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Given a compact self-adjoint operator $K$ mapping $L^2(\mathbb{R}^d) \rightarrow L^2(\mathbb{R}^d)$ as $f \rightarrow \int K(x,y) f(y) d\mu(y)$, let us define its eigenvalues $\lambda_i$ and eigen-functions $e^i(x)$.

Let us define a sequence of "partial sum" operators $K_N(x,y) = \sum_{i=1}^N \lambda_i e^i(x)e^i(y)$. Then this sequence converges to $K(x,y)$ in the product Hilbert space under the product measure.

More explicitly then,
$$\lim_{N\rightarrow \infty} \int \int |K(x,y) - \sum_{i=1}^N \lambda_i e^i(x)e^i(y)|^2 d\mu(x)d\mu(y) = 0.$$

But is there any example where for some specific $K$ someone has tried estimating this error-integrand $|K(x,y) - \sum_{i=1}^N \lambda_i e^i(x)e^i(y)|^2 $ ? I would be hapy to find any reference along these lines..

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2 Answers 2

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I'm following Szego's book on orthogonal polynomials.

In chapter III he considers $f \in L^2\left((a,b),d\alpha \right)$, with $-\infty \leq a < b \leq \infty $. There exist a set of orthogonal polynomials $\{p_n \}_{n=0}^{\infty}$ w.r. to the measure $d\alpha$.

We consider the kernel $K_n(t,x) := \sum\limits_{\nu=0}^{n} \bar{p_{\nu} (t)} p_{\nu} (x) $. Then one can show that $\int\limits_{a}^{b} K_n (t,x) f(t) \, d\alpha (t) = \sum\limits_{\nu = 0}^{n} \hat{f}(n)p_n (x) $, the spectral expension of order $n$.

So, if we settled that the polynomial expansion of $f$ is an integral transform, there's a detailed discussion in Szego's book, Davis "approximation theory" and many other books about its $L^2$ error.

Edit: If we look in Aronszajn (1950), we can have by Section 9 that the kernels converge in L^2 norm.

Edit 2: I recommend you to read the aforementioned paper, as it deals with the much broader class of reproducing kernels, and give a lot of conditions and results that are relevant to your question.

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  • $\begingroup$ Well, I am looking to estimate things like the value of what you call $K_n (t,x)$. You think Szego's book does this? $\endgroup$
    – gradstudent
    Commented Apr 22, 2016 at 6:33
  • $\begingroup$ You mean section 5.7 in Szego's book? $\endgroup$
    – gradstudent
    Commented Apr 22, 2016 at 6:35
  • $\begingroup$ 5.7 deals with Hermite and Laguerre polynomials, but for more general information about it look at chapters 14-15. What kind of error norm are you interested in? $\endgroup$
    – Amir Sagiv
    Commented Apr 24, 2016 at 5:55
  • $\begingroup$ I am interested in trying to estimate the difference between the Gaussian kernel and a finite truncation of its spectral expansion which is in terms of Hermite polynomials. At least this is one special case I am immediately interested in. (...I vaguely understand that these spectral expansions are somehow special because the sequence of partial sums here converge uniformly on compact sets - this is some special property for which I don't have much intuition as of now..) $\endgroup$
    – gradstudent
    Commented Apr 24, 2016 at 17:36
  • $\begingroup$ To be honest, I'm not sure that $K_N(t,x) \to K(t,x)$ pointwise, and not only in $w*$ topology. I'll ask about it in a different question. $\endgroup$
    – Amir Sagiv
    Commented Apr 24, 2016 at 22:25
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Actually, the presumption that the partial sums of the kernel converge to it in $L^2$ of the product is not quite right: for example, mapping $\ell^2\to \ell^2$ by $e_n\to \lambda_n\cdot e_n$ for a sequence of real numbers $\lambda_n$ going to $0$, but slowly, gives kernel $K=\sum_n \lambda_n\cdot e_n\otimes e_n$. For $\lambda_n$ going to $0$ slowly enough, certainly $\sum_n |\lambda_n|^2=+\infty$, so the kernel cannot be in $L^2$ of the product. This obstacle exists prior to talking about spaces of functions on reasonable physical spaces.

In general, asking about pointwise (much less uniform pointwise) convergence is somewhat worse, when the Hilbert spaces are concrete spaces of functions.

So maybe this is not quite an "answer to the question", but a reaction and request for clarification.

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  • $\begingroup$ So for what kind of operators can I assume this kind of a convergence in L^2 of the partial sums? (I am assuming that its a compact self-adjoint operator on a Hilbert space) If the operator is compact then the spectral theorem holds (theorem 2.5, page 20 here people.math.ethz.ch/~kowalski/spectral-theory.pdf) . In that case are there examples of being able to estimate the error in partial sums in the equation 2.9? $\endgroup$
    – gradstudent
    Commented Apr 22, 2016 at 4:08
  • $\begingroup$ @gradstudent, are you familiar with Hilbert-Schmidt operators? $\endgroup$
    – Uri Bader
    Commented Apr 22, 2016 at 6:38
  • $\begingroup$ Yes. These kernels are Hilbert-Schmidt operators. $\endgroup$
    – gradstudent
    Commented Apr 22, 2016 at 6:54
  • $\begingroup$ Are "reproducing kernels" the same as Hilbert-Schmidt operators? $\endgroup$
    – gradstudent
    Commented Apr 22, 2016 at 7:00
  • $\begingroup$ Hilbert-Schmidt operators are a proper sub-class of compact operators, namely, exactly the class of compact operators whose naturally-plausible kernel does converge in $L^2$. "Reproducing kernels" are kernels giving the identity map on some function-space, and possessing some pointwise or uniformly locally pointwise convergence ($L^2$ convergence for the identity map could happen only on a finite-dimensional space...) $\endgroup$
    – paul garrett
    Commented Apr 22, 2016 at 12:35

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