# Methods for finding decompositions of Hilbert-Schmidt integral operators

$$(Kf)(x) = \int_Y k(x,y)f(y) dy$$

a decomposition (called Hilbert-Schmidt decomposition) of the following form exists:

$$k(x,y) = \sum_n \sigma_n u_n(x)v_n(y)$$

where the functions (which we call "modes") $u_n(x)$ are orthonormal on the domain $X$ and $v_n(x)$ are orthonormal on the domain $Y$. I don't actually know which theorem states this (I think it's the Hilbert-Schmidt theorem, see also Spectral Theorem for bounded compact, self-adjoint operators as corollary of Hilbert-Schmidt theorem), but I know that it is true (although I may have forgotten some restrictions for it to apply).

@Mark pointed out in his comments and answer that this is equivalent to finding the eigenfunctions:

$\lambda u\left(x\right)=\int_{a_{0}}^{a_{1}}k_{1}\left(x,y\right)u\left(y\right)dy$ , $\lambda$ a scalar, $u \in L^2([a_0,a_1])$

where $k_{1}\left(y,s\right):=\int_{b_{0}}^{b_{1}}k\left(x,y\right)k\left(x,s\right)dx$.

So essentially what I am asking is for a reference to do eigenvalue decompositions of integral operators in practice.

Well, I'm a physicist and these types of operators pop up everywhere, in my particular case it is the context of optics. Hilbert-Schmidt decompositions are particularly useful due to the orthogonality of the modes, which is essential in statistical mechanics applications, inverse problems and few mode approximations (to only name a few).

Searching through the literature I found a lot of resources treating these operators from a what seems to me very pure mathematical viewpoint. I.e. general theorems about existence etc. are proven and so on. What I am interested in however is finding the actual decomposition of a given operator. We usually do this numerically (i.e. by discretizing the function domains and performing an SVD on the resulting matrix), but some of the operators seem quite "nice" to me (e.g. convolution-type operators are a common theme), so I was wondering if we could find analytical expressions for the modes. I couldn't find anything in the literature, so my question here is if anyone has could point me to:

• general methods for calculating mode decompositions
• tables of operators where the decomposition is known
• any other sources that would be useful in this context

EDIT:

Following @Mark's helpful comments, here is an example of the kind of operator I am dealing with

$$(Kf)(\beta) = \int_{a_0}^{a_1} \underbrace{ sinc\left( \frac{q_x w}{2} \right)}_{B} \times \underbrace{\frac{\sin\left( \frac{q_x d N}{2} \right)}{\sin\left( \frac{q_x d}{2} \right)}}_{C} \times f(\alpha) d\alpha,$$

where $q_x = k \left( \alpha + \beta \right)$. $k, w, d, a_0, a_1$ are some constants that characterize the physical dimensions of the system. $\beta$ is the also restricted to a domain $[b_0; b_1]$. Useful approximations are omitting the term $A$ or the term $B$ completely, so any of those decompositions would be useful. This one is a convolution operator, so I'm hopeful the solution can be found analytically.

A more complicated version would have varying $k$

$$(Kf)(\beta) = \int_{k_0}^{k_1} \int_{a_0}^{a_1} \underbrace{ sinc\left( \frac{q_x w}{2} \right)}_{B} \times \underbrace{\frac{\sin\left( \frac{q_x d N}{2} \right)}{\sin\left( \frac{q_x d}{2} \right)}}_{C} \times f(\alpha, k) d\alpha dk.$$

I am not asking to solve any of this in particular, rather searching for general methods and to see what is possible. If you know the decomposition for one of them it would be very helpful though ;)

• Have you looked at Reed & Simon's books on functional analysis? You'll probably find quite a bit of info there. Also look for "Mercer's theorem". – Mark Jul 1 '16 at 11:48
• @Mark thanks! i'll have a look at the book. Mercer's theorem is just an existence theorem though, isn't it? I'm looking for the actual decompositions. – Wolpertinger Jul 1 '16 at 12:42
• I am only familiar with the case $X=Y$ where $u_n = v_n$. In this case the functions $u_n$ are eigenfunctions of the operator $K$ so to find them you need to solve an integral equation of the form $(Kf)(x) = \lambda f(x) \forall x$. By differentiating you can turn it into a differential equation, and in some cases you can find its solutions explicitly. – Mark Jul 1 '16 at 13:20
• Also, note that a Hilbert-Schmidt operator has $X=Y$ by definition (look at your link). If your $X$ and $Y$ are just subsets of $\mathbb{R}^n$ (with the same $n$ for both) you can replace them both with $\mathbb{R}^n$ by defining $k(x,y)$ to be zero in the complement of their union. – Mark Jul 1 '16 at 13:33
• Nice Q and comments!. Any reference on general analytical techniques to solve this kind of eigenfunction decomposition? i.e. not only for the particular case – JuanPi Feb 24 '18 at 16:09

This answer is in regards to the two examples you've posted.

