Proof. We only have to prove the implication $(i)\Rightarrow (ii)$, the other one is obvious. Let $x'=\alpha x$ and $y'=\beta y$. Then $(i)$ takes the form $\|x'+y'\|=\|x'\|+\|y'\|$ so, by Lemma 2.1, $\|{1}/{|\alpha |}x'+({1}/{|\beta |})y'\|={1}/{|\alpha |}\|x'\|+{1}/{|\beta |}\|y'\|$. In other words,

\[ \left\|\frac{\alpha}{|\alpha|}x+\frac{\beta}{|\beta|}y\right\|=\|x\|+\|y\|. \]

Proof. We only prove $(v)\Rightarrow (i)$, the rest can be concluded from [Reference Kittaneh13, Proposition 3.3], [Reference Arambašić and Rajić2, Theorem 2.1], and [Reference Nakamoto and Takahasi18, Theorem 1]). To this end, it is enough to show that if $a$ and $b$ are two positive elements of a $C^{*}$-algebra $\mathcal {A}$, and $\|a\|+\|b\|\in V(a+b)$, then $\|a+b\|=\|a\|+\|b\|$:

Let $\varphi \in S(\mathcal {A})$ be such that $\varphi (a+b)=\|a\|+\|b\|$. Then

\[ \|a\|+\|b\|=\varphi(a+b)\le\|a+b\|. \]

We deduce immediately that $\|a+b\|=\|a\|+\|b\|$, as required.

Proof. By Proposition 2.4, condition $(ii)$ is equivalent to $\||x|^{2}+|y|^{2}\|=\|x\|^{2}+\|y\|^{2}$. Similarly, condition $(iii)$ can be replaced by:

\[ 2\||x|^{2}+|y|^{2}\|=\||x+y|^{2}+|x-y|^{2}\|=\|x+y\|^{2}+\|x-y\|^{2}. \]

The conclusion then follows easily.

Proof. It has been indicated in [Reference Lance14, p. 37] that, for any given state $\varphi$ of $\mathcal {A}$ and $x\in \mathscr {E}$ with $\varphi (|x|^{2})=1$, the map $s\mapsto \varphi (\langle {x,sx}\rangle )$ is a state of $\mathcal {L}(\mathscr {E})$. In addition, for any adjointable operator $s$ on $\mathscr {E}$,

(2)\begin{equation} \|s\|^{2}=\sup_{\phi(|x|^{2})=1}\varphi(|sx|^{2}). \end{equation}

$(i)\Rightarrow (ii)$. Let us consider, in view of (2), a sequence $(\varphi _n)_{n\ge 0}$ of states on $\mathcal {A}$ and a sequence $(x_n)_{n\ge 0}$ of elements in $\mathscr {E}$ such that $\varphi _n(|x_n|^{2})=1,\ n\ge 0$ and

\[ \varphi_n(|(s+t)x_n|^{2})\xrightarrow{n\to\infty}\|s+t\|^{2}. \]

We note that, for any $n\ge 0$,

\begin{align*} \varphi_n(|(s+t)x_n|^{2})& =\varphi_n(|sx_n|^{2})+\varphi_n(|tx_n|^{2})+\varphi_n(\langle{sx_n,tx_n}\rangle)+\varphi_n(\langle{tx_n,sx_n}\rangle)\\ & \le\|s\|^{2}+\|t\|^{2}+\|s^{*}t\|+\|t^{*}s\|\\ & \le(\|s\|+\|t\|)^{2}. \end{align*}

We pass to limit (as $n\to \infty$) to deduce, by $(i)$, that

\[ \varphi_n(\langle{sx_n,tx_n}\rangle)\xrightarrow{n\to\infty}\|s\|\|t\|, \]

which proves $(ii)$.

The implication $(ii)\Rightarrow (iii)$ is obvious.

$(iii)\Rightarrow (i)$. Let $(\varphi _n)_{n\ge 0}$, and let $(x_n)_{n\ge 0}$ be sequences as in $(iii)$. By passing to limit (as $n\to \infty$) in the inequalities

\[ |\varphi_n(\langle{sx_n,tx_n}\rangle)|\le\varphi_n(|sx_n|^{2})^{1/2}\varphi_n(|tx_n|^{2})^{1/2}\le\|s\|\|t\|\varphi_n(|x_n|^{2})\le\|s\|\|t\|,\quad n\ge 0, \]

we obtain that

\[ \varphi_n(|sx_n|^{2})\xrightarrow{n\to\infty}\|s\|^{2}\text{ and }\varphi_n(|tx_n|^{2})\xrightarrow{n\to\infty}\|t\|^{2}. \]

