Dual space of a finite dimensional Hilbert space I have a statement given that the normed vector space $(L(\mathcal H), ||.||_{\text{1}})$ is the dual space of $(L(\mathcal H), ||.||_{\text{op}})$. Here $L(\mathcal H) := L(\mathcal H, \mathcal H)$ denotes the vector space of linear operators from a vector space $\mathcal H$ to a vector space $\mathcal H$. The $||.||_1$ norm describes the trace norm and $||.||_{\text{op}}$ denotes the operator/spectral norm.
I'm pretty new to the topic and don't know how to show that this statement holds.
 A: I will denote $\mathcal B :=  (L(\mathcal H),\|\cdot\|_{\mathrm{op}})$, $\mathcal B_1 := (L(\mathcal H),\|\cdot\|_1)$, $\mathcal B_2 := (L(\mathcal H),\|\cdot\|_2)$.
First of all, note that every operator $A$ with finite trace norm leads to a corresponding element of $\mathcal B^*$, since we can define $\phi_A(B) = \langle A,B \rangle$ (and make use of the inequality $|\langle A,B\rangle| \leq \|A\|_1 \cdot \|B\|_{\mathrm{op}}$). It remains to be see that for every $\phi \in \mathcal B^*$, there is an $A \in \mathcal B_1$ with $\phi = \phi_A$.
First, we note that restricting $\phi$ (with $\|\phi\|_{\mathrm{op}}$ finite) to the elements of $\mathcal B$ that are in $\mathcal B_2$ gives us a bounded linear map over $\mathcal B_2$ since
$$
|\phi(B)| \leq \|\phi\|_{\mathrm{op}} \|B\|_{\mathrm{op}} \leq \|\phi\|_{\mathrm{op}}\|B\|_2.
$$
Because $\mathcal B_2$ is a Hilbert space, there necessarily exists a unique $A \in \mathcal B_2$ for which $\phi = \phi_A$, but it remains to be seen that $A \in \mathcal B_1$. One argument that this holds is as follows. Let $A = U|A|$ be a polar decomposition.
For every projection operator $P$ of finite rank, we have $\|P\|_{\operatorname{op}} = 1$ and hence
$$
\|\phi_A\|_{\mathrm{op}} \geq |\phi_A(PU^*)| = \langle A,PU^*\rangle = 
\operatorname{tr}(P|A|).
$$
It follows that we must have $\|\phi_A\|_{\mathrm{op}} \geq \operatorname{tr}(|A|) = \|A\|_1$, which is what we wanted.
