Let $X$ be a locally compact Hausdorff space. I want to show that $M(C_0(X)) \cong C_b(X)$ as $C^*$-algebras where
$$M(C_0(X)) = \mathcal{L}(C_0(X))$$ are the adjointable operators $C_0(X) \to C_0(X).$ To do this, I defined the mapping
$$C_b(X) \to M(C_0(X)): f \mapsto (g \mapsto fg).$$
I managed to show that this is an isometric $*$-morphism, but I can't show that it is surjective.
Given $t \in M(C_0(X))$, I thought to define
$$f: X \to \mathbb{C}: x \mapsto t(g)(x)$$
where $g\in C_0(X)$ is a function with $g(x) = 1$ and check that $t$ is given by multiplication by $g$, but I did not succeed in proving this.
Best Answer
We adopt your definition: $T\in M(C_{0}(X))$ if $T:C_{0}(X)\rightarrow C_{0}(X)$ is a map and there exists another map $T^{\ast}:C_{0}(X)\rightarrow C_{0}(X)$ such that $T(f)\bar{g}=f\overline{T^{\ast}(g)}$ for any $f,g\in C_{0}(X)$.
Claim 1: $T$ and $T^{\ast}$ are linear.
Proof of Claim 1: Let $f_{1},f_{2}\in C_{0}(X)$, $\alpha\in\mathbb{C}$. Let $x\in X$. By Urysohn lemma, we can choose $g\in C_{0}(X)$ such that $g(x)\neq0$. By assumption, \begin{eqnarray*} T(\alpha f_{1}+f_{2})\bar{g} & = & (\alpha f_{1}+f_{2})\overline{T^{\ast}(g)}\\ & = & \alpha f_{1}\overline{T^{\ast}(g)}+f_{2}\overline{T^{\ast}(g)}\\ & = & \alpha T(f_{1})\bar{g}+T(f_{2})\bar{g}. \end{eqnarray*} Evaluate the above at $x$, then $T(\alpha f_{1}+f_{2})(x)\bar{g}(x)=\alpha T(f_{1})(x)\bar{g}(x)+T(f_{2})(x)\bar{g}(x)$. Note that $\bar{g}(x)=\overline{g(x)}\neq0$. Dividing both sides with $\bar{g}(x)$, we obtain $T(\alpha f_{1}+f_{2})(x)=\alpha T(f_{1})(x)+T(f_{2})(x).$ Since $x\in X$ is arbitrary, we have that $T(\alpha f_{1}+f_{2})=\alpha T(f_{1})+T(f_{2})$. That is, $T$ is linear.
Observe that $T^{\ast}(g)\overline{f}=g\overline{T(f)}$ for any $f,g\in C_{0}(X)$. By the same reasoning, $T^{\ast}$ is linear.
Claim 2: For any $x\in X$, $\frac{Tf_{1}(x)}{f_{1}(x)}=\frac{Tf_{2}(x)}{f_{2}x)}$ whenever $f_{1},f_{2}\in C_{0}(X)$ such that $f_{1}(x)\neq0$ and $f_{2}(x)\neq0$.
Proof of Claim 2: Let $x\in X$. Let $f_{1},f_{2}\in C_{0}(X)$ be such that $f_{1}(x)\neq0$ and $f_{2}(x)\neq0$. By Urysohn lemma, we may choose $g\in C_{0}(X)$ such that $g(x)=1$. We have that $$ Tf_{1}(x)=Tf_{1}(x)\bar{g}(x)=f_{1}(x)\overline{T^{\ast}(g)}(x) $$ and $$ Tf_{2}(x)=Tf_{2}(x)\bar{g}(x)=f_{2}(x)\overline{T^{\ast}(g)}(x). $$ It follows that $$ \frac{Tf_{1}(x)}{f_{1}(x)}=\overline{T^{\ast}(g)}(x)=\frac{Tf_{2}(x)}{f_{2}(x)}. $$
Claim 3: Define $\theta:X\rightarrow\mathbb{C}$ by $\theta(x)=\frac{Tf(x)}{f(x)}$, where $f$ is any element in $C_{0}(X)$ such that $f(x)\neq0$. Then $\theta$ is continuous.
