Showing that $\alpha\geq \beta$

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 $$\alpha\geq \beta.$$

• Your title should be more descriptive. – Umberto P. May 21 at 19:12

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