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In this question $\mathcal{H}$ stands for a Hilbert space over $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$, with inner product $\langle\cdot\;| \;\cdot\rangle$ and the norm $\|\cdot\|$ and let $\mathcal{B}(\mathcal{H})$ the algebra of all bounded linear operators from $\mathcal{H}$ to $\mathcal{H}$.

We recall that for $T\in \mathcal{B}(\mathcal{H})$, the norm of $T$ is given by $$\|T\|=\sup_{\|x\|=1}\|Tx\|$$ Are the following equalities true $$(\displaystyle \sup_{\|x\|=1}\|Tx\|)^2=\displaystyle \sup_{\|x\|=1}\|Tx\|^2,\;\;\;(\displaystyle \sup_{\|x\|=1}\|Tx\|)^{1/2}=\displaystyle \sup_{\|x\|=1}\|Tx\|^{1/2}\;??$$ It is clear that $\{\|Tx\|;\;\|x\|=1\}\subseteq B(0,\|T\|)$, so it is bounded.

Thank you..

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Consider an increasing and left continuous function $$f : \mathbb{R} \cup \{ \pm \infty \}\longrightarrow \mathbb{R} \cup \{ \pm \infty\}.$$ Then for any collection of elements $\mathcal{A} \subseteq \mathbb{R}\cup \{ \pm \infty \}$ $$ f(\sup \{\mathcal{A} \}) = \sup f(\mathcal{A}). $$

Indeed $$f(\sup \{\mathcal{A} \}) \ge f(a), \ \ \forall a \in \mathcal{A}$$ which gives one inequality. The second inequality follows from left continuity by taking a sequence $$a_k \nearrow \sup \{\mathcal{A} \}$$ so that $$ f(\sup \{\mathcal{A} \}) = \lim f(a_k) \le \sup f(\mathcal{A}). $$

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