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](https://digitalcommons.lsu.edu/cgi/viewcontent.cgi?article=1112&context=cosa) 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.**



The space of linear operators between Hilbert spaces has martingale type 2
As noted by Mikael de la Salle, $L(H,K)$ contains an isometric copy of $\ell^\infty$, which is not of martingale type $p$ for any $p\in(1,2]$, and hence $L(H,K)$ is not of martingale type $p$ for any $p\in(1,2]$.




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The answer to your highlighted **Problem** is also negative. Indeed, suppose that $H=K=\ell^2$ and $A$ is a random linear operator such that $P(A=A\_k)=p\_k$ for integers $k\ge2$, where
$$A\_k e\_j=b\_j\,1(j=k)$$
for $j=1,2,\dots$, the $e\_j$'s are the standard basis vectors of $\ell^2$, $b\_j:=\ln j$, $p\_k=\frac c{k\ln^2k}\,1(k\ge2)$, and $c:=1/\sum\_{k\ge2}\frac 1{k\ln^2k}$, so that $\sum\_{k\ge2}p\_k=1$.


Then for any $x=(x\_1,x\_2,\dots)\in H$ and any integer $m\ge2$
$$E\|Ax\|^m=\sum\_{k\ge2}p\_k\|A\_kx\|^m=\sum\_{k\ge2}p\_k b\_k^m |x\_k|^m
\le C\_m\|x\|^m,$$
where
$$C\_m=\max\_{k\ge2}p\_k b\_k^m=\max\_{k\ge2}\frac c{k\ln^{2-m}k}<\infty.$$
On the other hand, $\|A\_k\|=b\_k$ and hence
$$E\|A\|^m=\sum\_{k\ge2}p\_k\|A\_k\|^m=\sum\_{k\ge2}p\_k b\_k^m
=\sum\_{k\ge2}\frac c{k\ln^{2-m}k}=\infty$$
for all $m\ge1$.


