Question about norm in trace class I'm having troubles to proof the next inequality 
$$\|A\|\leq\|A\|_1$$
where A is an operator in the trace class and $\|A\|_1=Tr|A|$. And $\|A\|$ is the norm of operators.
I just got this $\|A\|\leq\||A|\|$ but tell me nothing. Can you gime some hint? Please.
 A: Although this question is two years old, I figured I'd give it a go. The way to prove this heavily depends on how you define the trace-class. There are two different ways to introduce the trace-class (which unsurpisingly turn out to be equivalent in the end). Note that my scalar product will be antilinear in the first argument as usual in physics. Everything important below obviously stays the same if the scalar product is antilinear in the second argument as usual in mathematics.


*

*Each compact operator $A$ between infinite-dimensional complex Hilbert spaces $\mathcal H,\mathcal G$ has a Schmidt decomposition $A=\sum_{n=0}^\infty s_n\langle e_n,\cdot\rangle f_n$ where $(s_n)_{n\in\mathbb N}$ is a unique decreasing null sequence in $[0,\infty)$ (called singular values) and $(e_n)_{n\in\mathbb N},(f_n)_{n\in\mathbb N}$ are orthonormal systems in $\mathcal H,\mathcal G$ respectively. Now one defines the trace class as the subset of all compact operators with $\sum_{n=0}^\infty s_n(A)<\infty$. This is a Banach space under the trace norm $\Vert A\Vert_1=\sum_{n=0}^\infty s_n(A)$. Now $s_0(A)=\Vert A\Vert$ for any compact operator as a simple consequence of the construction of the Schmidt decomposition (since the largest eigenvalue of $A^\dagger A$ has norm $\Vert A\Vert^2$). Thus $s_n\geq 0$ for all $n\in\mathbb N_0$ implies $\Vert A\Vert\leq\Vert A\Vert_1$ for all trace-class operators $A$. This whole approach can be found in the book "Introduction to Functional Analysis" (1997) by Meise & Vogt, Chapter 16.

*Another way to introduce the trace-class is as the subset of bounded operators $A$ on an infinite-dimensional complex Hilbert spaces $\mathcal H$ which satisfy $\operatorname{tr}|A|<\infty$ where $|A|=\sqrt{A^\dagger A}$. This again forms a Banach space using the trace norm $\Vert A\Vert_1=\operatorname{tr}|A|$. This approach can for example be found in the book "Methods of Modern Mathematical Physics. I: Functional Analysis" (1980) by Reed & Simon. To show $\Vert A\Vert\leq\Vert A\Vert_1$ for any trace-class operator $A$, we first need this short Lemma on positive semi-definite operators. 
Lemma. Let a bounded positive semi-definite operator $B$ be given. Then $\operatorname{tr}(B^2)\leq(\operatorname{tr}B)^2$.
Proof. Let $(e_i)_{i\in I}$ be any orthonormal basis of $\mathcal H$. Since $B\geq 0$ implies $B$ being hermitian we get
$$
\operatorname{tr}(B^2)=\sum_{i\in I}\langle B e_i,B e_i\rangle=\sum_{i\in I}\Big\langle \sum_{j\in I}\langle e_j,B e_i\rangle e_j,B e_i\Big\rangle=\sum_{i,j\in I} |\langle e_j,Be_i\rangle|^2.
$$
In the second step, we just used the expansion of $Be_i$ in the orthonormal basis $(e_j)_{j\in I}$. Now
$$
|\langle e_j,Be_i\rangle|^2=|\langle \sqrt Be_j,\sqrt Be_i\rangle|^2  \leq \langle \sqrt Be_j,\sqrt Be_j\rangle\langle \sqrt Be_i,\sqrt Be_i\rangle =  \langle e_j,Be_j\rangle\langle e_i,Be_i\rangle
$$
as a simple consequence of Cauchy-Schwarz which directly implies $\operatorname{tr}(B^2)\leq(\operatorname{tr}B)^2$. $\tag*{$\blacksquare$}$
Theorem. For any trace-class operator $A$ we have $\Vert A\Vert\leq\Vert A\Vert_1$.
Proof. Let $x\in\mathcal H\backslash\lbrace0\rbrace$ be arbitrary. Extend the orthonormal system $\lbrace\frac{x}{\Vert x\Vert}\rbrace$ to an orthonormal basis $(f_i)_{i\in I}$ of $\mathcal H$. Since $A^\dagger A\geq 0$  we get
$$
\Vert Ax\Vert^2=\Vert x\Vert^2\Big\langle \frac{x}{\Vert x\Vert},A^\dagger A\frac{x}{\Vert x\Vert}\Big\rangle \leq \Vert x\Vert^2\sum_{i\in I}\langle f_i,A^\dagger Af_i\rangle=\Vert x\Vert^2\operatorname{tr}(A^\dagger A).
$$
Applying the previous Lemma to $B=\sqrt{A^\dagger A}=|A|\geq 0$ yields
$$
\Vert Ax\Vert^2 \leq \Vert x\Vert^2\operatorname{tr}(A^\dagger A)=\Vert x\Vert^2\operatorname{tr}(|A|^2)\leq \Vert x\Vert^2(\operatorname{tr}|A|)^2=\Vert A\Vert_1^2\Vert x\Vert^2.
$$
Taking the square root implies $\Vert A\Vert\leq\Vert A\Vert_1$ by the definition of the operator norm. $\tag*{$\blacksquare$}$ Note that since the Hilbert-Schmidt norm is given by $\Vert A\Vert_2^2=\operatorname{tr}(A^\dagger A)$ we actually showed $\Vert A\Vert\leq\Vert A\Vert_2\leq\Vert A\Vert_1$ for any trace class operator.
A: Perhaps a slightly simplified version of the above argument goes like this: 
Let $A=V\left|A\right|=\sum\lambda_{k}\left|u_{k}\left\rangle \right\langle v_{k}\right|$
be a trace class operator. Then $\left\Vert A\right\Vert =\left\Vert A^{*}A\right\Vert ^{1/2}=\lambda_{1}$
and $\left\Vert A\right\Vert _{1}=\sum\lambda_{k}$, so $\left\Vert A\right\Vert \leq\left\Vert A\right\Vert _{1}$. 
