Return to Question

I am trying to prove whether the space $L(H,K)$ has martingale type 2 for Hilbert spaces $H,K$. It is known that Hilbert spaces have martingale type 2, so I was wondering whether the space of bounded linear operators also enherit the property. We recall the definition here

Definition. Let $X$ be a Banach space and $(M_n)_{n=0}^N$ be a martingale with values in $X$. We say $X$ has martingale type $p\in [1,2]$ if for some $m\in (1,\infty)$, there exists a constant $C=C(m,p,X)$ such that \begin{align*} \mathbb E[\Vert M_N\Vert_X^m]\leq C\, \mathbb E\left[ \left\vert \Vert M_0\Vert_X^p+ {\sum}_n^N\Vert \Delta M_n\Vert_X^p\right\vert^{m/p}\right], \end{align*} where $\Delta M_n$ is the martingale difference.

Since $H$ and $K$ both have type $2$ a first line of applying that to linear operators leads to the following problem:

Problem. Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space and $H,K$ be separable Hilbert spaces. Consider a random bounded linear operator $A:\Omega\to L(H,K)$. Assume for any integer $m\geq 2$ there exists $C=C(m,K,H)$ such that for all $0\neq x\in H$ we have \begin{align*}\mathbb{E}[\Vert Ax\Vert_K^m] \leq C \Vert x\Vert_H^m \end{align*} There exists $\Gamma=\Gamma(m,K,H)$ such that $$\mathbb E[\Vert A\Vert_{L(H,K)}^m]\leq C\Gamma.$$

The only attempt I currently have is to show the statement through Kolmogorov continuity criterion for Hilbert spaces. I do not know whether such statement exists, but I was willing to use the paper sample properties of random fields to get the result. Note that if I believe Kolmogorov continuity holds for Hilbert spaces, then I get that the Holder seminorm $\vert A\vert_{C^\alpha}$ bounded in $L^m(\mathbb P)$ which together with the fact that $A$ is linear bounded operator gives the result. This is too overkill.

Context.

I am trying to prove whether the space $L(H,K)$ has martingale type 2 for Hilbert spaces $H,K$. It is known that Hilbert spaces have martingale type 2, so I was wondering whether the space of bounded linear operators also enherit the property. We recall the definition here

Definition. Let $X$ be a Banach space and $(M_n)_{n=0}^N$ be a martingale with values in $X$. We say $X$ has martingale type $p\in [1,2]$ if for some $m\in (1,\infty)$, there exists a constant $C=C(m,p,X)$ such that \begin{align*} \mathbb E[\Vert M_N\Vert_X^m]\leq C \operatorname{\mathbb E}\left[ \left\vert \Vert M_0\Vert_X^p+ {\sum}_n^N\Vert \Delta M_n\Vert_X^p\right\vert^{m/p}\right], \end{align*} where $\Delta M_n$ is the martingale difference.

Since $H$ and $K$ both have type $2$ a first line of applying that to linear operators leads to the following problem:

Problem. Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space and $H,K$ be separable Hilbert spaces. Consider a random bounded linear operator $A:\Omega\to L(H,K)$. Assume for any integer $m\geq 2$ there exists $C=C(m,K,H)$ such that for all $0\neq x\in H$ we have \begin{align*}\mathbb{E}[\Vert Ax\Vert_K^m] \leq C \Vert x\Vert_H^m \end{align*} There exists $\Gamma=\Gamma(m,K,H)$ such that $$\mathbb E[\Vert A\Vert_{L(H,K)}^m]\leq C\Gamma.$$

The only attempt I currently have is to show the statement through Kolmogorov continuity criterion for Hilbert spaces. I do not know whether such statement exists, but I was willing to use the paper sample properties of random fields to get the result. Note that if I believe Kolmogorov continuity holds for Hilbert spaces, then I get that the Holder seminorm $\vert A\vert_{C^\alpha}$ bounded in $L^m(\mathbb P)$ which together with the fact that $A$ is linear bounded operator gives the result. This is too overkill.