Proof of Claim 3: Given $x\in X$, by Urysohn lemma, there exists $f\in C_{0}(X)$ such that $f(x)\neq0$. By Claim 2, $\theta(x)$ is well-defined. Hence, we obtain a map $\theta:X\rightarrow\mathbb{C}$. Next, we show that $\theta$ is continuous. Let $x\in X$. Let $(x_{\alpha})$ be a net in $X$ such that $x_{\alpha}\rightarrow x$. By Urysohn lemma, there exists $f\in C_{0}(X)$ such that $f(x)\neq0$. Since $f$ is continuous, we can choose an open neighborhood $U$ of $x$ such that $f(y)\neq0$ for any $y\in U$. Choose $\alpha_{0}$ such that $x_{\alpha}\in U$ whenever $\alpha\succeq\alpha_{0}$. For any $\alpha\succeq\alpha_{0}$, \begin{eqnarray*} \theta(x_{\alpha}) & = & \frac{Tf(x_{\alpha})}{f(x_{\alpha})}\\ & \rightarrow & \frac{Tf(x)}{f(x)}\\ & = & \theta(x) \end{eqnarray*} because $Tf(x_{\alpha})\rightarrow Tf(x)$ and $f(x_{\alpha})\rightarrow f(x)$. This shows that $\theta$ is a continuous function.
Claim 4: $T$ and $T^{\ast}$ are bounded.
Proof of Claim 4: For each $x\in X$, choose a continuous function $g_{x}:X\rightarrow[0,1]$ such that $g_{x}(x)=1$ and $g_{x}$ has compact support. (This is possible by Urysohn lemma). Define $\theta_{x}:C_{0}(X)\rightarrow C_{0}(X)$ by $\theta_{x}(f)=f\overline{T^{\ast}(g_{x})}$. Clearly $\theta_{x}$ is linear and bounded. Consider the family of bounded linear operators $\{\theta_{x}\mid x\in X\}$. Let $f\in C_{0}(X)$ be arbitrary. We assert that $\{\theta_{x}(f)\mid x\in X\}$ is a bounded subset of $C_{0}(X)$. Observe that \begin{eqnarray*} \theta_{x}(f) & = & f\overline{T^{\ast}(g_{x})}\\ & = & Tf\cdot\overline{g_{x}}\\ & = & Tf\cdot g_{x}. \end{eqnarray*} It follows that \begin{eqnarray*} \sup_{x\in X}||\theta_{x}(f)||_{\infty} & = & \sup_{x\in X}||Tf\cdot g_{x}||_{\infty}\\ & \leq & \sup_{x\in X}||Tf||_{\infty}\cdot||g_{x}||_{\infty}\\ & = & ||Tf||_{\infty}\\ & < & \infty. \end{eqnarray*} By Uniform Boundedness Principle, it follows that $\sup_{x\in X}||\theta_{x}||<\infty$. Let $M=\sup_{x\in X}||\theta_{x}||$. For each $x\in X$, we have that \begin{eqnarray*} \left|Tf(x)\right| & = & \left|Tf(x)\overline{g_{x}}(x)\right|\\ & = & \left|\left(Tf\overline{g_{x}}\right)(x)\right|\\ & = & \left|\left(f\overline{T^{\ast}(g_{x})}\right)(x)\right|\\ & = & \left|\theta_{x}(f)(x)\right|.\\ & \leq & ||\theta_{x}(f)||_{\infty}\\ & \leq & M||f||_{\infty}. \end{eqnarray*} Hence $||Tf||_{\infty}\leq M||f||_{\infty}$. This shows that $T$ is bounded. We can prove similarly that $T^{\ast}$ is bounded.
Claim 5: $\theta$ defined in Claim 3 is bounded.
Let $x\in X$. Choose a continuous function $f:X\rightarrow[0,1]$ such that $f(x)=1$ and $f$ has compact support. Note that $f\in C_{0}(X)$ and $||f||_{\infty}=1$. We have that $|\theta(x)|=|\frac{Tf(x)}{f(x)}|=|Tf(x)|\leq||T||\,||f||_{\infty}=||T||$. Hence $\theta$ is bounded with $||\theta||_{\infty}\leq||T||$.
Claim 6: For each $f\in X$, $Tf=\theta f$.
Let $f\in X$ and $x\in X$. If $f(x)\neq0$, then $\theta(x)=\frac{Tf(x)}{f(x)}$, so $Tf(x)=\theta(x)f(x)$. Now, consider that case that $f(x)=0$. Choose $g\in C_{0}(X)$ such that $g(x)\neq0$. We have that $Tf\cdot\overline{g}=f\cdot\overline{T^{\ast}g}$. Evaluate it at $x$, then we obtain $Tf(x)=0$. It follows that $Tf(x)=\theta(x)f(x)$ because both sides are zero. Hence, $Tf=\theta f$.