The trace class operators are the dual of the compact operators I know that the map from the trace class operators $L_1(H)$ to the dual of the compact operators $K'(H)$ given by $A \mapsto tr( \cdot A)$ is an isometric isomorphism.  Linearity is obvious by the properties of trace, and injectivity will be proven by the isometry property.  However, I need someone to please provide a proof of the surjectivity and isometry.  I found a couple proofs online but they either were above my level or provided no explanation.  Thus, I would really appreciate if you would provide the proof here so I can ask questions if I need to.  Thanks!
Note an operator is trace class if the sum of the eigenvalues of $(A^*A)^{1/2}$ converge.
 A: (this is likely not the shortest proof, but it has the advantage of being fairly explicit)
Isometry: 
$$
\|\text{tr}\,(\cdot A)\|=\sup\{|\text{tr}(KA)|:\ K\in K(H), \|K\|=1\}
$$
Using the polar decomposition $A=U|A|$, we get 
$$
|\text{tr}(KA)|=|\text{tr}(KU|A|)|\leq\|KU\|\,\text{tr}(|A|)\leq\text{tr}(|A|)=\|A\|_1,
$$
so $\text{tr}(|A|)\leq\|\text{tr}(\cdot A)\|$.
For the reverse inequality, by the compacity of $\|A|$ we can find finite-rank projections $P_n$ with $\text{tr}(P_n|A|)\to\text{tr}(|A|)$. So $\text{tr}(P_nU^*A)=\text{tr}(P_n|A|)\to\text{tr}(|A|)$, so $\text{tr}(|A|)\geq\|\text{tr}(\cdot A)\|$.
Surjectivity: Let $\varphi\in K'(H)$. Fix an orthonormal basis $\{e_j\}$ of $H$ and write $\{E_{kj}\}$ for the corresponding matrix units.  Let 
$$
A=\sum_{k,j}\varphi(E_{jk})E_{kj}.
$$
I'm not saying right now that the sums converge in any sense. Simply, that 
$$
\langle Ae_k,e_j\rangle=\varphi(E_{kj})
$$
and that we extend by linearity.
Now we need to see that this defines a bounded operator. Given two finite combinations $x=\sum_sx_se_s$, $y=\sum_ty_te_t$, 
$$
|\langle Ax,y\rangle|=\left|\sum_{s,t}x_s\bar{y_t}\langle Ae_s,e_t\rangle \right|
=\left|\sum_{s,t}x_s\bar{y_t}\,\varphi(E_{st}) \right|
=\left|\varphi\left(\sum_{s,t}x_s\bar{y_t}\,E_{st}\right) \right|\\
\leq\|\varphi\|\,\left\|\sum_{s,t}x_s\bar{y_t} E_{st} \right\|
=\|\varphi\|\,\left\|\sum_{s}x_sE_{s1}\,\left(\sum_t{y_t} E_{t1}\right)^* \right\|\\
\leq\|\varphi\|\,\left\|\sum_{s}x_sE_{s1}\right\|\,\left\|\sum_t{y_t} E_{t1} \right\|
=\|\varphi\|\,\|x\|\,\|y\|.
$$
So $A$ is bounded on the set of bounded linear combinations of elements of the basis, and thus it extends uniquely to a bounded operator in all of $H$, with $\|A\|\leq\|\varphi\|$.
For any $X\in K(H)$, 
$$
\text{tr}(AX)=\sum_{k,j}\varphi(E_{jk})\text{tr}(E_{kj}X)=\sum_{k,j}\varphi(E_{jk})\text{tr}(E_{kj}E_{jj}XE_{kk})\\
=\sum_{k,j}\varphi(E_{jk})x_{jk}\text{tr}(E_{kk})
=\sum_{k,j}\varphi(E_{jk})x_{jk}=\sum_{k,j}\varphi(E_{jk}x_{jk})\\
=\varphi(\sum_{k,j}E_{jk}x_{jk})=\varphi(X).
$$
The equality also implies, using the first part of the proof, that $\|A\|_1=\|\varphi\|$, i.e. $A$ is trace-class.
A: It's been a few years since this was posted but, for the isometry part, isn't $$||A||_1 := tr(|A|) \ \ ?$$
So wouldn't the first part actually just be that 
$$||tr(\cdot A)|| \leq ||A||_1$$
still? 
Then we'd still need to show the reverse inequality that $ ||A||_1 \leq ||tr(\cdot A)||$. I thought the idea was generally correct still though. Following from Martin's idea:
Let $\{\lambda_j\}$ be the singular values of $A$ and $\{\phi_j\}$ the associated orthonormal set. Extend it to an orthonormal basis $\{ \hat \phi_n\}$ of $H$. The finite rank projections $P_N$ are defined as
$$ P_N(\phi_n) = sgn(\lambda_n)\phi_n \ \ \forall n \leq N$$
when $\phi_n$ is one of the associated eigenvectors, and zero on the other basis vectors. This way,
$$ \lim_{N \rightarrow \infty } tr(P_N A) \rightarrow tr(|A|) = ||A||_1 $$
so in the sup we'll get $||tr(\cdot A)|| \geq ||A||_1$.
Does this argument perhaps work ?
Upon re-reading I think this is what Martin meant just with the details filled in?
