Dual of injective tensor norm is not projective tensor norm

Question

Let $A$, $B$ are two Banach space, on the algebraic tensor space $A$ $\odot$ $B$, we can define the projection（maximal） tensor norm $\gamma$ and injective（minimal） tensor norm $\lambda$. For the algebraic tensor space $A*$$\odot$$B*$, we have $\gamma*$=$\lambda$, but $\lambda$* maybe is not $\gamma$.

How to explain that the dual of injective tensor norm maybe is not projective tensor norm? Any simple counterexample?

What are the cirle with the dot, $,^,A\circ B,A\circ B,\gamma,\lambda,\gamma $ and $\lambda^$? — superAnnoyingUser, May 06 '14 at 06:30
You can see wiki here:http://en.wikipedia.org/wiki/Topological_tensor_product — Strongart, May 06 '14 at 06:49
π and ε are γ and λ, two different symbols, I follow the book Theory of operator algebra I by Takesaki. — Strongart, May 06 '14 at 06:53
You should make you questions self contained. I still don't know what those symbols mean. — superAnnoyingUser, May 06 '14 at 07:19

score 3 · Accepted Answer · edited Apr 13 '17 at 12:58

We claim that $$ (c_0\otimes_\varepsilon c_0)^*\neq c_0\otimes_\pi c_0 $$ From this answer we know that $c_0\otimes_\varepsilon c_0=c_0$, so $(c_0\otimes_\varepsilon c_0)^*=c_0^*=\ell_1$ By result of Kothe the only infinite dimensional complemented subspce of $l_1$ is $l_1$.

Now, for any Banach spaces $X$, $Y$ the space $X$ is complemened in $X\otimes_\pi Y$ (see ex 2.1 in Introduction to tensor products of Banach spaces. R. A. Ryan). Thus $c_0\otimes_\pi c_0 $ have complemented subspace isomorphic to $c_0$.

Since $c_0$ is not isomorphic to $\ell_1$ ($c_0$ is not a dual space while $\ell_1=c_0^*$), then $(c_0\otimes_\varepsilon c_0)^*$ is not isomorphic to $c_0\otimes_\pi c_0$.

It is a nice counterexample, thanks. I also want to see some theoretical explaination for this fact. — Strongart, May 06 '14 at 06:28

Dual of injective tensor norm is not projective tensor norm

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