What is a bounded operator on a Hilbert space that does not attain its norm? An example in $L^2$ or $l^2$ would be preferred.
All of the simple examples I have looked at (the identity operator, the shift operator) attain their respective norms.
Functional Analysis – Bounded Operator That Does Not Attain Its Norm
functional-analysishilbert-spacesoperator-theory
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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.
Hint: Consider iterated powers of the unilateral left shift operator $S\colon \ell^2(\mathbb N)\to \ell^2(\mathbb N)$ given by $$(Sa)(n)=a(n+1).$$ Can you compute $\lVert S^ka\rVert$ in terms of the entries of $a$? On the other hand, what is the operator norm of $S^k$?
Best Answer
For an example in $L^2[0,1]$, consider the operator of multiplication by $x$, i.e. $(Tf)(x) = x f(x)$.