Trace-norm bound for exterior powers of trace-class operators

If $n=0$, [A11] gives ${\mathrm{\Lambda }}^{0}H=\mathbb{C}$ and [A3] gives ${\mathrm{\Lambda }}^{0}T=I_{\mathbb{C}}$. Its range has the one-element ordered orthonormal basis $(1)$, so it is compact by [A14]. [A16] gives $I_{\mathbb{C}}^{\ast }=I_{\mathbb{C}}$; the identity is positive and its positive square root is itself, so [A15] gives $\mathrm{\mid }{I}_{\mathbb{C}}\mathrm{\mid }=I_{\mathbb{C}}$. Thus its only positive singular value is $1$, with the remaining sequence entries zero by [A8]. By [A6], $I_{\mathbb{C}}$ is trace class and has trace norm $1$. Now the rank-one nuclear representation $I_{\mathbb{C}}x=⟨x,1⟩1$ has scalar trace sum $⟨1,1⟩=1$, so [A10] gives $\mathrm{tr}({\mathrm{\Lambda }}^{0}T)=1$.

Henceforth let $n\ge 1$. If $J_T=\mathrm{\varnothing }$, then $T=0$ by [A4], so ${\mathrm{\Lambda }}^{n}T=0$; the singular-value product family is empty and the trace norm and bound are both zero.

Suppose $J_T\mathrm{\ne }\mathrm{\varnothing }$, put $K:=(\ker T{)}^{\perp }$, and define ${\mathcal{I}}_n:=\{(j_1,\dots ,j_n)\in J_T^n:j_1<\cdots <j_n\}$. For each $J=(j_1<\cdots <j_n)\in {\mathcal{I}}_n$ set ${\eta }_J:=e_{j_1}\wedge \cdots \wedge e_{j_n}$, ${\theta }_J:=f_{j_1}\wedge \cdots \wedge f_{j_n}$, and ${\mu }_J:={\prod }_{k=1}^n{s}_{j_k}(T)$. The Gram determinant in [A1] makes both families orthonormal.

Let $F$ be any finite subset of ${\mathcal{I}}_n$ and choose $N$ at least every index appearing in $F$. Expanding $({\sum }_{j=1}^N{s}_j(T){)}^n={\sum }_{(i_1,\dots ,i_n)\in \{1,\dots ,N{\}}^n}{s}_{i_1}(T)\cdots s_{i_n}(T)$ shows it is at least $n!{\sum }_{J\in F}{\mu }_J$, because every increasing tuple in $F$ contributes its $n!$ distinct permutations and all other terms are nonnegative. The left side is at most $\mathrm{\parallel }T{\mathrm{\parallel }}_1^n$ by [A6]. Taking the supremum over finite $F$ in [A9] proves ${\sum }_{J\in {\mathcal{I}}_n}{\mu }_J\le \mathrm{\parallel }T{\mathrm{\parallel }}_1^n\mathrm{/}n!$.

From the SVD expansion, $(e_j)$ is total in $K$: if $x\in K$ is orthogonal to every $e_j$, then $Tx=0$, so $x\in K\cap \ker T=\{0\}$. Thus finite linear combinations of $(e_j)$ are dense in $K$. Functoriality and $T=TP$ give $S:={\mathrm{\Lambda }}^{n}T=S{\mathrm{\Lambda }}^{n}P$; the Gram identity and self-adjointness of $P$ show ${\mathrm{\Lambda }}^{n}P$ is self-adjoint, while functoriality and $P^2=P$ show it is idempotent. Its range is the closed span $M$ of the ${\eta }_J$: on dense decomposable wedges, ${\mathrm{\Lambda }}^{n}P$ gives wedges of vectors in $K$, and approximating each such vector by finite linear combinations of $(e_j)$, then expanding by multilinearity, places that wedge in $M$. Continuity follows from the antisymmetrizer tensor construction in [A2]. Conversely, every ${\eta }_J$ is fixed by ${\mathrm{\Lambda }}^{n}P$. Thus ${\mathrm{\Lambda }}^{n}P$ is the orthogonal projection onto $M$ and $S$ vanishes on $M^{\perp }$.