Context.

I am trying to prove whether the space $L(H,K)$ has martingale type 2 for Hilbert spaces $H,K$. It is known that Hilbert spaces have martingale type 2, so I was wondering whether the space of bounded linear operators also enherit the property. We recall the definition here

Definition. Let $X$ be a Banach space and $(M_n)_{n=0}^N$ be a martingale with values in $X$. We say $X$ has martingale type $p\in [1,2]$ if for some $m\in (1,\infty)$, there exists a constant $C=C(m,p,X)$ such that \begin{align*} \mathbb E[\Vert M_N\Vert_X^m]\leq C \mathbb E\left[ \left\vert \Vert M_0\Vert_X^p+ {\sum}_n^N\Vert \Delta M_n\Vert_X^p\right\vert^{m/p}\right], \end{align*} where $\Delta M_n$ is the martingale difference.

Since $H$ and $K$ both have type $2$ a first line of applying that to linear operators leads to the following problem:

Problem. Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space and $H,K$ be separable Hilbert spaces. Consider a random bounded linear operator $A:\Omega\to L(H,K)$. Assume for any integer $m\geq 2$ there exists $C=C(m,K,H)$ such that for all $0\neq x\in H$ we have \begin{align*}\mathbb{E}[\Vert Ax\Vert_K^m] \leq C \Vert x\Vert_H^m \end{align*} There exists $\Gamma=\Gamma(m,K,H)$ such that $$\mathbb E[\Vert A\Vert_{L(H,K)}^m]\leq C\Gamma.$$

The only attempt I currently have is to show the statement through Kolmogorov continuity criterion for Hilbert spaces. I do not know whether such statement exists, but I was willing to use the paper sample properties of random fields to get the result. Note that if I believe Kolmogorov continuity holds for Hilbert spaces, then I get that the Holder seminorm $\vert A\vert_{C^\alpha}$ bounded in $L^m(\mathbb P)$ which together with the fact that $A$ is linear bounded operator gives the result. This is too overkill.

Context.

I am trying to prove whether the space $L(H,K)$ has martingale type 2 for Hilbert spaces $H,K$. It is known that Hilbert spaces have martingale type 2, so I was wondering whether the space of bounded linear operators also enherit the property. We recall the definition here

Definition. Let $X$ be a Banach space and $(M_n)_{n=0}^N$ be a martingale with values in $X$. We say $X$ has martingale type $p\in [1,2]$ if for some $m\in (1,\infty)$, there exists a constant $C=C(m,p,X)$ such that \begin{align*} \mathbb E[\Vert M_N\Vert_X^m]\leq C\, \mathbb E\left[ \left\vert \Vert M_0\Vert_X^p+ {\sum}_n^N\Vert \Delta M_n\Vert_X^p\right\vert^{m/p}\right], \end{align*} where $\Delta M_n$ is the martingale difference.

Since $H$ and $K$ both have type $2$ a first line of applying that to linear operators leads to the following problem:

Problem. Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space and $H,K$ be separable Hilbert spaces. Consider a random bounded linear operator $A:\Omega\to L(H,K)$. Assume for any integer $m\geq 2$ there exists $C=C(m,K,H)$ such that for all $0\neq x\in H$ we have \begin{align*}\mathbb{E}[\Vert Ax\Vert_K^m] \leq C \Vert x\Vert_H^m \end{align*} There exists $\Gamma=\Gamma(m,K,H)$ such that $$\mathbb E[\Vert A\Vert_{L(H,K)}^m]\leq C\Gamma.$$

The only attempt I currently have is to show the statement through Kolmogorov continuity criterion for Hilbert spaces. I do not know whether such statement exists, but I was willing to use the paper sample properties of random fields to get the result. Note that if I believe Kolmogorov continuity holds for Hilbert spaces, then I get that the Holder seminorm $\vert A\vert_{C^\alpha}$ bounded in $L^m(\mathbb P)$ which together with the fact that $A$ is linear bounded operator gives the result. This is too overkill.

