Skip to main content
deleted 67 characters in body
Source Link
ABB
  • 4.1k
  • 1
  • 11
  • 19

Let $A$ be a unital *-ring. Let us put, $A^+=\{\sum_1^N x^*_ix_i: x_i\in A , N\in \mathbb{N} \}$. We write $x\geq0$ if $x\in A^+$.

Suppose that

1- For for every $n\in \mathbb{N}$ and $a\in A$, there exsits $x\in A$ with $a=nx$.

2- For every $x\in A$ $$ a\geq0 \Rightarrow x^*ax\geq0.$$

Let us assume for natural numbers $m,n$ we have $mx^*x-n\geq0$. Can we conclude that $mxx^*-n\geq0$ as well?

Let $A$ be a unital *-ring. Let us put, $A^+=\{\sum_1^N x^*_ix_i: x_i\in A , N\in \mathbb{N} \}$. We write $x\geq0$ if $x\in A^+$.

Suppose that

1- For every $n\in \mathbb{N}$ and $a\in A$, there exsits $x\in A$ with $a=nx$.

2- For every $x\in A$ $$ a\geq0 \Rightarrow x^*ax\geq0.$$

Let us assume for natural numbers $m,n$ we have $mx^*x-n\geq0$. Can we conclude that $mxx^*-n\geq0$ as well?

Let $A$ be a unital *-ring. Let us put, $A^+=\{\sum_1^N x^*_ix_i: x_i\in A , N\in \mathbb{N} \}$. We write $x\geq0$ if $x\in A^+$.

Suppose that for every $n\in \mathbb{N}$ and $a\in A$, there exsits $x\in A$ with $a=nx$.

Let us assume for natural numbers $m,n$ we have $mx^*x-n\geq0$. Can we conclude that $mxx^*-n\geq0$ as well?

Source Link
ABB
  • 4.1k
  • 1
  • 11
  • 19

Contraction elements in unital *-rings

Let $A$ be a unital *-ring. Let us put, $A^+=\{\sum_1^N x^*_ix_i: x_i\in A , N\in \mathbb{N} \}$. We write $x\geq0$ if $x\in A^+$.

Suppose that

1- For every $n\in \mathbb{N}$ and $a\in A$, there exsits $x\in A$ with $a=nx$.

2- For every $x\in A$ $$ a\geq0 \Rightarrow x^*ax\geq0.$$

Let us assume for natural numbers $m,n$ we have $mx^*x-n\geq0$. Can we conclude that $mxx^*-n\geq0$ as well?