Hence,

\[ \varphi_n(|(s+t)x_n|^{2})=\varphi_n(|sx_n|^{2})+\varphi_n(|tx_n|^{2})+2\Re\varphi_n(\langle{sx_n,tx_n}\rangle)\xrightarrow{n\to\infty}(\|s\|+\|t\|)^{2}. \]

Letting again $n\to \infty$ in the inequalities $\varphi _n(|(s+t)x_n|^{2})\le \|s+t\|^{2}\le (\|s\|+\|t\|)^{2},\ n\ge 0$ we finally get the triangle ‘equality’ in $(i)$.

Proof. Let us firstly note that

(3)\begin{equation} \begin{aligned} \|x+y\|^{2} & =\||x|^{2}+2\Re(\langle{x,y}\rangle)+|y|^{2}\|\\ & \le\||x|^{2}+|y|^{2}\|\le\|x\|^{2}+\|y\|^{2}. \end{aligned} \end{equation}

If $(i)$ holds true, then the inequalities in (3) become equalities. Also, by [Reference Kittaneh13, Proposition 3.3], the triangle equality $\||x|^{2}+|y|^{2}\|=\|x\|^{2}+\|y\|^{2}$ can be written in the form $\||x||y|\|=\|x\|\|y\|$, which is exactly the last condition of $(ii)$. The converse follows the same path.

Proof. The implication $(i)\Rightarrow (ii)$ follows by [Reference Murphy17, Theorem 3.3.6].

$(ii)\Rightarrow (i)$. Conversely, if $\varphi$ is a state on $\mathcal {A}$ as in $(ii)$, then, by (3),

\[ \|x+y\|^{2}\le\|x\|^{2}+\|y\|^{2}=\varphi(|x+y|^{2})\le\||x+y|^{2}\|=\|x+y\|^{2}. \]

Consequently, $(i)$ holds true.

$(iii)\Rightarrow (ii)$. Let $\varphi$ be a state on $\mathcal {A}$ such that

\[ \varphi(|x|^{2})=\|x\|^{2},\quad \varphi(|y|^{2})=\|y\|^{2}\text{ and }\varphi(\Re(\langle{x,y}\rangle))=0. \]

Then

\[ \varphi(|x+y|^{2})=\varphi(|x|^{2})+2\varphi(\Re(\langle{x,y}\rangle))+\varphi(|y|^{2})=\|x\|^{2}+\|y\|^{2}. \]

$(ii)\Rightarrow (iii)$. Conversely, let $\varphi \in S(\mathcal {A})$ be a state which satisfies condition $(ii)$. Then

\begin{align*} \|x\|^{2}+\|y\|^{2} & =\varphi(|x|^{2})+2\varphi(\Re(\langle{x,y}\rangle))+\varphi(|y|^{2})\\ & \le\|x\|^{2}+2\varphi(\Re(\langle{x,y}\rangle))+\varphi(|y|^{2})\\ & \le\|x\|^{2}+\varphi(|y|^{2})\\ & \le\|x\|^{2}+\|y\|^{2}. \end{align*}

Thus, $\varphi (|x|^{2})=\|x\|^{2}$, $\varphi (|y|^{2})=\|y\|^{2}$ and $\varphi (\Re (\langle {x,y}\rangle ))=0$. The statement $(iii)$ is proved.

Proof. Under the assumption $\Re (\langle {x,y}\rangle )=0$, condition $(i)$ takes the form $\||x|^{2}+|y|^{2}\|=\|x\|^{2}+\|y\|^{2}$. The equivalences between $(i),\ (ii),$ and $(iii)$ are deduced from Proposition 2.4. By the same result, the statements are also equivalent to $S_{|x|^{2}}(\mathcal {A})\cap S_{|y|^{2}}(\mathcal {A})\ne \emptyset$. So $(iv)$ implies $(i)$. Finally, if $\|x+y\|^{2}=\|x\|^{2}+\|y\|^{2}$ (condition $(i)$ holds true), then, for any $\varphi \in S_{|x+y|^{2}}(\mathcal {A})$,

\begin{align*} \|x+y\|^{2}& =\varphi(|x+y|^{2})\\ & =\varphi(|x|^{2})+\varphi(|y|^{2})\\ & \le\|x\|^{2}+\|y\|^{2}. \end{align*}

Hence, $\varphi \in S_{|x|^{2}}(\mathcal {A})\cap S_{|y|^{2}}(\mathcal {A})$. In other words, $S_{|x+y|^{2}}(\mathcal {A})\subseteq S_{|x|^{2}}(\mathcal {A})\cap S_{|y|^{2}}(\mathcal {A})$. The converse inclusion is obvious, so the statement $(iv)$ is verified.

