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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.

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1 Answer 1

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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 $

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