On each basis wedge, $S{\eta }_J={\mu }_J{\theta }_J$. For each $N$ let $S_{N}x:={\sum }_{J\in {\mathcal{I}}_n,j_n\le N}{\mu }_J⟨x,{\eta }_J⟩{\theta }_J$; this is bounded and finite rank, hence compact by [A14]. By [step 2.1], both $S$ and $S_N$ vanish on $M^{\perp }$, and $({\eta }_J)$ is a complete orthonormal family in $M$, its closed span. For $x\in M$, Parseval [A13] and orthonormality of $({\theta }_J)$ give $\mathrm{\parallel }(S-S_N)x{\mathrm{\parallel }}^2={\sum }_{J:j_n>N}{\mu }_J^2\mathrm{\mid }⟨x,{\eta }_J⟩{\mathrm{\mid }}^2\le (s_{N+1}(T)s_1(T{)}^{n-1}{)}^2\mathrm{\parallel }x{\mathrm{\parallel }}^2$, because each omitted tuple has largest index greater than $N$. Decomposing a general vector into $M\oplus M^{\perp }$ therefore gives $\mathrm{\parallel }S-S_N\mathrm{\parallel }\le s_{N+1}(T)s_1(T{)}^{n-1}$ when $J_T$ is infinite. If $J_T$ is finite, $S_N=S$ for $N\ge \mathrm{\mid }{J}_T\mathrm{\mid }$. Therefore $S_N\to S$ in operator norm and [A7] makes $S$ compact. The target ${\mathrm{\Lambda }}^{n}H$ is Banach by its Hilbert construction [A2, A12].

Define $D{\eta }_J:={\mu }_J{\eta }_J$ on the orthonormal basis of $M$ and set $D=0$ on $M^{\perp }$; since $0\le {\mu }_J\le s_1(T{)}^n=\mathrm{\parallel }T{\mathrm{\parallel }}^n$, this diagonal rule extends boundedly. Its finite diagonal truncations $D_N$ (retaining only tuples with all indices at most $N$) have finite rank and are compact by [A14], and converge in norm by the coefficient estimate of [step 3.1] with output vectors ${\eta }_J$ instead of ${\theta }_J$. Thus [A7] makes $D$ compact; its real nonnegative diagonal coefficients make it positive and self-adjoint. By [step 2.1, step 3.1] and the adjoint identity [A16], for $x\in {\mathrm{\Lambda }}^{n}H$ the expansion of $Sx$ in the ${\theta }_J$ gives $⟨Sx,{\theta }_J⟩={\mu }_J⟨x,{\eta }_J⟩=⟨x,{\mu }_J{\eta }_J⟩$, hence $S^{\ast }{\theta }_J={\mu }_J{\eta }_J$ and $S^{\ast }S{\eta }_J={\mu }_J^2{\eta }_J=D^2{\eta }_J$; both operators vanish on $M^{\perp }$, so $S^{\ast }S=D^2$. Uniqueness of the compact positive square root in [A15] yields $D=\mathrm{\mid }S\mathrm{\mid }$ by the definition [A8].

By [step 4.1], the positive eigenvalues of $\mathrm{\mid }S\mathrm{\mid }$, counted with multiplicity, are exactly the values ${\mu }_J$ over $J\in {\mathcal{I}}_n$: the ${\eta }_J$ form a basis of its support $M$ and $D$ is diagonal there. By the compact-operator singular-value definition [A8], these are precisely the positive singular values of $S={\mathrm{\Lambda }}^{n}T$; the remaining entries of its singular-value sequence are zero padding.

The tuple index set ${\mathcal{I}}_n$ is countable by [A9], and [step 5.1] identifies its nonnegative family, with multiplicities, with the singular values of $S$. Thus [step 1.4] says their singular-value series converges, so $S$ is trace class by [A6] and its trace norm equals that sum. This proves the equality and factorial bound in the statement. If $n$ exceeds the finite rank of $T$, ${\mathcal{I}}_n=\mathrm{\varnothing }$ and $S=0$; if $n=1$, the tuples are single indices and the sum is exactly $\mathrm{\parallel }T{\mathrm{\parallel }}_1$.