Proof. The equivalence $(i)\Leftrightarrow (ii)$ has been obtained by Arambašić and Rajić in [Reference Arambašić and Rajić3, Theorem 2.7] (see also [Reference Bhattacharyya and Grover8, Theorem 4.4]).

Clearly, $(ii)$ is a consequence of $(iii)$. Conversely, if $\varphi$ is a state of $\mathcal {A}$ which verifies $\varphi (|x|^{2})=\|x\|^{2}$ and $\varphi (\langle {x,y}\rangle )=0$ (by $(i)$), then, for every $\lambda \in \mathbb {C}$,

\begin{align*} \|x+\lambda y\|^{2}& \ge\varphi(|x+\lambda y|^{2})\\ & =\varphi(|x|^{2})+\bar{\lambda}\varphi(\langle{x,y}\rangle)+\lambda\varphi(\langle{y,x}\rangle)+|\lambda|^{2}\varphi({|y|^{2}})\\ & =\|x\|^{2}+|\lambda|^{2}\varphi({|y|^{2}})\\ & \ge\|x\|^{2}+|\lambda|^{2}m(|y|^{2}). \end{align*}

Condition $(iii)$ is proved.

Proof. We firstly observe that, since $\lim _{|\alpha |\to \infty }\|x+\alpha y\|=\infty$,

\[ \inf\{\|x+\alpha y\|:\alpha\in\mathbb{C}\}=\inf\{\|x+\alpha y\|:|\alpha|\le\delta\} \]

for a certain $\delta >0$. In addition, as the map $\alpha \mapsto \|x+\alpha y\|$ is continuous on the compact set $\{|\alpha |\le \delta \}$, it attains its minimum at some point $\alpha _0\in \mathbb {C}$, that is, $\|x+\alpha _0y+\lambda y\|\ge \|x+\alpha _0y\|$ for every $\lambda \in \mathbb {C}$. Formula (5) then follows by Theorem 4.6 $(iii)$. If, for some $\alpha _1\in \mathbb {C}$,

\[ \|x+\alpha_1y+\lambda y\|^{2}\ge\|x+\alpha_1y\|^{2}+|\lambda|^{2}m(|y|^{2}),\quad\lambda\in\mathbb{C} \]

then, by taking $\lambda =\alpha _0-\alpha _1$, we obtain

\[ m(|y|^{2})|\alpha_1-\alpha_0|^{2}\le\|x+\alpha_0y\|^{2}-\|x+\alpha_1y\|^{2}\le 0, \]

so, $\alpha _1=\alpha _0$. The final statement is a consequence, in view of the uniqueness of $\alpha _0$, of Theorem 4.6 ($(i)\Leftrightarrow (iii$)).

Proof. The direct implication is obvious. Conversely, if $(ii)$ holds true, then, by Theorem 4.6 $(iii)$,

\begin{align*} \|x+\lambda y\|^{2}& \ge\|x\|^{2}+|\lambda|^{2}m(|y|^{2})\\ & =\|x\|^{2}+\alpha|\lambda|^{2}\\ & =\|x\|^{2}+|\lambda|^{2}\|y\|^{2}, \quad\lambda\in\mathbb{C}. \end{align*}

In view of the parallelogram law, the inequalities above become equalities. Hence $x\perp _P y$, as required.

Proof. We may assume, without loss of generality, that $\alpha =1$ ($B$ can be replaced by $\left({1}/{\alpha}\right)B$, if necessary).

Let us now observe that, for every $n\ge 0$, the following inequalities hold true:

\begin{align*} M_{A,A+B}(\xi_n)& =\|[(1+\lambda_0)A+\lambda_0B]\xi_n\|^{2}-\frac{|\langle{[(1+\lambda_0)A+\lambda_0B]\xi_n,(A+B)\xi_n}\rangle|^{2}}{\|(A+B)\xi_n\|^{2}}\\ & \le\|[(1+\lambda_0)A+\lambda_0B]\xi_n\|^{2}-\frac{|\langle{[(1+\lambda_0)A+\lambda_0B]\xi_n,(A+B)\xi_n}\rangle|^{2}}{\|A+B\|^{2}}\\ & \le\|[(1+\lambda_0)A+\lambda_0B]\xi_n\|^{2}\\ & \le\|[(1+\lambda_0)A+\lambda_0B]\|^{2}=(1+\lambda_0)^{2}\|A\|^{2}+\lambda_0^{2}\|B\|^{2}. \end{align*}

