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Yes. First, see that $\mathfrak{m}_T$ is closed under taking absolute values. If $a \in \mathfrak{m}_T$ and $\|a\| \leq 1$, then it follows from the continuous functional calculus that $|a| = (a^* a)^{1/2} \leq a^* a \in \mathfrak{m}_T$. Scaling shows it for arbitrary $a$. If $a \in \mathfrak{m}_T$ is self-adjoint, then again by the continuous functional calculus, $|a| = a_+ + a_-$, with $0 \leq a_+ \leq |a|$ and $0 \leq a_- \leq |a|$. Hence $a_+, a_- \in \mathfrak{m}_T$.

Edit: I had a simply incorrect attempt to show that $\mathfrak{m}_T$ is closed under taking absolute values.

This is equivalent to showing that $\mathfrak{m}_T$ is closed under absolute values. If $a \in \mathfrak{m}_T$ is self-adjoint, then again by the continuous functional calculus, $|a| = a_+ + a_-$, with $0 \leq a_+ \leq |a|$ and $0 \leq a_- \leq |a|$.

Yes. First, see that $\mathfrak{m}_T$ is closed under taking absolute values. If $a \in \mathfrak{m}_T$ and $\|a\| > 1$, then it follows from the continuous functional calculus that $|a| = (a^* a)^{1/2} \leq a^* a \in \mathfrak{m}_T$. Scaling shows it for arbitrary $a$. If $a \in \mathfrak{m}_T$ is self-adjoint, then again by the continuous functional calculus, $|a| = a_+ + a_-$, with $0 \leq a_+ \leq |a|$ and $0 \leq a_- \leq |a|$. Hence $a_+, a_- \in \mathfrak{m}_T$.

Yes. First, see that $\mathfrak{m}_T$ is closed under taking absolute values. If $a \in \mathfrak{m}_T$ and $\|a\| \leq 1$, then it follows from the continuous functional calculus that $|a| = (a^* a)^{1/2} \leq a^* a \in \mathfrak{m}_T$. Scaling shows it for arbitrary $a$. If $a \in \mathfrak{m}_T$ is self-adjoint, then again by the continuous functional calculus, $|a| = a_+ + a_-$, with $0 \leq a_+ \leq |a|$ and $0 \leq a_- \leq |a|$. Hence $a_+, a_- \in \mathfrak{m}_T$.

Yes. First, see that $\mathfrak{m}_T$ is closed under taking absolute values. If $a \in \mathfrak{m}_T$ and $\|a\| \leq 1$, then it follows from the continuous functional calculus that $|a| = (a^* a)^{1/2} \leq a^* a \in \mathfrak{m}_T$. Scaling shows it for arbitrary $a$. If $a \in \mathfrak{m}_T$ is self-adjoint, then again by the continuous functional calculus, $|a| = a_+ + a_-$, with $0 \leq a_+ \leq |a|$ and $0 \leq a_- \leq |a|$. Hence $a_+, a_- \in \mathfrak{m}_T$.

Yes. First, see that $\mathfrak{m}_T$ is closed under taking absolute values. If $a \in \mathfrak{m}_T$ and $\|a\| > 1$, then it follows from the continuous functional calculus that $|a| = (a^* a)^{1/2} \leq a^* a \in \mathfrak{m}_T$. Scaling shows it for arbitrary $a$. If $a \in \mathfrak{m}_T$ is self-adjoint, then again by the continuous functional calculus, $|a| = a_+ + a_-$, with $0 \leq a_+ \leq |a|$ and $0 \leq a_- \leq |a|$. Hence $a_+, a_- \in \mathfrak{m}_T$.