# Singular-value inequalities

This is my question: Is the following statement true ?

Let $H$ be a real or complex Hilbertspace and $R,S:H \to H$ compact operators. For every $n\in\mathbb{N}$ the following inequality holds:

$$\sum_{j=1}^n s_j(RS) \leq \sum_{j=1}^n s_j(R)s_j(S)$$

Note: $s_j(R)$ denotes the j-th singular value of the opeartor $R$. The sequence of the singular values falls monotonically to zero.

With best regards, mat

Edit: I found out, that the statement is true for products instead of sums. By that I mean:

Let $H$ be a $\mathbb{K}$-Hilbertspace and $R,S: H \to H$ compact operators. For every $n\in\mathbb{N}$ we have:

$$\Pi_{j=1}^n s_j(RS) \leq \Pi_{j=1}^n s_j(R)s_j(S)$$

Is it possible to derive the statement for sums from this?

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## 1 Answer

The statement is true. It is a special case of a result by Horn (On the singular values of a product of completely continuous operators, Proc. Nat.Acad. Sci. USA 36 (1950) 374-375).

The result is the following. Let $f:[0,\infty)\rightarrow \mathbb{R}$ with $f(0)=0$. If $f$ becomes convex following the substitution $x=e^t$ with $-\infty\leq t<\infty$, then for any linear completely continuous operators $R$ and $S$, $$\sum_{j=1}^n f(s_j(RS))\leq \sum_{j=1}^n f(s_j(R)s_j(S)).$$

The function $f(x)=x$ falls in the scope of the theorem and the proof follows from the theorem you stated about products.

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Thank you very much for your answer ! With best regards, Mat –  Mat Sep 24 '12 at 12:12
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