What are the eigenfunctions and the eigenvalues of the exponential kernel?

The exponential kernel is defined as $$k(x,x')=\sigma^2\exp\left(\frac{||x-x'||}{l}\right)$$ where both $\sigma>0$ and $l>0$.

Mercers theorem tell us that for every kernel function $k(x,x')$ there exists a decomposition in eigenfunctions $\phi_i(x)$ and corresponding eigenvalues $\lambda_i$ such that

$$k(x,x')=\sum_{i=1}^\infty \lambda_i \phi_i(x)\phi_i(x)$$

The Fourier transform $$\mathcal{F}(k)(\omega)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} k(r) e^{i\omega r}dr$$ of the function $$k(r)=\sigma^2 \exp\left(\frac{||r||}{l}\right)$$ with $r=x-x'$ is $$\mathcal{F}(k)(\omega)=\frac{\sqrt{\frac{2}{\pi }} \sigma^2 l}{l^2 \omega ^2+1}.$$ How to proceed from here?

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    $\begingroup$ Some information may be found in Gaussian Processes for Machine Learning, section 4.3, "Eigenfunction Analysis of Kernels." $\endgroup$
    – Sycorax
    Jan 13, 2016 at 19:15
  • $\begingroup$ Is this a machine learning question or a pure math question? $\endgroup$ Jan 13, 2016 at 20:28
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    $\begingroup$ @gung It could be both. I was however hoping that the solution to this problem is better known in the machine learning problem. That's why I posted it here. $\endgroup$ Jan 14, 2016 at 12:14
  • $\begingroup$ @JulianKarch Did you find an answer to it? $\endgroup$
    – newbie
    Aug 18, 2021 at 5:32

2 Answers 2


First of all, your question is not quite well-posed. The reason is that Mercer's theorem only applies for the case of a kernel defined on a finite measure space. Practically, this means that in order to apply the theorem, the eigenfunctions $\phi_i$ are in fact taken with respect to the operator $$K_{\mu}(f)= \left( x\mapsto \int_{\mathbb{R}} K(x,y)f(y)\mu(dy)\right)$$ where $\mu(dy)=p(y)dy$ is a probability measure. The $\phi_i$ are then orthonormal wrt to the inner product defined by $<f,g>=\int f(x)g(x)\mu(dx)$.

It is simple to see that the condition $\mu(\mathbb{R})<\infty$ is necessary for Mercer's theorem to hold. Consider the identity:

$$\int_{\mathbb{R}} e^{-(x-y)^2}xdx=\sqrt{\pi}y$$

This shows that the function $f(x)=x$ is an eigfunction of the operator $\int K(x,y)f(y)dy$. But evidently $\int_{\mathbb{R}} f(x)^2dx=\infty$, which shows that it is not possible to construct an orthonormal basis of $K$, without introducing a weighting function $p(y)$.

Secondly, I am assuming there should be a minus sign in the definition of the kernel $e^{-|x-y|}$, otherwise the resulting kernel fails to be positive definite.

  • $\begingroup$ Aren't you quoting the solution to a different problem? The question concerns an exponential kernel, not a Gaussian one. $\endgroup$
    – whuber
    Jan 17, 2022 at 18:45
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    $\begingroup$ @whuber, oops, you are right, I misread the question. but the first point still stands $\endgroup$ Jan 17, 2022 at 18:48

Assuming the Hamiltonian of your system is the |x> operator, what you are really trying to find is the reciprocal space of x. The easiest way to do this is to take the fourier transform of k(x,x') which by definition is a linear combination of of the k(x,x') states. You can read more here: https://en.wikipedia.org/wiki/Fourier_transform

I also recommend you switch variables to r = x-x' to simplify the math.

  • $\begingroup$ Hey thanks for your comment. I added your suggestion to my original question. How to proceed from there? $\endgroup$ Jan 14, 2016 at 12:29
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    $\begingroup$ Actually I should take a step back. What operator are you investigating? To find eigenvalues and eigenfunctions you need to define this first. I realized I was assuming it was the Fourier Transform operator. $\endgroup$ Jan 15, 2016 at 19:11

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