# A necessary and sufficient criterion for an element of a multiplier $C^{*}$-algebra to be positive.

I am trying to find a reference for the following assertion:

Let $A$ be a $C^{*}$-algebra, and let $M(A)$ denote its multiplier algebra. Then $m$ is a positive element of $M(A)$ if and only if $a^{*} m a$ is a positive element of $A$ for all $a \in A$. In other words, $$m \in M(A)_{\geq} \iff (\forall a \in A)(a^{*} m a \in A_{\geq}).$$

The forward implication is trivial because if $m \in M(A)_{\geq}$, then there exists an $n \in M(A)$ such that $m = n^{*} n$, so $$\forall a \in A: \quad a^{*} m a = a^{*} n^{*} n a = (n a)^{*} (n a) \in A_{\geq}.$$ Note: $A$ is an ideal of $M(A)$, so $n a \in A$ for all $a \in A$.

I have absolutely no idea how to prove the backward implication though.

Thank you very much for your gracious help.

Oh well, how silly of me not to have thought of the following argument sooner.

In what follows, $(e_{i})_{i \in I}$ is a self-adjoint approximate identity of $A$ that is bounded in norm by $1$. It is $C^{*}$-folklore that such an approximate identity exists.

Lemma. If $(x_{i})_{i \in I}$ is a convergent net in $A$ whose limit is $x$, then $\displaystyle \lim_{i \in I} e_{i} x_{i} = x$.

Proof of Lemma

By the Triangle Inequality, \begin{align} \forall i \in I: \quad \| e_{i} x_{i} - x \| & = \| e_{i} x_{i} - e_{i} x + e_{i} x - x \| \\ & \leq \| e_{i} x_{i} - e_{i} x \| + \| e_{i} x - x \| \\ & \leq \| e_{i} \| \| x_{i} - x \| + \| e_{i} x - x \| \\ & \leq \| x_{i} - x \| + \| e_{i} x - x \|. \end{align} As $\displaystyle \lim_{i \in I} x_{i} = x$ and $\displaystyle \lim_{i \in I} e_{i} x = x$, the lemma follows immediately. $\quad \blacksquare$

Proposition 1. Let $m \in M(A)$. If $a^{*} m a \in A_{\geq}$ for all $a \in A$, then $m$ is self-adjoint.

Proof of Proposition 1

Let $m \in M(A)$, and suppose that $a^{*} m a \in A_{\geq}$ for all $a \in A$. As positive elements of a $C^{*}$-algebra are by definition self-adjoint, we have $$\forall a \in A: \quad a^{*} m^{*} a = (a^{*} m a)^{*} = a^{*} m a.$$ It follows readily that $e_{i} m^{*} e_{i} = e_{i} m e_{i}$, or equivalently, $e_{i} (m^{*} - m) e_{i} = 0$ for all $i \in I$.

Let $a,b \in A$. By the previous paragraph, $$\forall i \in I: \quad a e_{i} (m^{*} - m) e_{i} b = 0.$$ As $$\lim_{i \in I} (m^{*} - m) e_{i} b = (m^{*} - m) b,$$ the lemma yields $a (m^{*} - m) b = 0$. Then as (i) $a$ and $b$ are arbitrary and (ii) $A$ is an essential ideal of $M(A)$, we obtain $m^{*} - m = 0$, which proves that $m$ is self-adjoint. $\quad \blacksquare$

Proposition 2. Let $m \in M(A)$. If $a^{*} m a \in A_{\geq}$ for all $a \in A$, then $m$ is positive.

Proof of Proposition 2

Let $m \in M(A)$, and suppose that $a^{*} m a \in A_{\geq}$ for all $a \in A$. By Proposition 1, $m$ is self-adjoint, so it remains to show that $\sigma(m) \subseteq [0,\infty)$.

By way of contradiction, suppose that there exists a $\lambda \in \sigma(m) \cap (- \infty,0)$. Consider the continuous function $f: \Bbb{R} \to \Bbb{R}$ defined by $$\forall x \in \Bbb{R}: \quad f(x) \stackrel{\text{df}}{=} \begin{cases} x & \text{if  x < 0 }; \\ 0 & \text{if  x \geq 0 }. \end{cases}$$ Applying $f$ to $m$ via the continuous functional calculus (a valid step because $m$ is self-adjoint) gives us a non-zero element $f(m) \in M(A)$.

Claim 1. For all $n \in \Bbb{N}$ and all $x \in A$, we have $\displaystyle \lim_{i \in I} (e_{i} m e_{i})^{n} x = m^{n} x$.

