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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.

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  • $\begingroup$ It seems 99% of mathematicians are oblivious to such conspicuous typographical differences at that between the following two examples: $$ C\mathbb E \left[ \vphantom{\sum_n^N} \cdots\cdots \right. $$ $$ C\operatorname{\mathbb E} \left[ \vphantom{\sum_n^N} \cdots\cdots \right. $$ This is one of the things that cause you to perceive one font as looking much better than another without noticing specifically what the differences in the design of the characters are. (The second example above is by civilized standards the correct one.) $\endgroup$ Commented Jan 31, 2023 at 18:17
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    $\begingroup$ @MichaelHardy Thank you very much! I changed it accordingly! $\endgroup$
    – Shashi
    Commented Jan 31, 2023 at 18:19
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    $\begingroup$ To be a bit more specific: The use of \operatorname results in context-dependent horizontal spacing. $\endgroup$ Commented Jan 31, 2023 at 18:20
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    $\begingroup$ (And so do such things as \cos, \log, \det, \max, etc., as opposed to \text{cos} or \mathrm{cos}, which do not result in context-dependent horizontal spacing.) $\endgroup$ Commented Jan 31, 2023 at 18:26
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    $\begingroup$ $L(H,K)$ contains an isometric copy of $\ell_\infty$ (eg the diagonal operators if $H=K=\ell_2$), and therefore of every separable Banach space. $\endgroup$ Commented Jan 31, 2023 at 19:33

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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 $p\in(1,2]$.


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 $m\ge1$.

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  • $\begingroup$ Thank you very much for the answer! $\endgroup$
    – Shashi
    Commented Feb 1, 2023 at 9:41

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