Assume $1\leq p <\infty$. It is known that for all $q>p$ $$ S^{p} \subset S^{p,\infty} \subset S^{q}. $$ Out by curioisity, it is true that $\cap_{q>p} S^{q}=S^{p,\infty}$ ?
Recall that $$ S^{p,\infty} =\Big\{T \in S^\infty : s_n(T)=O_{n \to \infty}\Big(n^{-\frac{1}{p}}\Big) \Big\}. $$
No. The diagonal operator with $\ln(n)n^{-\frac{1}{p}}$ along the diagonal, for example, is in $S^q$ for all $q > p$ but not in $S^{p, \infty}$.