These aren't convolution operators. In a convolution operator, $k(\alpha,\beta)$ ($k$ is my notation for the kernel here, not the constant in your examples) depends only on $\alpha -\beta$. In your case they depend on $\alpha + \beta$. This is sometimes called cross-correlation, which is similar to but not the same as convolution.

We can get a convolution operator if we replace the kernel $k(\alpha,\beta)$ with $\tilde{k}\left(\alpha,\beta\right):=k\left(\alpha,-\beta\right)$ but then this is a new operator (which spits out functions defined on $[-b_1,-b_0]$) and it will have different eigenfunctions $u_n, v_n$ for it.

Convolution operators are diagonalized by the Fourier transform- this is what makes them easier to deal with than general integral operators. Operators such as yours, on the other hand, are diagonalized by the Laplace transform (after making some slight adjustments, like enlarging the domains).

Unfortunately I only have a vague familiarity with the Laplace transform. In the case of a convolution operator $Kf := k*f$ the classic theorems imply (under some minimal assumptions) that the eigenvalues are $\hat{k} (\xi)$ (where $\hat{k}$ is the Fourier transform of $k$), the eigenfunctions of $K$ are the complex exponentials $e^{2 \pi i \xi x}$ (this would be your $u_n, v_n$), and these are no other eigenfunctions or eigenvalues.

I would imagine that similar results hold for the Laplace transform, but I don't know where you should look. As far as integral transforms go, I've mostly used the Fourier transform.

EDIT: You are right in that your decomposition resembles the SVD more than an eigendecomposition. I wasn't aware of this, but apparently there is an analogous SVD for compact linear transformations between (different) Hilbert spaces, which can be applied to the operator $K$. The functions $u_n$ are then eigenvectors of $K^{*}K$ and the $v_n$ are eigenvectors of $K^{*}K$. These operators are also integral operators: $K^{*}K$ is an operator on $L^2 ([a_0,a_1])$ (domain and range) with kernel $k_{1}\left(y,s\right):=\int_{b_{0}}^{b_{1}}k\left(x,y\right)k\left(x,s\right)dx$ and $KK^{*}$ is an operator on $L^2([b_0,b_1])$ (domain and range) with kernel $k_{2}\left(x,t\right):=\int_{a_{0}}^{a_{1}}k\left(x,y\right)k\left(t,y\right)dy$, where $k$ is the kernel of $K$. The singular values $\sigma_n$ are the eigenvalues of either of these two operators, like in the classical SVD.

Consequently, your problem is equivalent to finding the eigenfunctions of these two operators. There are various methods and theories about how to do that. However, in this case the kernel is pretty complicated, so I doubt that an explicit analytic solution is aviailable (there may be effective numeric methods). Basically what you want is to solve the following integral equations:

$\lambda u\left(x\right)=\int_{a_{0}}^{a_{1}}k_{1}\left(x,y\right)u\left(y\right)dy$ , $\lambda$ a scalar, $u \in L^2([a_0,a_1])$.

$\mu v\left(x\right)=\int_{b_{0}}^{b_{1}}k_{2}\left(x,y\right)v\left(y\right)dy$ , $\mu$ a scalar, $v \in L^2([b_0,b_1])$.

This is an interesting topic but it is outside the scope of my knowledge. Perhaps you should post a new question about this.

• thanks for your answer and the comments, it's been very helpful! I don't think the Hilbert-Schmidt decomposition can be found the way you say though for several reasons. Most simply I've solved the problem numerically and the functions don't look like Fourier or Laplace basis functions at all. On a more fundamental level I don't understand how the eigenfunctions after "some sleight adjustments" will be the same as the Hilbert Schmidt functions. Also note that my simulations show that $u_n$ is not $v_n$ (even after domain modification). – Wolpertinger Jul 3 '16 at 11:18
• that's just a guess, but I think this might be like an SVD vs eigenvalue decomposition in the matrix case. And I need the SVD (Hilbert Schmidt decomposition), not the e-value decomposition – Wolpertinger Jul 3 '16 at 11:18
• @Numrok: Good point. You are right that "some slight adjustments" won't do here. The only compact (and in particular, Hilbert-Schmidt) operator on $L^2 (\mathbb{R})$ is the zero operator, if I'm not mistaken. So it doesn't look like your problem can be reduced to convolution operators on the real line. The Fourier and Laplace transforms can't be applied in an obvious way here. – Mark Jul 3 '16 at 15:09