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I have a linear operator $A\in\mathcal{L}(X,Y)$ where $X$ and $Y$ are some Banach spaces (or Hilbert spaces would also do, if that simplifies the answer.). The operator norm of $A$ is given by $$ \|A\|=\sup_{x\in X} \|Ax\|_Y/\|x\|_X. $$ Now, if the operator is of finite rank (makes things a little easier), I can view it as an element of the tensor product space $X^*\otimes Y$, where $X^*$ is the continuous dual of $X$, such that $$ A=\sum_{i=1}^n u_i^*\otimes v_i $$ with $u_i^*\in X^*$ and $v_i\in Y$. Now, there are quite a few tensor norms that I can define on $X^*\otimes Y$ in order to complete the space, e.g. the projective norm $$ \epsilon(A)=\inf\left(\sum_{i=1}^n \|u_i^*\|_{X^*} \|v_i\|_Y\right) $$ where the infimum goes over all such decompositions, and (as I take from R. Ryan, Introduction to Tensor Products of Banach Spaces) 13 more norms you can sensibly define on a tensor product space. Now my question is: is there any of these tensor norms that coincides with the operator norm? The only thing I could show is, that the projective norm if always larger or equal to the operator norm. However, that doesn't help me much, I would need equality.

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up vote 13 down vote accepted

Your question is a very natural one (if I understand it correctly) and at the same time it raises a rather difficult question.

As you say, it makes sense to identify the operators $A : X \to Y$ of finite rank with elements of $X^{\ast} \otimes Y$. Now you want the operator norm of $A$ to coincide with its norm in $X^{\ast} \otimes Y$ with respect to some tensor norm. I leave it to you to check that the injective tensor norm defined for $\omega = \sum x_{i}^{\ast} \otimes y_{i}$ by $$\Vert \omega \Vert_{\varepsilon} = \sup_{\substack{\Vert \phi \Vert_{X^{\ast\ast}} \leq 1 \\\ \Vert \psi \Vert_{Y^{\ast}} \leq 1}}{\left\vert \sum \phi(x_{i}^{\ast}) \, \psi(y_{i})\right\vert}$$ does what you want (it is independent of the representation of $\omega$ as a finite sum of elementary tensors and $\|A\| = \|A\|_{\varepsilon}$ for operators of finite rank). Edit: To see the second claim, use Goldstine's theorem that allows you to replace the supremum over $\phi \in X^{\ast\ast}$ with $\Vert\phi\Vert_{X^{\ast\ast}} \leq 1$ by the supremum over $\operatorname{ev}_{x}$ with $x \in X$ and $\Vert x \Vert_{X} \leq 1$.

The projective tensor norm of a finite rank operator is usually much larger than its operator norm (see the discussion on pp.41ff in Ryan, for example).

Given this, we can identify the completion $X^{\ast} \otimes_{\varepsilon} Y$ of $X^{\ast} \otimes Y$ with respect to the injective tensor norm with a space $K_{0}(X,Y) \subset L(X,Y)$ of operators $X \to Y$ and we will freely do so from now on. Note that $K_{0}(X,Y)$ is nothing but the closure of the operators of finite rank in $L(X,Y)$.

Now the question is: What are the operators lying in $K_{0}(X,Y)$ ?

This is really difficult and I'll outline the closest I know to an answer to that.

As a first observation note that the compact operators $K(X,Y) \subset L(X,Y)$ are a closed subspace of $L(X,Y)$ containing $X^{\ast} \otimes_{\varepsilon} Y = K_{0}(X,Y)$. Looking at the examples of Hilbert spaces or the classical Banach spaces one finds out that quite often $K(X,Y) = K_{0}(X,Y)$ holds. However, it may fail in general, and that's where the famous Approximation Property comes in. I'll refrain from delving into the numerous equivalent formulations and use it as a black box. We have the following theorem due to Grothendieck:

Theorem. The Banach space $X^{\ast}$ has the approximation property if and only if $K_{0}(X,Y) = K(X,Y)$ holds for all Banach spaces $Y$.

Edit 2: (in response to a comment of the OP) It follows that for a reflexive Banach space $X$ with the approximation property we have $K(X,Y) = X^{\ast} \otimes_{\varepsilon} Y$ for all Banach spaces $Y$.

Now most of the Banach spaces you'll run into have the approximation property, e.g. $L^{p}$, $C(X)$ and so on. However, P. Enflo (in a veritable tour de force) has shown that there exist Banach spaces failing the approximation property. An explicit example (identified by Szankowski) is the space $L(H,H)$ of a separable Hilbert space. Note that this space is the dual space of the trace class operators. A famously open question is whether the space $H^{\infty}(D)$ of bounded holomorphic functions on the open unit disk has the approximation property.

I hope this answers your question. The approximation property is discussed in detail in any book that treats the tensor products of Banach spaces. In particular, this is well treated in Ryan's book.

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Thanks a lot for that quick and exhaustive question. It just took me a while to think it through. –  Elmar Zander Apr 28 '11 at 11:04
    
I just have one question left: when I check that both norms are the same I need to replace the supremum over the bidual $X^{**}$ with the supremum over $X$ itself. On what grounds can I do that? Do I need that the canonical embedding is dense in the (?) weak-* norm? My functional analysis knowledge doesn't go that far. (Currently, the spaces I'm interested in are reflexive, but I'd really like to know that in general.) –  Elmar Zander Apr 28 '11 at 11:13
    
@Elmar: You're welcome. Concerning your question, that's precisely the way to go about it. The result you're looking for is called Goldstine's theorem and is not very difficult to prove. –  t.b. Apr 28 '11 at 11:17
    
@Elmar: I should have said that there is no such thing as a weak$^{\ast}$-norm (except in the finite-dimensional case), but density of the unit ball of $X$ in the one of $X^{\ast\ast}$ in the weak$^{\ast}$-topology is good enough. –  t.b. Apr 28 '11 at 11:19
    
@Elmar: A last remark: Note that for $X$ reflexive it follows that $X^{\ast}$ has the approximation property provided that $X$ has the approximation property. This means that if $X$ is reflexive and has the approximation property then $X^{\ast} \otimes_{\varepsilon} Y = K(X,Y)$, or in words, every compact operator $X \to Y$ where $Y$ is arbitrary can be (uniquely) identified with an element of $X^{\ast} \otimes_{\varepsilon} Y$ and vice versa. –  t.b. Apr 28 '11 at 11:33

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