Context.

I am trying to prove whether the space $L(H,K)$ has martingale type 2 for Hilbert spaces $H,K$. It is known that Hilbert spaces have martingale type 2, so I was wondering whether the space of bounded linear operators also enherit the property. We recall the definition here

Definition. Let $X$ be a Banach space and $(M_n)_{n=0}^N$ be a martingale with values in $X$. We say $X$ has martingale type $p\in [1,2]$ if for some $m\in (1,\infty)$, there exists a constant $C=C(m,p,X)$ such that \begin{align*} \mathbb E[\Vert M_N\Vert_X^m]\leq C \mathbb E\left[ \left\vert \Vert M_0\Vert_X^p+ {\sum}_n^N\Vert \Delta M_n\Vert_X^p\right\vert^{m/p}\right], \end{align*} where $\Delta M_n$ is the martingale difference.

Since $H$ and $K$ both have type $2$ a first line of applying that to linear operators leads to the following problem:

Problem. Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space and $H,K$ be separable Hilbert spaces. Consider a random bounded linear operator $A:\Omega\to L(H,K)$. Assume for any integer $m\geq 2$ there exists $C=C(m,K,H)$ such that for all $0\neq x\in H$ we have \begin{align*}\mathbb{E}[\Vert Ax\Vert_K^m] \leq C \Vert x\Vert_H^m \end{align*} There exists $\Gamma=\Gamma(m,K,H)$ such that $$\mathbb E[\Vert A\Vert_{L(H,K)}^m]\leq C\Gamma.$$

The only attempt I currently have is to show the statement through Kolmogorov continuity criterion for Hilbert spaces. I do not know whether such statement exists, but I was willing to use the paper sample properties of random fields to get the result. Note that if I believe Kolmogorov continuity holds for Hilbert spaces, then I get that the Holder seminorm $\vert A\vert_{C^\alpha}$ bounded which together with the fact that $A$ is linear bounded operator gives the result. This is too overkill.

Context.

I am trying to prove whether the space $L(H,K)$ has martingale type 2 for Hilbert spaces $H,K$. It is known that Hilbert spaces have martingale type 2, so I was wondering whether the space of bounded linear operators also enherit the property. We recall the definition here

Definition. Let $X$ be a Banach space and $(M_n)_{n=0}^N$ be a martingale with values in $X$. We say $X$ has martingale type $p\in [1,2]$ if for some $m\in (1,\infty)$, there exists a constant $C=C(m,p,X)$ such that \begin{align*} \mathbb E[\Vert M_N\Vert_X^m]\leq C \mathbb E\left[ \left\vert \Vert M_0\Vert_X^p+ {\sum}_n^N\Vert \Delta M_n\Vert_X^p\right\vert^{m/p}\right], \end{align*} where $\Delta M_n$ is the martingale difference.

Since $H$ and $K$ both have type $2$ a first line of applying that to linear operators leads to the following problem:

Problem. Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space and $H,K$ be separable Hilbert spaces. Consider a random bounded linear operator $A:\Omega\to L(H,K)$. Assume for any integer $m\geq 2$ there exists $C=C(m,K,H)$ such that for all $0\neq x\in H$ we have \begin{align*}\mathbb{E}[\Vert Ax\Vert_K^m] \leq C \Vert x\Vert_H^m \end{align*} There exists $\Gamma=\Gamma(m,K,H)$ such that $$\mathbb E[\Vert A\Vert_{L(H,K)}^m]\leq C\Gamma.$$

The only attempt I currently have is to show the statement through Kolmogorov continuity criterion for Hilbert spaces. I do not know whether such statement exists, but I was willing to use the paper sample properties of random fields to get the result. Note that if I believe Kolmogorov continuity holds for Hilbert spaces, then I get that the Holder seminorm $\vert A\vert_{C^\alpha}$ bounded in $L^m(\mathbb P)$ which together with the fact that $A$ is linear bounded operator gives the result. This is too overkill.

Context.