Proof of Claim 1

We proceed by induction. The case $n = 1$ is an immediate consequence of the lemma above. Hence, let $x \in A$ and suppose that $\displaystyle \lim_{i \in I} (e_{i} m e_{i})^{k} x = m^{k} x$ for some $k \in \Bbb{N}$. Observe that $$\forall i \in I: \quad (e_{i} m e_{i})^{k + 1} x = (e_{i} m e_{i}) (e_{i} m e_{i})^{k} x.$$ Both the induction hypothesis and the lemma imply that $$\lim_{i \in I} e_{i} (e_{i} m e_{i})^{k} x = m^{k} x,$$ which, in turn, implies that $$\lim_{i \in I} m e_{i} (e_{i} m e_{i})^{k} x = m^{k + 1} x.$$ Hence, by the lemma once more, $$\lim_{i \in I} (e_{i} m e_{i})^{k + 1} x = m^{k + 1} x.$$ By mathematical induction, the claim is therefore true. $\quad \square$

Claim 2. For all $a \in A$, we have $\displaystyle \lim_{i \in I} a^{*} f(e_{i} m e_{i}) a = a^{*} f(m) a$.

Proof of Claim 2

If $a = 0$, then there is nothing to prove, so suppose that $a \in A \setminus \{ 0 \}$.

Fix an $\epsilon > 0$. By the Stone-Weierstrass Theorem, we can find a polynomial function $p$ such that $$(\clubsuit) \qquad \max_{x \in [- \| m \|,\| m \|]} |f(x) - p(x)| < \frac{\epsilon}{3 \| a \|^{2}}.$$ As $e_{i} m e_{i} \in A_{\geq}$ and $\| e_{i} m e_{i} \| \leq \| m \|$ for all $i \in I$, we have $\sigma(e_{i} m e_{i}) \subseteq [0,\| m \|]$, so by $(\clubsuit)$, $$(\spadesuit) \qquad \begin{cases} \| f(m) - p(m) \| < \dfrac{\epsilon}{3 \| a \|^{2}}, \\ \| f(e_{i} m e_{i}) - p(e_{i} m e_{i}) \| < \dfrac{\epsilon}{3 \| a \|^{2}}. \end{cases}$$ Now, \begin{align} \forall i \in I: \qquad & ~ \| a^{*} f(e_{i} m e_{i}) a - a^{*} f(m) a \| \\ \leq & ~ \| a^{*} f(e_{i} m e_{i}) a - a^{*} p(e_{i} m e_{i}) a \| + \\ & ~ \| a^{*} p(e_{i} m e_{i}) a - a^{*} p(m) a \| + \\ & ~ \| a^{*} p(m) a - a^{*} f(m) a \| \qquad (\text{By the Triangle Inequality.}) \\ \leq & ~ \| a^{*} \| \| f(e_{i} m e_{i}) - p(e_{i} m e_{i}) \| \| a \| + \\ & ~ \| a^{*} p(e_{i} m e_{i}) a - a^{*} p(m) a \| + \\ & ~ \| a^{*} \| \| p(m) - f(m) \| \| a \| \\ = & ~ \| a \|^{2} \| f(e_{i} m e_{i}) - p(e_{i} m e_{i}) \| + \\ & ~ \| a^{*} p(e_{i} m e_{i}) a - a^{*} p(m) a \| + \\ & ~ \| a \|^{2} \| p(m) - f(m) \| \\ < & ~ \frac{\epsilon}{3} + \| a^{*} p(e_{i} m e_{i}) a - a^{*} p(m) a \| + \frac{\epsilon}{3}. \qquad (\text{By $(\spadesuit)$.}) \end{align} By Claim 1, we can choose an index $i_{0} \in I$ so that $$\forall i \in I_{\geq i_{0}}: \quad \| a^{*} p(e_{i} m e_{i}) a - a^{*} p(m) a \| < \frac{\epsilon}{3}.$$ Therefore, $$\forall i \in I_{\geq i_{0}}: \quad \| a^{*} f(e_{i} m e_{i}) a - a^{*} f(m) a \| < \epsilon,$$ which proves Claim 2. $\quad \square$

For each $i \in I$, as $e_{i} m e_{i} \in A_{\geq}$, we have $f(e_{i} m e_{i}) = 0$, so $a^{*} f(e_{i} m e_{i}) a = 0$ for all $a \in A$. We thus obtain $a^{*} f(m) a = 0$ for all $a \in A$. Arguing as in the proof of Proposition 1, we get $a f(m) b = 0$ for all $a,b \in A$. We then use the fact that $A$ is an essential ideal of $M(A)$ to obtain $f(m) = 0$, which is a contradiction of an earlier statement.

Therefore, $\sigma(m) \cap (- \infty,0) = \varnothing$, and we conclude that $m \in M(A)_{\geq}$. $\quad \blacksquare$