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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 $k$.

Questions:

  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?

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 $k$.

Questions:

  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.

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 $k$.

Questions:

  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?

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Measurability of eigenvalues-eigenvectoseigenvectors of a positive compact operator

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 $k$.

Questions:

  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.

Measurability of eigenvalues-eigenvectos of a positive compact operator

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, 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 $k$.

Questions:

  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.

Measurability of eigenvalues-eigenvectors of a positive compact operator

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 $k$.

Questions:

  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.

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Measurability of eigenvalues-eigenvectos of a positive compact operator

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, 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 $k$.

Questions:

  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.