Show that if $\mu,\eta\in I\implies \dfrac{\mu+\eta}{2}\in I$ and $\{0,1,1/2\}\subset I$ implies $[0,1]\subset I$

Question

Let $R_1,R_2,S_1,S_2,T_1,T_2$ be positive operators such that $[R_1,R_2]=[S_1,S_2]=[T_1,T_2]=0$, and $R_1\ge S_1+T_1$ $,R_2\ge S_2+T_2$ then for all $0\le t\le 1$, \begin{align} R_1^tR_2^{1-t}\ge S_1^tS_2^{1-t}+T_1^tT_2^{1-t}\tag{A6.8}\label{A6.8} \end{align} is true as a matrix inequality.

This is given as a Lemma in proving Lieb's theorem, please check my reference, Equation A6.8, Appendix 6: Proof of Lieb’s theorem, Page 646, Quantum Computation and Quantum Information by Nielsen and Chuang.

---

Let $I$ be the set of all $t$ such that the equation $R_1^tR_2^{1-t}\ge S_1^tS_2^{1-t}+T_1^tT_2^{1-t}$ holds. Then we can easily prove in few steps that, $0,1/2,1$ are elements of $I$. ie., $$ R_1\ge S_1+T_1\\\\ R_2\ge S_2+T_2\\\\ R_1^{1/2}R_2^{1/2}\ge S_1^{1/2}S_2^{1/2}+T_1^{1/2}T_2^{1/2} $$

Having done that, suppose $\mu$ and $\eta$ are any two elements of $I$ so that $$ R_1^\mu R_2^{1-\mu}\ge S_1^\mu S_2^{1-\mu}+T_1^\mu T_2^{1-\mu}\\ R_1^\eta R_2^{1-\eta}\ge S_1^\eta S_2^{1-\eta}+T_1^\eta T_2^{1-\eta}\\ $$ where $R_1^\mu R_2^{1-\mu},S_1^\mu S_2^{1-\mu},T_1^\mu T_2^{1-\mu},R_1^\eta R_2^{1-\eta},S_1^\eta S_2^{1-\eta},T_1^\eta T_2^{1-\eta}$ are positive operators such that $[R_1^\mu R_2^{1-\mu},R_1^\eta R_2^{1-\eta}]=[S_1^\mu S_2^{1-\mu},S_1^\eta S_2^{1-\eta}]=[T_1^\mu T_2^{1-\mu},T_1^\eta T_2^{1-\eta}]=0$

We have already proven the $t=1/2$ case, therefore, $$ (R_1^\mu R_2^{1-\mu})^{1/2}(R_1^\eta R_2^{1-\eta})^{1/2}\ge (S_1^\mu S_2^{1-\mu})^{1/2}(S_1^\eta S_2^{1-\eta})^{1/2}+(T_1^\mu T_2^{1-\mu})^{1/2}(T_1^\eta T_2^{1-\eta})^{1/2} $$ Using the commutativity assumptions $[R_1,R_2]=[S_1,S_2]=[T_1,T_2]=0$ and $\nu=\dfrac{\mu+\eta}{2}$, $$ R_1^\nu R_2^{1-\nu}\ge S_1^\nu S_2^{1-\nu}+T_1^\nu T_2^{1-\nu} $$ $\implies $ whenever $\mu$ and $\eta$ are in $I$, so is $\dfrac{\mu+\eta}{2}$.

---

This much is fine. But now it went on to say that,

Since $0$ and $1$ are in $I$, it is easy to see that any number $t$ between $0$ and $1$ with a finite binary expansion must be in $I$. Thus $I$ is dense in $[0, 1]$. The result now follows from the continuity in $t$ of the conclusion, (\ref{A6.8}).

How do I justify this?

