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In order to find an orthogonal basis of eigenvectors of the Fourier transform operator $F:L_2(\mathbb{R})\to L_2(\mathbb{R})$, $f\mapsto\lim_{N\to\infty}\int_{[-N,N]}f(x)e^{-i\lambda x}d\mu_x$ for Euclidean separable space $L_2(\mathbb{R})$, so that $F$ would be represented by an infinite diagonal matrix, A.N. Kolmogorov and S.V. Fomin, in their Элементы теории функций и функционального анализа (Elements of the theory of functions and functional analysis), uses functions that are, up to a constant factor, the Hermite functions, which I know to constitute an orthogonal basis of $L_2(\mathbb{R})$.

While looking for such an orthogonal basis of eigenvectors, Kolmogorov and Fomin search for Schwartz functions, belonging to $S_\infty\subset L_2(\mathbb{R})$, in the form $w(x)e^{-x^2/2}$ where $w$ is a polynomial, satisfying the equation$$\frac{d^2f}{dx^2}-x^2f=\mu g\quad\quad\text{equation }(3)$$where $\mu$ is a constant. Such an equation change into $\frac{d^2g}{d\lambda^2}-\lambda^2g=\mu f$ when acted upon by operator $F$, $f\mapsto F[f]=g$. It is shown that the polynomials $w$ of such (Hermite up to a constant factor) functions satisfy equation (3) for $\mu=-(2n+1)$ when $\deg w=n$ (let us call such a polynomial $P_n$) and they have non-null coefficients $a_k$ of the variable $x^k$ only for the $k$'s of the same oddity of $n$. It had also previously proved (p. 401) that $P_n$ is, up to a constant, $(-1)^ne^{x^2}\frac{d^n e^{-x^2}}{dx^n}$ (I am writing this in the case it can be used to prove that $P_n(x)e^{-x^2/2}$ defines an eigenvector of $F$).

Kolmogorov-Fomin's says that the fact that the $P_ne^{-x^2/2}$ are eigenvectors of $F$ and their eigenvalues are $\pm\sqrt{2\pi}$, $\pm i\sqrt{2\pi}$ derives from the following fact:

  1. Equation (3) is invariant with respect to transformation $F$.
  2. Equation (3) has got, up to a constant factor, one solution of the form $P_ne^{-x^2/2}$.
  3. The Fourier transform maps $x^ne^{-x^2/2}$ to $i^n\sqrt{2\pi}\frac{d^n}{dx^n}e^{-x^2/2}$ (as I knew, by using the fact that $F[e^{-x^2/2}](\lambda)=\sqrt{2\pi}e^{-\lambda^2/2}$).

The proof of this derivation contained in the book only says that $F^4[P_n e^{-x^2/2}]=4\pi^2 P_n e^{-x^2/2}$, but I cannot see this and, even proved this, I could not see how this is related to the fact that the eigenvalues are the fourth roots of $4\pi^2$ (since I do not think that if the fourth power of matrix $A$ is diagonal $A^4=\text{diag}(4\pi^2,4\pi^2,\ldots)$ then $A=\text{diag}((4\pi^2)^{1/4},(4\pi^2)^{1/4},\ldots)$). I have found some resources talking about the topic on line, but nothing using Kolmogorov-Fomin's argument, which I would like to understand... Does anybody understand and can explain it? I heartily thank you!

P.S.: If it were useful to understand what Kolmogorov-Fomin's says, I know that $\forall f\in S_\infty$, $k\in\mathbb{N}$ $F[f^{(k)}](\lambda)=(i\lambda)^kF[f](\lambda)$ and $\frac{d^k}{d\lambda^k}F[f](\lambda)=(-i)^kF[x^k f](\lambda)$.

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The Fourier transform is unitary on $L^{2}(\mathbb{R})$ if you choose the correct normalization: $$ Ff = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-ist}f(t)\,dt. $$ Some people instead absorb the constant into the exponent, but this is what I'm used to seeing because this $F$ is directly linked to the spectral resolution of the operator $A=\frac{1}{i}\frac{d}{dt}$ on the domain consisting of all $f\in L^{2}$ which are absolutely continuous on $\mathbb{R}$ with $f' \in L^{2}$. This operator $A$ is selfadjoint.

For the $F$ as stated, it is true that $F : L^{2}\rightarrow L^{2}$ is unitary with unitary inverse $$ F^{-1}g = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{ixs}g(s)\,ds. $$ The fact that $F^{-1}$ is related to $F$ is where you get the general identity that $F^{4}=I$. Indeed, $(F^{2}f)(x) = f(-x)$ is a reflection. That's how you get $F^{4}=I$ on $L^{2}$. That alone is enough to imply that the spectrum of $F$ is $\{ 1,i,-1,-i\}$. To see why, let these roots be identified with $\lambda_{n}=i^{n-1}$ for $n=1,2,3,4$, and form the Lagrange interpolating polynomials $$ p_{k}(\lambda)=\frac{\prod_{n\ne k}(\lambda-\lambda_{n})}{\prod_{n\ne k}(\lambda_{k}-\lambda_{n})},\;\;\; k =1,2,3,4. $$ The polynomial $p_{k}$ is $1$ on $\lambda_{k}$ and $0$ on the other $\lambda_{n}$'s. Hence, $1=p_{1}+p_{2}+p_{3}+p_{4}$ because this is a third order polynomial that is $1$ at each of $4$ distinct points. It follows that $$ I=p_{1}(F)+p_{2}(F)+p_{3}(F)+p_{4}(F). $$ That means that each $P_{k}=p_{k}(F)$ is a projection. To see why, first observe that $p_{j}(F)p_{k}(F)=0$ for $j \ne k$ because $\lambda^{4}-1$ divides $p_{k}p_{j}$ in that case. Applying $p_{k}(F)$ to the above equation gives $$ p_{k}(F) = p_{k}(F)^{2}. $$ Therefore each $P_{k}=p_{k}(F)$ is a projection; these projections are disjoint; and their sum is the identity $I$ on $L^{2}$.

