How can you show that the trace class norm $\|A\|_1:=\mathrm{Tr}(|A|)$ satisfies the triangle inequality?

Question

Exact wording of my question is a bit oxymoronic, since a norm by definition is a metric, and thus requires proper context. Let $H$ be a separable Hilbert space over the field $\mathbb{K}$. I am aware that a bounded linear operator $A\in\mathcal{B}(H)$ is said to be of trace class if $\mathrm{Tr}(|A|) < \infty$ for $|A|$ the square root of $A^*A$ and

$$\mathrm{Tr}(A):= \sum_{k=1}^\infty\left<Ae_k, e_k\right>$$

for a Hilbert basis $\{e_k\}$ for $H$. Then, the $p$th Schatten norm for $p\in [1,\infty)$ is defined as

$$\|A\|_p := \left(\mathrm{Tr}(|A|^p)\right)^{1/p}$$

and as a special case, the trace class norm is given by

$$\|A\|_1 := \mathrm{Tr}(|A|)$$

None of the references I have seen anywhere actually prove that what we call the Schatten norm is actually a norm, and the same goes for the trace norm. The main difficulty I have with proving that Schatten norm is actually a norm is already present in the case of the trace norm, and hence this question focuses on that.

What I struggle most with showing that for any two trace class operators $A, B$ we have

$$\|A + B\|_1 = \mathrm{Tr}(|A+B|)\leq \mathrm{Tr}(|A|) + \mathrm{Tr}(|B|)$$

is that I don't know what I know/can say about the operator $|A + B|$, that is the square root of $C := (A + B)^*(A + B)$. And by this I mean that nothing comes to my mind on how I could estimate e.g. that

$$\left<|A + B|e_k, e_k\right>\leq \left<|A|e_k, e_k\right> + \left<|B|e_k, e_k\right>$$

pointwise. Some sources define both the Schatten norm and the trace norm as the $l^p$ (resp. $l^1$) norm of the singular values of an operator. But the (or any) relationship between the singular values of two bounded linear operators $A, B$ and their sum operator $C = A + B$ is an even bigger mystery to me.

So how can I conclude that the trace class norm is actually a norm by showing that it satisfies the triangle inequality? Other properties of norm are quite easy to verify after this.

Answer 1

I do not know a "nice" proof of this! Here is an outline of how it is done in the first volume of Reed and Simon (Section VI.6). If you don't already know that $\operatorname{Tr}$ does not depend on the basis that is used to compute it, this is a relatively straightforward exercise, and something I use below.

Recall that if $T = U|T|$ is the polar decomposition of the operator $T \in \mathcal{B}(H)$, then $U^* T = |T|$ (e.g. because the operator $U^* U$ is projection onto the orthocomplement of the kernel of $|T|$, and the orthocomplement of the kernel of a positive operator is the closure of its range).

With that in mind, suppose that $A+B = U|A+B|$, $A = V|A|$, and $B = W|B|$ are the polar decompositions of $A$, $B$, and $A+B$, respectively, and that $A$ and $B$ are trace class, and that $(e_n)$ is an orthonormal basis, we have (using the observation just made with $T = A + B$) $$ \sum_n \langle |A+B|e_n,e_n\rangle = \sum_n \langle U^*(A+B) e_n, e_n\rangle = \sum_n (\langle U^* V|A| e_n, e_n\rangle + \langle U^* W |B| e_n, e_n\rangle), $$ so to show that $A+B$ is trace class and prove the triangle inequality for $\|\cdot\|_1$ it would suffice to show that $\sum_n |\langle U^* V |A| e_n, e_n\rangle| \leq \|A\|_1$ whenever $A$ is trace class, $(e_n)$ is an orthonormal basis, and $U$, $V$ are partial isometries.

