Let $H$ be a separable Hilbert space over $\mathbb{R}$. Let ${A} = \{a\colon H\to H\,|\,a\text{ is a positive, compact linear operator}\}$.

By the spectral theorem, given $a \in A$, there are scalars $\lambda_{1,a}\geqslant \lambda_{2,a}\geqslant\cdots\geqslant0$ and an orthonormal basis $\Phi_a = \{\varphi_{1,a},\varphi_{2,a},\dots\}$ of $H$ such that $a(\varphi_{k,a}) = \lambda_{k,a}\varphi_{k,a}$ for all $k$.


  1. Are the mappings $a\mapsto \lambda_{k,a}$ measurable?
  2. Is it possible to choose $\Phi_a$ such that the mappings $a\mapsto\varphi_{k,a}$ are measurable?

Remark: on $\mathbb{R}$ and $H$ I'm considering the Borel $\sigma$-fields. On $A$ I can accept any “reasonable” $\sigma$-field.

Since the general case above seems to be difficult to prove, let me adapt the question to a simpler scenario: consider a measurable space $\mathcal{Z}$ and, for each $z\in\mathcal{Z}$, assume $k_z\colon[0,1]\times[0,1]\to\mathbb{R}$ is a continuous symmetric positive-definite kernel. By Mercer's Theorem, there exists a sequence of real numbers $\lambda_{z1}\ge\lambda_{z2}\ge\cdots\ge0$ and measurable, square-integrable functions $\varphi_{z1}(\cdot),\varphi_{z2}(\cdot),\dots$ on $[0,1]$ such that

$$k_z(x,y) = \sum_{j=1}^\infty \lambda_{zj}\varphi_{zj}(x)\varphi_{zj}(y)$$ where the convergence is absolute and uniform in $(x,y)\in[0,1]^2$. In particular, it holds that $\int_0^1 k_z(x,y)\varphi_{zj}(y)\mathrm{d}y = \lambda_{zj}\varphi_{zj}(x)$, $x\in[0,1]$, for all $j\ge1$, and the functions $\varphi_{zj}(\cdot)$ are seen to be continuous whenever $\lambda_{zj}>0.$ Now, assuming either that

  1. for each $(x,y)\in[0,1]^2$ the mapping $z\mapsto k_z(x,y)$ is measurable from $\mathcal{Z}$ to $\mathbb{R}$, or;
  2. the mapping $z\mapsto k_z$ is measurable from $\mathcal{Z}$ to $C([0,1]^2)$ (the space of continuous functions on the unit square), or;
  3. the mapping $z\mapsto k_z$ is measurable from $\mathcal{Z}$ to $L^2([0,1]^2)$

is it true that the mappings $z\mapsto \lambda_{zj}$ are measurable? Additionally, is it possible to choose the $z\mapsto\varphi_{zj}(\cdot)$ in a measurable manner?

  • 1
    $\begingroup$ The map $T\mapsto \sigma(T)$ is continuous in the Hausdorff-metric, hence the eigenvalue maps are continuous. $\endgroup$
    – Echo
    Dec 10, 2022 at 6:16
  • 1
    $\begingroup$ @Echo: I'm having difficulties to follow your first comment. The ranks of the spectral projections can jump, so they can't be continuos. To get continuity one would need to decompose the spectral projections associated to non-simple eigenvalues into rank-1 projections in a continuous way. I don't see why this is possible. $\endgroup$ Dec 10, 2022 at 10:20
  • 1
    $\begingroup$ @Echo: Re your 2nd comment: I don't see how continuity with respect to the Hausdorff metric alone can give the continuity of the eigenvalue maps: one can't reconstruct the list of eigenvalues of $T$ from the set $\sigma(T)$ since $\sigma(T)$ does not contain information about eigenvalue multiplicity. $\endgroup$ Dec 10, 2022 at 10:31
  • 3
    $\begingroup$ @JochenGlueck: The point is well taken, but I think the argument should work anyway since one can also remove degeneracies by arbitrarily small (in operator norm) perturbations. But min-max gives this more conveniently and it also shows somewhat more quantitatively that $\|\lambda_k(S)-\lambda_k(T)\|\le \|S-T\|$. $\endgroup$ Dec 10, 2022 at 17:25
  • 1
    $\begingroup$ The answer here by user2048 seems to suggest a way to answer both questions. $\endgroup$
    – John
    Mar 23 at 0:53


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