---

Using the argument whenever $\mu$ and $\eta$ are in $I$, so is $\dfrac{\mu+\eta}{2}$, we can easily show that $1/2,1/4,1/8,\cdots,1/2^n\in I$, similarly infinite number of other elements in $I$, but how do I prove that $t$ spans the entire range $[0,1]$?

What about irrational numbers in the range $[0,1]$?

Answer 1

Part 1

A number $t_n\in[0,1]$ with finite binary expansion can be written as \begin{align} t_n&=0.\phi_1\phi_2\phi_3\cdots\phi_n=\frac{\phi_1}{2^1}+\frac{\phi_2}{2^2}+\frac{\phi_3}{2^3}+\cdots+\frac{\phi_n}{2^n}\\ &=\frac{1}{2}\Bigg[\phi_1+\Bigg(\frac{\phi_2}{2^1}+\frac{\phi_3}{2^2}+\frac{\phi_4}{2^3}+\cdots+\frac{\phi_n}{2^{n-1}}\Bigg)\Bigg]\\ &=\frac{1}{2}\Bigg[\phi_1+\frac{1}{2}\Bigg(\phi_2+\frac{\phi_3}{2^1}+\frac{\phi_4}{2^2}+\frac{\phi_5}{2^3}+\cdots+\frac{\phi_n}{2^{n-2}}\Bigg)\Bigg]\\ &=\frac{1}{2}\Bigg[\phi_1+\frac{1}{2}\Bigg(\phi_2+\frac{1}{2}\bigg(\phi_3+\frac{\phi_4}{2^1}+\frac{\phi_5}{2^2}+\frac{\phi_6}{2^3}+\cdots+\frac{\phi_n}{2^{n-3}}\bigg)\Bigg)\Bigg]\\ &=\frac{1}{2}\Bigg[\phi_1+\frac{1}{2}\Bigg(\phi_2+\frac{1}{2}\bigg(\phi_3+\cdots\phi_{n-3}+\Big(\frac{1}{2}\big(\phi_{n-2}+\frac{1}{2}(\phi_{n-1}+\frac{\phi_n}{2^1})\big)\Big)\bigg)\Bigg)\bigg]\\ \end{align}

where $\phi_1,\phi_2,\phi_3,\cdots,\phi_n\in\{0,1\}$

Looking at the inner most term, $\frac{1}{2}(\phi_{n-1}+\frac{\phi_n}{2^1})$ in which $\phi_{n-1}=0$ or $1$ and $\phi_n/2=0$ or $1/2$, ie., $\phi_{n-1},\phi_n/2\in I$, and therefore $\frac{1}{2}(\phi_{n-1}+\frac{\phi_n}{2^1})\in I$. In the next term $\frac{1}{2}\big(\phi_{n-2}+\frac{1}{2}(\phi_{n-1}+\frac{\phi_n}{2^1})\big)$ has $\frac{1}{2}(\phi_{n-1}+\frac{\phi_n}{2^1})\in I$ (as shown in the last step) and $\phi_{n-2}\in I$, and therefore $\frac{1}{2}\big(\phi_{n-2}+\frac{1}{2}(\phi_{n-1}+\frac{\phi_n}{2^1})\big)\in I$. Continuing this way we can prove that any number $t_n=0.\phi_1\phi_2\phi_3\cdots\phi_n=\frac{\phi_1}{2^1}+\frac{\phi_2}{2^2}+\frac{\phi_3}{2^3}+\cdots+\frac{\phi_n}{2^n}=\frac{k}{2^n}$ with finite binary expansion (dyadic rationals) is a member of I.

---

Definition: Let $A\subset\mathbb{R}$ be a set that is bounded above. Then a number $\alpha\in\mathbb{R}$ is called the supremum (least upper bound) of A, and is denoted by $\sup A$ iff:

• $x\le\alpha$ for all $x\in A$
• if $y$ is a upper bound for $A$, then $y\ge \alpha$

The Completeness Axiom: Every nonempty subset $A$ of $\mathbb{R}$ that is bounded above has a least upper bound. That is, $\sup A$ exists and is a real number.