Turning to that goal, we have $$ \sum_n |\langle U^* V|A| e_n, e_n\rangle| = \sum_n |\langle |A|^{1/2} e_n, |A|^{1/2} V^* U e_n\rangle| \leq \sum_n \||A|^{1/2} e_n\| \cdot \||A|^{1/2} V^* U e_n\|, $$ where by the Cauchy-Schwarz inequality in $\ell^2(\mathbb{R})$ (if we know it is applicable; or, alternatively, taking the sums only over $n$ in some finite set, using Cauchy-Schwarz there, and later taking a sup only at the end) we deduce $$ \sum_n |\langle U^* V|A| e_n, e_n\rangle| \leq \left(\sum_n \||A|^{1/2} e_n\|^2 \right)^{1/2} \left(\sum_n \||A|^{1/2} V^* U e_n\|^2\right)^{1/2} $$ Note that the summand $\||A|^{1/2} e_n\|^2$ inside the first square root is $\langle |A|^{1/2} e_n, |A|^{1/2} e_n\rangle = \langle |A| e_n, e_n\rangle$, so the sum inside the first square root is at most $\operatorname{Tr} |A|$, while the summand $\||A|^{1/2} V^* U e_n\|^2$ inside the second square root is $\langle |A|^{1/2} V^* U e_n, |A|^{1/2} V^* U e_n\rangle = \langle U^* V |A| V^* U e_n, e_n\rangle$, so the sum inside the second square root is at most $\operatorname{Tr}(U^* V |A| V^* U)$, and we'd be done if we can show that $\operatorname{Tr}(U^* V |A| V^* U) \leq \operatorname{Tr} |A|$.

... And that follows from two applications of a much simpler exercise: if $A$ is trace class and $U$ is a partial isometry, then $\operatorname{Tr}(U^* |A|U) \leq \operatorname{Tr} |A|$ (just compute the trace using an orthonormal basis where each element is in $\ker U$ or $(\ker U)^{\perp}$, and use the fact that the partial isometry $U$ maps orthonormal vectors in the latter space to an orthonormal set).

I expect that at its heart, this approach is the same as the Conway approach indicated by Chad K in comments above. At least, the above could conceivably be simplified considerably if one developed or already knew basic facts about the Hilbert-Schmidt norm and/or the relation between trace class operators and Hilbert Schmidt operators (note, for example, the implicit appearance of $\| |A|^{1/2}\|_2$ and $\||A|^{1/2} V^* U\|_2$ up above).

Answer 2

A more or less direct proof depends on the inequality $$\tag1 \|AB\|_1\leq\|A\|\,\|B\|_1. $$ To show $(1)$, we note that $$\tag2 |AB|=(B^*A^*AB)^{1/2}\leq(B^*\|A\|^2B)^{1/2}=\|A\|\,|B|. $$ The proof of $(2)$ depends on three facts:

• if $T$ is positive then $STS^*$ is positive (straightforward calculation)

• $A^*A\leq \|A\|^2\, I$ (straightforward computation: $\langle A^*Ax,x\rangle\leq \|A^*Ax\|\,\|x\|\leq \|A\|^2\|x\|^2=\langle \|A\|^2x,x\rangle$)

• The square root preserves operator inequalities.

Having $(2)$ available, we get $\def\tr{\operatorname{Tr}}$ $$\tag3 \tr(|AB|)\leq \|A\|\,\tr(|B|). $$ We can then obtain$\def\abajo{\\[0.2cm]}$ $$\tag4 |\tr(A)|\leq \tr(|A|). $$ Indeed, writing the polar decomposition $A=V|A|$, \begin{align} |\tr(A)| &=\tr(V|A|)=\tr(V|A|^{1/2}|A|^{1/2})\abajo &\leq \tr(|A|^{1/2}V^*V|A|^{1/2})^{1/2}\tr(|A|)^{1/2}\tag5\abajo &\leq\tr(|A|). \end{align}

And now, using the polar decomposition,$A+B=V|A+B|$ and $(5)$, \begin{align} \|A+B\|_1 &=\tr(|A+B|) =\tr(V^*(A+B)\abajo &=\tr(V^*A+V^*B)\abajo &=\tr(V^*A)+\tr(V^*B)\abajo &\leq \tr(|A|)+\tr(|B|). \end{align}

Usually inequalities $(1)$ to $(5)$ are developed as one starts to consider the trace-norm, so by the time you want to prove the triangle inequality, they are all available. So the proof consists of just this last inequality.