$P_{k}$ is the projection onto the eigenspace of $F$ with eigenvalue $\lambda_{k}$ because $(F-\lambda_{k}I)P_{k}=0$. This is a spectral decomposition for $F$: $$ P_{k}^{2}=P_{k},\;\;\; P_{k}P_{j}=0,\;\; j \ne k,\\ I = P_1+P_2+P_3+P_4,\\ F = \lambda_1 P_1 + \lambda_2 P_2 + \lambda_3 P_3 + \lambda_4 P_4. $$ Notice that these arguments don't require inner products or norms. These arguments work for any linear operator $F$ on a complex linear space for which $F^{4}-I=0$. The same thing works for the unnormalized $F$ that you have, too.

Hermite Functions: The Hermite polynomials, as you have defined them, are $$ P_{n}(x) = e^{x^{2}}\frac{d^{n}}{dx^{n}}e^{-x^{2}}. $$ The Hermite functions $H_{n}(x)=e^{-x^{2}/2}P_{n}$ are solutions of the ODE $$ \left[-\frac{d^{2}}{dx^{2}}+x^{2}\right]H_{n}=\lambda_{n}H_{n}. $$ These are the unique $L^{2}(\mathbb{R})$ eigenfunctions of the differential equation and $\{ H_{n} \}_{n=0}^{\infty}$ forms a complete orthonormal basis of $L^{2}(\mathbb{R})$. Knowing this, when you Fourier transform the Hermite equation, you get $$ s^{2}\widehat{H_{n}}(s)-\frac{d^{2}}{ds^{2}}\widehat{H_{n}}(s)=\lambda_{n}\widehat{H_{n}}(s). $$ Because of the uniqueness of such solutions, it follows that $\widehat{H_{n}}$ must be a constant multiple of $H_{n}$. That constant multiple is an eigenvalue of the Fourier transform, which must be either $1$, $i$, $-1$ or $-i$. (Remember, I use the normalized Fourier transform with $1/\sqrt{2\pi}$ in front of it so that $\|Ff\|=\|f\|$.)

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    $\begingroup$ A projection is a linear operator satisfying $P^{2}=P$. Notice that $P(P-I)=0$. So the spectrum is contained in $\{ 0, 1\}$. Using the same construction the linear space decomposes into a sum of two subspaces $X=PX\oplus (I-P)X$. On the first subspace, $P=I$ and on the second $P=0$. If you have a matrix with $P^{2}=P$, then there is a basis for $P$ where the matrix is diagonal with $1$'s and $0$'s along the diagonal. $\endgroup$ – Disintegrating By Parts Nov 27 '14 at 9:36
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    $\begingroup$ If $P(P-1)=0$, or equivalently, $P^{2}=P$, then every $x=(I-P)x+Px$, and $P(Px)=(Px)$, $P((I-P)x)=0$. So every $x$ decomposes into components on which $P$ is either $0$ or $I$. For $F^{4}=I$, I showed you that $x=x_{1}+x_{i}+x_{-1}+x_{-i}$ where $Fx_{1}=x_{1}$, $Fx_{i}=ix_{i}$, $Fx_{-1}=-x_{-1}$ and $Fx_{-i}=-ix_{-i}$. This decomposition is unique. $\endgroup$ – Disintegrating By Parts Nov 27 '14 at 10:52
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    $\begingroup$ @DavideZena : Notice that these ideas extend to any linear operator $L$ for which $m(L)=0$ for some polynomial $m$ with distinct linear factors. It is this algebraic trick that forces a diagonalization of $L$, regardless of dimension. In finite dimensional spaces, $L$ is similar to a diagonal matrix iff the minimal polynomial has no repeated linear factors. The algebra of this is very strong because it leads to projections through the interpolating polynomials $p_k$, which decomposes the space into a direct sum of eigenspaces. The $P_{k}$ project to the component associated with $\lambda_k$. $\endgroup$ – Disintegrating By Parts Nov 27 '14 at 11:06
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    $\begingroup$ Because $(\lambda - \lambda_{k})p_{k}(\lambda)$ is a multiple of $\lambda^{4}-1$, then $(F-\lambda_{k}I)p_{k}(F)=0$. That's how the $p_{k}$ were designed: they're missing just one factor so that when you apply that last operator $(F-\lambda_{k}I)$ to $p_{k}(F)$, you get $0$, which means the range of $p_{k}(F)$ is contained in the null space of $F-\lambda_{k}I$. $\endgroup$ – Disintegrating By Parts Nov 27 '14 at 12:07
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    $\begingroup$ @DavideZena : I've added more about the Hermite functions. $\endgroup$ – Disintegrating By Parts Nov 27 '14 at 14:17

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