $1/\sqrt{x}$ as an eigenfunction because it is difficult to formulate the Laplace transform $\mathscr{L}$ in such a way that it maps some space to itself. It is more natural to think of $\mathscr{L} : L^{2}[0,\infty) \rightarrow H^{2}(\Pi_{+})$ where $H^{2}(\Pi_{+})$ is the Hardy space of holomorphic functions on the right half-plane $\Pi_{+}$ with $L^{2}$ boundary values; this correspondence is also unitary from one Hilbert space to a different Hilbert space, but, in this context, eigenfunction doesn't make sense.