Definition: Let $A\subset\mathbb{R}$ be a set that is bounded below. Then a number $\beta\in\mathbb{R}$ is called the infimum (greatest lower bound) of A, and is denoted by $\inf A$ iff:

• $x\ge\beta$ for all $x\in A$
• if $y$ is a lower bound for $A$, then $y\le \beta$

The Archimedean property states that for any $x\in\mathbb{R}$ there exists $n\in\mathbb{N}$ such that $n>x$.

Equivalently, for any $x\in\mathbb{R}$ with $x>0$ there exists $n\in\mathbb{N}$ such that $1/n<x$.

Proof:

Let $x\in\mathbb{R}$. Suppose the converse is true, ie., there doesn't exists $n\in\mathbb{N}$ such that $n>x$. That means, $n\le x$ for all $n\in\mathbb{N}$. \begin{align} &\implies \mathbb{N}\subset\mathbb{R}\text{ is bounded above by }x\\ &\implies \mathbb{N}\text{ has a least upper bound}, \alpha=\sup(\mathbb{N})\in\mathbb{R} \text{ (by the Completeness Axiom)}\\ &\implies \alpha-1<\alpha,\text{ therefore } \alpha-1\text{ is not an upper bound for }\mathbb{N}\\ &\implies \text{There exists }m\in\mathbb{N}\text{ such that } \alpha-1<m\le\alpha\\ &\implies \alpha<m+1\in\mathbb{N} \end{align} which contradicts the fact that $\alpha$ is an upper bound of $\mathbb{N}$.

Lemma 1: For $\delta>0$ there exists $n\in\mathbb{N}$ such that $\frac{1}{2^n}<\delta$

Proof:

Substitute $x=\delta$ in the Archimedean Property, obtains $\frac{1}{2^n}<\frac{1}{n}<\delta$

---

Part 2

$X\subset Y$ is dense in $Y$ if for any $\delta>0$ and $y\in Y$, there is some $x\in X$ such that $|y-x|<\delta$.

We have to prove that the set $\{\frac{k}{2^n}:k\in\mathbb{W},n\in\mathbb{N},0\le k\le 2^n\}$ of dyadic rationals, ie., numbers with binary expansion in $[0,1]$, are dense in $[0,1]$.

Let's take any $\delta>0$ then from lemma 1 there exists $n\in\mathbb{N}$ such that $\frac{1}{2^n}<\delta$. Let $x\in[0,1]$ and $k=\lfloor x.2^n\rfloor$ correspond to the greatest integer less than or equal to $x.2^n$ such that $0<k\le 2^n$. \begin{align} &k=\lfloor x.2^n\rfloor\le x.2^n\le k+1=\lfloor x.2^n\rfloor+1\\ &\implies \frac{k}{2^n}\le x\le \frac{k+1}{2^n}\\ &\implies 0\le x-\frac{k}{2^n}\le \frac{1}{2^n}<\delta \quad\Big(\text{Lemma 1}\Big) \end{align} $\therefore$ for any $\delta>0$ and $x\in[0,1]$ there is some $\frac{k}{2^n}\in\{\frac{k}{2^n}:k\in\mathbb{W},n\in\mathbb{N},0\le k\le 2^n\}$ such that $|x-\frac{k}{2^n}|<\delta$

$\implies$ The set $\{\frac{k}{2^n}:k\in\mathbb{W},n\in\mathbb{N},0\le k\le 2^n\}$ is dense in $[0,1]$

$\implies$ $I$ is dense in $[0,1]$

Part 1 proves that dyadic rationals are members of $I$, and in part 2 we have proven that the dyadic rationals are dense in $[0,1]$, ie., we can approach any $t\in[0,1]$ by a sequence in our already established $I$ which contains the dyadic rationals. Therefore, the validity of the equations for $t\in[0,1]$ is proven.