On the isometry between bounded linear operators and the dual of nuclear linear operators Let $H$ be a separable Hilbert space. Let $(e_i)_i$ be an orthonormal basis. For any bounded linear map $T$ we write, whenever possible
$$\operatorname{tr} T := \sum_{i}^{\infty} \langle T e_i, e_i \rangle$$
Now let $L(H)$ be the set of bounded operators, $N(H)$ the set of nuclear operators. We want to show
$$i: L(H) \rightarrow N(H)' , \quad T \mapsto ( S \mapsto \operatorname{tr}(TS))$$
is an isometric isomorphism.
By the ideal property of the nuclear operators, it is easy to see that $i$ is well-defined, linear and bounded with $|i| \leq 1$.
Recall that the operators $\langle \cdot, e_i \rangle e_j$ form a basis of $N(H)$. You easily see the operator is injective (if the induced functional is zero, it's preimage must have been zero,too) and surjective (simply build up an operator $T \in L(H)$ which induces a given functional in $N(H)$).
Question:
How can I finish the proof with showing that $i$ is an isometry?
 A: I would do it as follows:


*

*For $(x,y) \in H \times H$ define an element of $N(H)$ by $x \,\tilde{\otimes}\, y = \langle \cdot , y \rangle x$ and note that $\|x \, \tilde{\otimes}\,y\|_{1} = \|x\|\,\|y\|$.

*For a functional $\varphi \in N(H)'$ define a sesquilinear form on $H$ by $B_{\varphi}(x,y) = \varphi(x \,\tilde{\otimes}\, y)$. The map $\varphi \mapsto B_{\varphi}$ is clearly linear and of norm $\leq 1$ since $|B_{\varphi}(x,y)| = |\varphi(x \, \tilde{\otimes}\,y)| \leq \|\varphi\|\,\|x\|\,\|y\|$ by 1.

*Use the version of the Riesz representation theorem stating that every bounded sesquilinear form $B$ is of the form $B(x,y) = \langle x,T y \rangle$ for a unique $T$. Moreover, $\|B\| = \|T\|$. In other words, the map $B \mapsto T$ from bounded sesquilinear forms to $L(H)$ is a linear isometry.

*Define $j(\varphi)$ by $\varphi \mapsto B_{\varphi} \mapsto T_{\varphi}$ by combining the maps from 2. and 3. Verify that $ji = 1_{L(H)}$ and $ij = 1_{N(H)'}$. Since both $i$ and $j$ have norm $\leq 1$, they must be isometries.

A: I think it helps to think of the commutative analogue of this, which is the fact that $\ell^\infty$ is the dual of $\ell^1$.  If $(a_n)_n\in\ell^\infty$, then the corresponding functional is $(x_n)_n\mapsto\sum_n a_nx_n$.  To show that the map from $\ell^\infty$ to the dual of $\ell^1$ does not decrease norm, let $\varepsilon\gt0$, take $m$ such that $|a_m|\gt \|(a_n)_n\|_\infty-\varepsilon$, and let $(x_n)_n\in\ell^1$ be such that $x_m=1$ and $x_n=0$ for $n\neq m$.  Then $|\sum_n a_nx_n|=|a_m|=|a_m|\|(x_n)_n\|_1 \geq(\|(a_n)_n\|_\infty-\varepsilon)\|(x_n)_n\|_1$. 
In the noncommutative case, you have a bounded operator $T$, and given small $\varepsilon$ you can find a unit vector $e\in H$ such that $\|Te\|\geq\|T\|-\varepsilon$, and a unit vector $f\in H$ such that $\langle Te,f\rangle = \|Te\|$.  Consider the rank one operator $S=\langle\cdot,f\rangle e$, which has norm $1$ in $N(H)$.  If my calculations are correct, the trace of TS is $\langle Te,f\rangle=\|Te\|$, showing that the norm of $i(T)$ is at least $\|T\|-\varepsilon$.
Note that on Hilbert space, nuclear operators are often called "trace class", and that a proof that $L(H)$ is the dual of the trace class operators can be found in standard texts on operator theory.  
