Pitt's theorem and reflexivity

Question

Does it follow from Pitt's theorem that the space of bounded operators from $\ell_2$ to $\ell_p$ ($p<2$) is actually reflexive? We have $$ \mathcal{B}(\ell_2, \ell_p) = \mathcal{K}(\ell_2, \ell_p) $$ so by standard duality also $$ \mathcal{K}(\ell_2, \ell_p)^{**} =\mathcal{B}(\ell_2, \ell_p) $$ Is this true?

Answer 1

Proof №1. In the article On the structre of the tensor products of $\ell_p$-spaces it is stated that $$ \ell_{p_1}\otimes_\pi\ldots\otimes_{\pi}\ell_{p_n} $$ is reflexive if $r(p_1,\ldots,p_n):=\min\left(1,\sum_{k=1}^n p_k^{-1}\right)^{-1}>1$.

Let $q$ be conjugate to $p$, then $q>2$ and respective $r(2,q)>1$. Hence $\ell_2\otimes_\pi \ell_q$ is refexive. Note that $$ \mathcal{K}(\ell_2,\ell_p)=\mathcal{B}(\ell_2,\ell_p)=\mathcal{B}(\ell_2,\ell_q^*)=(\ell_2\otimes_{\pi}\ell_q)^* $$ Hence $\mathcal{K}(\ell_2,\ell_p)$ is reflexive as dual of reflexive space.

Proof №2. In the article Remarks on Banach spaces of compact operators theorem 2 states that if $E$ and $F$ Banach spaces one of which have approximation property then the following are equivalent

• $\mathcal{K}(E,F)$ is reflexive
• $\mathcal{B}(E,F)$ is reflexive
• $E$ and $F$ are reflexive and $\mathcal{K}(E,F)=\mathcal{B}(E,F)$

Now take $E=\ell_2$, $F=\ell_p$. They are both reflexive. Since $E$ admits Shauder basis (any orthonormal basis will fit), then it have approximation property. By Pitt's theorem $\mathcal{K}(E,F)=\mathcal{B}(E,F)$. Hence third condition of previous theorem is satisfied, so $\mathcal{K}(E,F)$ is reflexive.

Remark One can easily see that the last approach can give more general result: if $1<p<q<\infty$ then $\mathcal{K}(\ell_p,\ell_q)$ is reflexive.