Let $X$ be a Banach space. A bounded linear map $u:X\to\ell_2$ is said to be $1$-summing if for all finite sequence $(x_i)\subseteq X$ there is a constant $C>0$ such that $\sum\|ux_i\|\leq C\sup\Big\{\sum|x^*(x_i)|_2:\|x^*\|_{X^*}\leq 1\Big\}.$ A Banach space is said to be satisfy Grothendieck's theorem (in short G. T. space) if any operator $u:X\to\ell_2$ is $1$-summing. My question is the following. Does there exist a G. T. space space whose dual is not a G. T. space?

## 1 Answer

The answer is yes (there exists a GT space whose dual is not a GT space), given by the very first test example that one might consider.

The form of Grothendieck's theorem that gives rise to the terminology "GT-space" is the fact that every bounded operator from $\ell_1$ to $\ell_2$ is 1-summing.

So the first test for your question would be:

is every bounded operator $\ell_\infty \to\ell_2$ 1-summing?

Now recall that every separable Banach space embeds isomorphically into $\ell_\infty$ by a Hahn-Banach argument (I think this embedding can be made isometric, but we dont need that). Also recall that the p-summing norm defines an operator ideal: if $S:X\to Y$ is p-summing then so is $RST:W\to Z$ for any bounded operators $T:W\to X$ and $R:Y\to Z$.

Consequently: if every bounded operator $\ell_\infty\to \ell_2$ is p-summing, for some $1\leq p<\infty$, then every $X\to \ell_2$ is p-summing for any choice of separable Banach space $X$. In particular the identity operator on $\ell_2$ would be p-summing. But it is known that on any infinite-dimensional Banach space the identity operator cannot be p-summing for any $1\leq p<\infty$.