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.
(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.