The space of linear operators between Hilbert spaces has martingale type 2

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 \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.

Answer 1

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 $m\ge1$.