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.
 A: 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 $