Letting $n\to \infty$ we conclude that (6) is equivalent with the following limit conditions:

(10)\begin{align} & \frac{\langle{[(1+\lambda_0)A+\lambda_0B]\xi_n,(A+B)\xi_n}\rangle}{\|(A+B)\xi_n\|}\xrightarrow{n\to\infty}0, \end{align}

(11)\begin{align} & \langle{[(1+\lambda_0)A+\lambda_0B]\xi_n,(A+B)\xi_n}\rangle\xrightarrow{n\to\infty}0 \end{align}

and

(12)\begin{equation} \|[(1+\lambda_0)A+\lambda_0B]\xi_n\|^{2}\xrightarrow{n\to\infty}(1+\lambda_0)^{2}\|A\|^{2}+\lambda_0^{2}\|B\|^{2}. \end{equation}

Following the notation of (7), one can write (11) as

\[ (1+\lambda_0)a^{2}+\lambda_0b^{2}+(1+\lambda_0)c+\lambda_0\bar{c}=0. \]

Similarly, (12) takes the form

\[ (1+\lambda_0)^{2}a^{2}+\lambda_0^{2}b^{2}+2\lambda_0(\lambda_0+1)\Re c=(1+\lambda_0)^{2}\|A\|^{2}+\lambda_0^{2}\|B\|^{2}. \]

Easy computations then show that (11) and (12) are actually equivalent with (8).

According to these remarks, in order to prove $(a)$, it only remains to let $n\to \infty$ into the formulas

\[ M_{A,A+B}(\xi_n)\le\min_{\lambda\in\mathbb{C}}\|(1+\lambda)A+\lambda B\|^{2}\le\|(1+\lambda_0)A+\lambda_0B\|^{2},\quad n\ge 0 \]

and

(13)\begin{equation} \begin{aligned} \|A+\lambda B\|^{2} & \ge\|(A+\lambda B)\xi_n\|^{2}\\ & =\|A\xi_n\|^{2}+2\Re\lambda\langle{A\xi_n,B\xi_n}\rangle+|\lambda|^{2}\|B\xi_n\|^{2},\quad\lambda\in\mathbb{C},\ n\ge 0. \end{aligned} \end{equation}

(b) As seen above, (11) and (12) are a consequence of (8). In addition,

\[ \|(A+B)\xi_n\|^{2}\xrightarrow{n\to\infty}\dfrac{a^{2}-[(1+\lambda_0)^{2}\|A\|^{2}+\lambda_0^{2}\|B\|^{2}]}{\lambda_0^{2}}. \]

Hence, $\lim _{n\to \infty }\|(A+B)\xi _n\|>0$, which shows that (10) also holds true. The proof is completed.

Proof. Let $x_1,x_2,y_1,y_2$ be non-null vectors in $\mathscr {H}$ such that $A+\alpha _1 B=x_1\otimes y_1$ and $A+\alpha _2 B=x_2\otimes y_2$. Then

\[ A+\lambda B=\frac{\lambda-\alpha_2}{\alpha_1-\alpha_2}x_1\otimes y_1+\frac{\alpha_1-\lambda}{\alpha_1-\alpha_2}x_2\otimes y_2,\quad\lambda\in\mathbb{C}. \]

We distinguish two cases:

$(i)$ $\{x_1,x_2\}$ are linearly independent. Since $\operatorname {rank}(A+\alpha _3 B)=1$, one can find $\beta _1,\beta _2\in \mathbb {C}$ (at least one of them is non-null) such that $.$. Then, for every $z\in \mathscr {H}$, there exists $\mu \in \mathbb {C}$ such that

\[ \frac{\alpha_3-\alpha_2}{\alpha_1-\alpha_2}\langle{z,y_1}\rangle=\mu\beta_1\text{ and }\frac{\alpha_1-\alpha_3}{\alpha_1-\alpha_2}\langle{z,y_2}\rangle=\mu\beta_2. \]

