Let $\mu$ be a finite, strictly positive measure on $\mathbb{R}$, and let $k$ be a measurable positive-definite kernel. Assume $k$ is bounded, and let $T:L^2(\mu)\rightarrow L^2(\mu)$ be defined by $$ Tf(x):=\int k(x,y) f(y)d\mu(y).$$

Does the representation $$k(x, y) = \sum_{j=1}^\infty \lambda_j \varphi_j(x) \varphi_j(y)$$ hold uniformly on $(x,y)$?

Here, the $(\lambda_j, \varphi_j)$ pairs are defined via $$T\varphi_j = \lambda_j\varphi_j.$$

  • $\begingroup$ when you say $\mu$ is finite you mean that $\int d\mu < \infty$ so that $T$ is bounded and compact ? $\endgroup$
    – reuns
    Jan 12, 2016 at 19:48
  • $\begingroup$ Am I right in assuming that the $\varphi_j$ are an orthonormal basis of eigenvectors of $T$? $\endgroup$
    – user159517
    Jan 12, 2016 at 20:18
  • $\begingroup$ user1952009: yes, exactly. In fact, $T$ is trace class. $\endgroup$
    – user127022
    Jan 12, 2016 at 21:02
  • $\begingroup$ user159517: yes, precisely! $\endgroup$
    – user127022
    Jan 12, 2016 at 21:03

2 Answers 2


I'm assuming that the functions $\varphi_j$ are real-valued. The Cauchy-Schwartz inequality on $\ell^2(\mathbb{N})$ gives

\begin{align*}\sum_{j=1}^{\infty} |\lambda_j \varphi_{j}(x) \varphi_{j}(y)| &= \sum_{j=1}^{\infty} |\sqrt{\lambda_j} \varphi_{j}(x) \sqrt{\lambda_j}\varphi_{j}(y)| \leq \sum_{j=1}^{\infty} \lambda_j \varphi_j(x)^2 \sum_{j=1}^{\infty}\lambda_j \varphi_j(y)^2 \\ &= k(x,x) k(y,y) \leq C^2\end{align*}

As $k(x,y)$ is bounded. This shows uniform convergence, assuming you know already that the equality holds pointwise.

EDIT: I made an implicit assumption in the following argument that I just became aware of now. Let's assume that the $\varphi_j$ are continuous (that would follow if we would assume $k$ to be continuous in both variables). Then we know the equality holds pointwise for some subsequence $\sum_{j=1}^{n_k} \lambda_j \varphi_j(x) \varphi_j(y)$. Applying the argument above shows uniform convergence for this subsequence. This implies that there is a $k_0$ such that $$\sum_{j=n_k}^{\infty} \|\lambda_j \varphi_j(.) \varphi_j(.)\|_{\infty} < \epsilon, \quad k\geq k_0$$ which implies the existence of a $N$ such that

$$ \sum_{j=N}^{\infty} \|\lambda_j \varphi_j(.) \varphi_j(.)\|_{\infty} < \epsilon, \quad n\geq N$$

Now, if the $\varphi_j$ are continuous, we can identify $(\sum_{j=1}^{n}\lambda_j \varphi_j(.)\varphi_j(.))_{n\in\mathbb{N}}$ as a Cauchy-sequence in $C([0,1]\times[0,1])$ with the supremum norm. As this is a complete space, uniform convergence follows.

  • $\begingroup$ The equality holds in $L^2(\mu\times\mu)$ and hence there is a subsequence that converges $\mu\times\mu$ almost everywhere. I wonder if the combined results imply uniform convergence of the whole sequence and not only of the subsequence. $\endgroup$
    – user127022
    Jan 12, 2016 at 21:01
  • 1
    $\begingroup$ I changed my mind, it actually follows that the whole sequence converges uniformly (the convergence of the subsequence implies that $\sum_{j=n_k}^{\infty} \|\lambda_j \varphi_j(.)\varphi_j(.)\|_\infty$ is smaller than $\epsilon$ for k big enough; This implies that the original sequence is a Cauchy-sequence w.r.t the $\infty-$norm and thus uniformly convergent. $\endgroup$
    – user159517
    Jan 12, 2016 at 22:47
  • $\begingroup$ I'm having second thoughts. So far I only know that there is a set $E_0\subset\mathbb{R}^2$ of total $\mu\times\mu$-measure such that, for $(x,y)\in E_0$ the equality $k(x,y) = \lim_{k\rightarrow\infty} \sum_{j=1}^{n_k}\lambda_j\varphi_j(x)\varphi_j(y)$ holds. How do I get from here to uniform convergence on $\mathbb{R}^2$? $\endgroup$
    – user127022
    Jan 14, 2016 at 1:10
  • $\begingroup$ @user127022 My new opinion on the matter is that it holds if $k$ is continuous, not sure if that assumption is necessary though. See my edited answer. $\endgroup$
    – user159517
    Jan 14, 2016 at 15:06
  • $\begingroup$ I agree that it holds for continuous $k$. Now, maybe it does not hold in general, but in fact I know that the functions $\varphi_j$ are càdlàg, that is, $\lim_{x_n\downarrow x}\varphi_j(x_n) = \varphi_j(x)$ and $\lim_{x_n\uparrow x}\varphi_j(x_n)$ exists. This implies that the set $V_j$ of discontinuity points of $\varphi_j$ is at most countable. I'm just not sure how to use this fact to establish pointwise convergence of a subsequence. $\endgroup$
    – user127022
    Jan 14, 2016 at 16:26

Uniform convergence is established in Theorem 3.a.1 in König's Eigenvalue Distribution of Compact Operators (DOI: 10.1007/978-3-0348-6278-3)

Theorem: Let $(\Omega, \mu)$ be a finite measure space and $k\in L^\infty(\Omega\times\Omega,\mu\times\mu)$ be a kernel such that $T_k\colon L^2(\Omega,\mu)\to L^2(\Omega,\mu)$ is positive. Then the eigenvalues $(\lambda_n)$ of $T_k$ are absolutely summable. The eigenfunctions $f_n\in L^2(\Omega,\mu)$ of $T_k$, associated with those $n$ such that $\lambda_n\neq 0$, and normalized by $\| f_n\|_2 = 1$, actually belong to $L^\infty(\Omega,\mu)$ with $\sup_n \|f_n\|_\infty <\infty$ and $$k(x,y) = \sum_{n\in\mathbb{N}} \lambda_n \overline{f_n(y)}f_n(x)$$ holds $\mu\times\mu$-a.e., where the series converges absolutely and uniformly $\mu\times\mu$-a.e.


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