How to prove $\|AB\|_1\leq\|A\|_2\|B\|_2$ (Trace-class and Hilbert-Schmidt operators) I'm trying to prove the next inequality
$$\|AB\|_1\leq\|A\|_2\,\|B\|_{2},$$
where $\|\cdot\|_1$ denotes the norm for trace-class operators, and $\|\cdot\|_2$ is the norm of Hilbert-Schmidt operators.
I'm stuck proving this; I've tried to use polar decomposition but I don't get any useful. In a book I've seen a proof using singular values but it uses many tools and corollaries.
Is there an easier proof of this?
 A: I don't know of an easy way around singular values. Henceforth, let $\mathcal F,\mathcal G,\mathcal H$ be arbitrary complex Hilbert spaces.
The only Lemma we need is the following (cf. Proposition 16.3 in "Introduction to Functional Analysis" by Meise & Vogt (1997)).

Lemma. Let $T\in\mathcal B(\mathcal H,\mathcal G)$ be compact. Then there exists a uniquely determined decreasing null sequence $(s_n(T))_{n\in\mathbb N_0}$ (singular values) in $[0,\infty)$ and orthonormal systems $(e_n)_{n\in\mathbb N_0}$ in $\mathcal H$ and $(f_n)_{n\in\mathbb N_0}$ in $\mathcal G$ such that
$$
T=\sum_{n=0}^\infty s_n(T)\langle e_n,\cdot\rangle f_n
$$
where the series converges in the operator norm.

Using this approach, the trace-class $\mathcal B^1(\mathcal H,\mathcal G)$ is given by all compact operators $T$ such that $\Vert T\Vert_1=\sum_{n=0}^\infty s_n(T)<\infty$. Analogously, the Hilbert-Schmidt class $\mathcal B^2(\mathcal H,\mathcal G)$ is given by all compact operators $T$ such that $\Vert T\Vert_2=\big(\sum_{n=0}^\infty s_n^2(T)\big)^{1/2}<\infty$.
In separable Hilbert spaces, this is equivalent to the usual definition of the trace-class $(\operatorname{tr}(\sqrt{T^\dagger T})<\infty)$ and Hilbert-Schmidt operators $(\sqrt{\operatorname{tr}(T^\dagger T)}<\infty)$, cf. Chapter 6.6 of "Methods of Modern Mathematical Physics. I: Functional Analysis" Reed & Simon (1980). Now we get the desired result somewhat easily.

Theorem. Let $A\in\mathcal B^2(\mathcal G,\mathcal H)$, $B\in\mathcal B^2(\mathcal F,\mathcal G)$. Then
$$
\Vert AB\Vert_1\leq\Vert A\Vert_2\Vert B\Vert_2\,.
$$

Proof. We orient ourselves towards the proof of Lemma 16.21 in the book of Meise & Vogt. Apply the above Lemma to $B$, e.g. $B=\sum_{n=0}^\infty s_n(B)\langle e_n,\cdot\rangle f_n$ and thus $AB=\sum_{n=0}^\infty s_n(B)\langle e_n,\cdot\rangle Af_n$. Note that $\Vert\langle x,\cdot\rangle y\Vert_1=\Vert x\Vert\Vert y\Vert$ for any $x\in\mathcal G,y\in\mathcal F$ (e.g. see edit below) so by the triangle inequality
$$
\Vert AB\Vert_1\leq\sum_{n=0}^\infty s_n(B)\Vert \langle e_n,\cdot\rangle Af_n\Vert_1=\sum_{n=0}^\infty s_n(B)\Vert Af_n\Vert\,.
$$
This is just the $\ell_2$ scalar product of $(s_n(B))_n$ and $(\Vert Af_n\Vert)_n$ so the Cauchy-Schwartz inequality yields
$$
\Vert AB\Vert_1\leq\Big( \sum_{n=0}^\infty s_n(B)^2 \Big)^{1/2}\Big( \sum_{n=0}^\infty \Vert Af_n\Vert^2 \Big)^{1/2}=\Vert B\Vert_2\Big( \sum_{n=0}^\infty \langle f_n,A^\dagger Af_n\rangle \Big)^{1/2}.
$$
As $(f_n)_{n\in\mathbb N_0}$ is an orthonormal system in $\mathcal F$ one has $\sum_{n=0}^\infty \langle f_n,A^\dagger Af_n\rangle \leq\operatorname{tr}(A^\dagger A)$ as all summands are non-negative (this can be seen by extending  $(f_n)_{n\in\mathbb N_0}$ to an orthonormal basis of $\mathcal F$ with respect to which one then computes the trace). In total we get
$$
\Vert AB\Vert_1\leq\Vert B\Vert_2\Big( \sum_{n=0}^\infty \langle f_n,A^\dagger Af_n\rangle \Big)^{1/2}\leq(\operatorname{tr}(A^\dagger A))^{1/2}\Vert B\Vert_2=\Vert A\Vert_2\Vert B\Vert_2.\tag*{$\square$}
$$

Edit: To see $\Vert\langle x,\cdot\rangle y\Vert_1=\Vert x\Vert\Vert y\Vert$, consider
$$
T:=\langle x,\cdot\rangle y=\| x\|\| y\|\Big\langle\frac{x}{\| x\|},\cdot\Big\rangle\frac{y}{\| y\|}
$$
where the latter is a Schmidt decomposition of $T$ (as $\frac{x}{\| x\|}$ and $\frac{y}{\| y\|}$ obviously are orthonormal systems and $\| x\| \| y\|\geq 0$). Now
$$
\Vert\langle x,\cdot\rangle y\Vert=\Vert T\Vert_1=\sum_{n} s_n(T)=\Vert x\Vert\Vert y\Vert
$$
as $T$ only has one non-vanishing singular value. Alternatively, one can just compute
$$
\operatorname{tr}(\sqrt{(\langle y,\cdot\rangle x)(\langle x,\cdot\rangle y)})=\| y\|\operatorname{tr}(\sqrt{\langle x,\cdot\rangle x})=\Vert x\Vert\| y\|\underbrace{\operatorname{tr}\Big(\Big\langle \frac{x}{\Vert x\Vert},\cdot\Big\rangle \frac{x}{\Vert x\Vert}\Big)}_{=1}=\| x\|\| y\|.
$$