Therefore, $\beta _1,\beta _2$ are both non-null and ${\bar {\beta _2}(\bar {\alpha _3}-\bar {\alpha _2})}/{\bar {\alpha _1}-\bar {\alpha _2}}y_1-{\bar {\beta _1}(\bar {\alpha _1}-\bar {\alpha _3})}/{\bar {\alpha _1}-\bar {\alpha _2}}y_2=0$, so $\{y_1,y_2\}$ are linearly dependent. In other words,

\[ A+\lambda B=\left[\frac{\lambda-\alpha_2}{\alpha_1-\alpha_2}x_1+\frac{\beta_2(\alpha_3-\alpha_2)(\alpha_1-\lambda)}{\beta_1(\alpha_1-\alpha_2)(\alpha_1-\alpha_3)}x_2\right]\otimes y_1,\quad\lambda\in\mathbb{C}. \]

$(ii)$ $\{x_1,x_2\}$ are linearly dependent. In this case, there exists a complex number $\beta \ne 0$ such that $x_2=\beta x_1$. We conclude that

\[ A+\lambda B=x_1\otimes\left[\frac{\bar{\lambda}-\bar{\alpha_2}}{\bar{\alpha_1}-\bar{\alpha_2}}y_1+\frac{\bar{\beta}(\bar{\alpha_1}-\bar{\lambda})}{\bar{\alpha_1}-\bar{\alpha_2}}y_2\right],\quad\lambda\in\mathbb{C}. \]

Since $\{A,B\}$ are linearly independent (as assumed earlier; this also implies that $\{y_1,y_2\}$ are linearly independent), we deduce that $A+\lambda B$ has rank one for every $\lambda \in \mathbb {C}$.

Proof. $(i)\Rightarrow (ii)$. The parallelogram law is obviously weaker than (or, at most equivalent to) Pythagoras orthogonality.

Our next aim is to prove the limit conditions of $(ii)$. Since $\Re (2\alpha _2A^{*}B)\ge 0$ and $\Re (3\alpha _2A^{*}B)\ge 0$ and, by Lemma 4.11, at least one of the operators $A+\alpha _2 B$, $A+2\alpha _2 B$ and $A+3\alpha _2 B$ has rank strictly greater than one, so we can assume that $\alpha _1=\alpha _2=\alpha$. As we have previously done, we can also assume that $\alpha =1$. We firstly observe that, by $(i)$,

\[ \|(1+\lambda)A+\lambda B\|^{2}=|1+\lambda|^{2}\|A\|^{2}+|\lambda|^{2}\|B\|^{2},\quad\lambda\in\mathbb{C}. \]

An easy computation then shows that

(14)\begin{equation} \min_{\lambda\in\mathbb{C}}\|(1+\lambda)A+\lambda B\|^{2}=\frac{\|A\|^{2}\|B\|^{2}}{\|A\|^{2}+\|B\|^{2}}, \end{equation}

which is attained for $\lambda _0=-{\|A\|^{2}}/{\left(\|A\|^{2}+\|B\|^{2}\right)}$. It follows from Theorem 4.9 and (14) that there exists a sequence $(\xi _n)_{n\ge 0}$ of unit vectors in $\mathscr {H}$ such that

(15)\begin{equation} M_{A,A+B}(\xi_n)\xrightarrow{n\to\infty}\frac{\|A\|^{2}\|B\|^{2}}{\|A\|^{2}+\|B\|^{2}}. \end{equation}

We may suppose, eventually on a subsequence, that $(A+B)\xi _n\ne 0$ for every $n\ge 0$. Indeed, if, otherwise, $(A+B)\xi _n=0$ for every $n\ge n_0$ and for a certain $n_0\ge 0$, then (15) takes the form

(16)\begin{equation} \|A\xi_n\|^{2}=\|B\xi_n\|^{2}\xrightarrow{n\to\infty}\frac{\|A\|^{2}\|B\|^{2}}{\|A\|^{2}+\|B\|^{2}} \end{equation}

by the definition of $M_{A,A+B}$. As $(A\xi _n)_{n\ge 0}$ is a bounded sequence in $\mathscr {H}$, it contains a weakly convergent subsequence (denoted also by $(A\xi _n)_{n\ge 0}$) to a vector $w\in \mathscr {H}$. Obviously, $\operatorname {span}\{(A+B)^{*}w\}\subsetneq \overline {\operatorname {ran}(A+B)^{*}}$ as the rank of $(A+B)^{*}$ is strictly greater than $1$. Consequently, one can find a unit vector $e\in \overline {\operatorname {ran}(A+B)^{*}}$, which is orthogonal to $(A+B)^{*}w$. Then, by setting $u_n=\sqrt {{n}/({n+1})}\xi _n+\left({1}/{\sqrt {n+1}}\right)e,\ n\ge 0$, we have

