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Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $F$.

Let $A,B\in \mathcal{B}(F)$. Consider the following numbers: $$\alpha=\sup_{\substack{a,b\in \mathbb{C}^2,\\ |a|^2+|b|^2<1}}\sup\{|\mu|\,;\;\mu\in \sigma(aA+bB)\}.$$ $$\beta=\sup\{\lambda^{1/2}\,;\;\lambda\in \sigma(A^*A+B^*B)\},$$ where $\sigma(X)$ is the spectrum of an operator $X$.

I want to show that $$\alpha\geq \beta.$$

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    $\begingroup$ Your title should be more descriptive. $\endgroup$ – Umberto P. May 21 at 19:12
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The inequality does not hold. Take $A$ any nonzero nilpotent element, and $B=0$. Then $\alpha=0$ and $\beta=\|A\|$.

The reverse inequality does hold, though. We have \begin{align} \operatorname{spr}(aA+bB)^2 &\leq \|aA+bB\|^2\leq (|a|\,\|A\|+|b|\,\|B\|)^2. \end{align} Then \begin{align} \sup_{a,b}\operatorname{spr}(aA+bB)^2 &\leq\sup_{a,b}(|a|\,\|A\|+|b|\,\|B\|)^2=\max\{\|A\|^2,\|B\|^2\}\\ &=\max\{\|A^*A\|,\|B^*B\|\}\\ &\leq \|A^*A+B^*B\|\\ &=\sup\{\lambda:\ \lambda\in\sigma(A^*A+B^*B)\}\\ &=\sup\{\lambda^{1/2}:\ \lambda\in\sigma(A^*A+B^*B)\}^2\\ \end{align} So $$ \alpha\leq\beta. $$

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  • $\begingroup$ Thank you for the explanation. If $A$ and $B$ are commuting normal operator, I think that the inequality holds. $\endgroup$ – Schüler May 21 at 19:54
  • $\begingroup$ No, if you take $B=A$, the inequality is always strict $\endgroup$ – Martin Argerami May 21 at 21:34

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