\[ (A+B)u_n=\sqrt{\frac{n}{n+1}}(A+B)\xi_n+\frac{1}{\sqrt{n+1}}(A+B)e=\frac{1}{\sqrt{n+1}}(A+B)e\ne 0,\quad n\ge n_0, \]

since $e\perp \ker (A+B)$. Moreover, for $n\ge 0$,

(17)\begin{align} M_{A,A+B}(u_n) & =\|Au_n\|^{2}-\frac{|\langle{Au_n,(A+B)u_n}\rangle|^{2}}{\|(A+B)u_n\|^{2}}\notag\\ & =\frac{n}{n+1}\|A\xi_n\|^{2}+\frac{1}{n+1}\|Ae\|^{2}+2\frac{\sqrt{n}}{n+1}\Re\langle{A\xi_n,Ae}\rangle\notag\\ & \quad -\frac{\left|\left\langle\sqrt{\frac{n}{n+1}}A\xi_n+\frac{1}{\sqrt{n+1}}Ae,\frac{1}{\sqrt{n+1}}(A+B)e\right\rangle\right|^{2}}{\left\|\frac{1}{\sqrt{n+1}}(A+B)e\right\|^{2}} \notag\\ & =\frac{n}{n+1}\|A\xi_n\|^{2}+\frac{1}{n+1}\|Ae\|^{2}+2\frac{\sqrt{n}}{n+1}\Re\langle{A\xi_n,Ae}\rangle\notag\\ & \quad -\frac{\left|\sqrt{\frac{n}{n+1}}\langle A\xi_n,(A+B)e\rangle+\frac{1}{\sqrt{n+1}}\left\langle Ae,(A+B)e\right\rangle\right|^{2}}{\|(A+B)e\|^{2}}. \end{align}

In view of (16) and the observation that

\[ \langle A\xi_n,(A+B)e\rangle\xrightarrow{n\to\infty}\langle{w,(A+B)e}\rangle=\langle{(A+B)^{*}w,e}\rangle=0, \]

we deduce, by passing to limit in (17), that

\[ M_{A,A+B}(u_n)\xrightarrow{n\to\infty}\frac{\|A\|^{2}\|B\|^{2}}{\|A\|^{2}+\|B\|^{2}}. \]

One may consider, in this particular situation, the sequence $(u_{n+n_0})_{n\ge 0}$ which will be also denoted by $(\xi _n)_{n\ge 0}$.

The assumptions of Lemma 4.10 $(a)$ are verified. So, the limits (7) satisfy the conditions (equivalent with (8))

(18)\begin{equation} c=\frac{\|A\|^{2}(b^{2}-\|B\|^{2})}{\|B\|^{2}}=\frac{\|B\|^{2}(a^{2}-\|A\|^{2})}{\|A\|^{2}}. \end{equation}

We deduce that $c\le 0$. Also, by hypothesis (i.e., $\Re (A^{*}B)\ge 0$),

\[ c=\Re c=\lim_{n\to\infty}\Re\langle{\xi_n,A^{*}B\xi_n}\rangle=\lim_{n\to\infty}\langle{\xi_n,\Re(A^{*}B)\xi_n}\rangle\ge 0. \]

This forces $c=0$ and, by (18), $a=\|A\|$ and $b=\|B\|$.

$(ii)\Rightarrow (iii)$. Clearly, if a sequence $(\xi _n)_{n\ge 0}$ of unit vectors in $\mathscr {H}$ verifies $(ii)$, then

\begin{align*} & \|(A+\lambda B)\xi_n\|^{2}\\ & \quad=\|A\xi_n\|^{2}+2\Re\bar{\lambda}\langle{A\xi_n,B\xi_n}\rangle+|\lambda|^{2}\|B\xi_n\|^{2}\xrightarrow{n\to\infty}\|A\|^{2}+|\lambda|^{2}\|B\|^{2},\quad\lambda\in\mathbb{C}. \end{align*}

Hence $(\xi _n)_{n\ge 0}$ also verifies $(iii)$.

$(iii)\Rightarrow (i)$. Let $(\xi _n)_{n\ge 0}$ be a sequence of unit vectors in $\mathscr {H}$ such that $(iii)$ holds true. Then

\[ \|A+\lambda B\|^{2}\ge\|(A+\lambda B)\xi_n\|^{2}\xrightarrow{n\to\infty}\|A\|^{2}+|\lambda|^{2}\|B\|^{2}. \]

The proof is finished, as before, by the use of the parallelogram law.