[proofplan]
We use the double-centralizer model of the multiplier algebra. First we check that every bounded [continuous function](/page/Continuous%20Function) acts on $C_0(X)$ by bounded pointwise multiplication, giving an injective unital $*$-homomorphism $C_b(X)\to M(C_0(X))$. Conversely, from a multiplier $(L,R)$ we recover a pointwise function $g:X\to\mathbb C$ by testing $L$ against compactly supported functions nonzero at a point. The centralizer identity makes this definition independent of the [test function](/page/Test%20Function), local quotient formulas prove continuity, bump functions give boundedness, and density of $C_c(X)$ in $C_0(X)$ proves that the recovered $g$ gives the original multiplier.
[/proofplan]
[step:Realize bounded continuous functions as pointwise multipliers]
Let $A:=C_0(X)$. Define
\begin{align*}
C_c(X):=\{f:X\to\mathbb C : f \text{ is continuous and has compact support}\}.
\end{align*} By [citetheorem:8575], we use the double-centralizer model
\begin{align*}
M(A)=\{(L,R): L,R:A\to A \text{ are bounded linear maps satisfying } L(ab)=L(a)b,\ R(ab)=aR(b),\ aL(b)=R(a)b\}.
\end{align*}
The product and involution are the standard ones for double centralizers, and the unit is $(\operatorname{id}_A,\operatorname{id}_A)$.
For $g\in C_b(X)$, define bounded linear maps $L_g:A\to A$ and $R_g:A\to A$ by $L_g(f)=gf$ and $R_g(f)=fg$ for every $f\in A$.
If $f\in C_0(X)$ and $\varepsilon>0$, then
\begin{align*}
\{x\in X: |g(x)f(x)|\ge \varepsilon\}\subseteq \{x\in X: |f(x)|\ge \varepsilon/\|g\|_\infty\}
\end{align*}
when $g\ne 0$, and the right-hand set is compact; if $g=0$ the assertion is immediate. Hence $gf\in C_0(X)$. Also
\begin{align*}
\|gf\|_\infty\le \|g\|_\infty \|f\|_\infty,
\end{align*}
so $L_g$ and $R_g$ are bounded. Since multiplication in $C_0(X)$ is pointwise and commutative, the identities
\begin{align*}
L_g(fh)=L_g(f)h
\end{align*}
and
\begin{align*}
R_g(fh)=fR_g(h)
\end{align*}
and
\begin{align*}
fL_g(h)=R_g(f)h
\end{align*}
hold for all $f,h\in A$. Thus $(L_g,R_g)\in M(A)$.
The assignment $\Theta:C_b(X)\to M(A)$ defined by $\Theta(g)=(L_g,R_g)$ is a unital algebra homomorphism and satisfies $\Theta(\overline g)=\Theta(g)^*$ because pointwise multiplication by $\overline g$ is the adjoint multiplier of pointwise multiplication by $g$. It is injective: if $\Theta(g)=0$ and $x\in X$, choose $f\in C_c(X)$ with $f(x)=1$, which exists by the locally compact Hausdorff bump-function lemma for nonempty locally compact Hausdorff spaces; then
\begin{align*}
g(x)=g(x)f(x)=(L_g f)(x)=0.
\end{align*}
Therefore $g=0$.
[/step]
[step:Recover a function from an arbitrary multiplier]
Let $(L,R)\in M(A)$. For each $x\in X$, choose a function $f_x\in C_c(X)$ with $f_x(x)\ne 0$, possible by local compactness and the Hausdorff separation property. Define $g:X\to\mathbb C$ by
\begin{align*}
g(x)=\frac{(Lf_x)(x)}{f_x(x)}.
\end{align*}
We prove that this value is independent of the chosen test function. Suppose $f,h\in C_c(X)$ satisfy $f(x)\ne 0$ and $h(x)\ne 0$. Since $fh\in A$ and $L(fh)=L(f)h=L(h)f$, evaluating at $x$ gives
\begin{align*}
(Lf)(x)h(x)=(Lh)(x)f(x).
\end{align*}
Dividing by $f(x)h(x)\ne 0$ yields
\begin{align*}
\frac{(Lf)(x)}{f(x)}=\frac{(Lh)(x)}{h(x)}.
\end{align*}
Thus $g$ is well-defined.
[guided]
Fix a multiplier $(L,R)\in M(A)$. The goal is to extract from the operator $L$ a scalar value at each point $x\in X$. Since elements of $A=C_0(X)$ are functions, the natural idea is to divide $(Lf)(x)$ by $f(x)$ for a function $f$ that does not vanish at $x$.
For each $x\in X$, the locally compact Hausdorff bump-function lemma gives a compactly supported continuous function $f_x:X\to\mathbb C$ with $f_x(x)\ne 0$. Since $f_x\in C_c(X)\subseteq C_0(X)=A$, the value $(Lf_x)(x)$ is defined. We set $g:X\to\mathbb C$ by
\begin{align*}
g(x)=\frac{(Lf_x)(x)}{f_x(x)}.
\end{align*}
The only possible problem is dependence on the choice of $f_x$. Let $f,h\in C_c(X)$ be two admissible choices at the same point $x$, so $f(x)\ne 0$ and $h(x)\ne 0$. The double-centralizer identity for the left multiplier says that
\begin{align*}
L(ab)=L(a)b
\end{align*}
for all $a,b\in A$. Applying this first with $a=f$ and $b=h$, and then with $a=h$ and $b=f$, gives
\begin{align*}
L(fh)=L(f)h
\end{align*}
and
\begin{align*}
L(hf)=L(h)f.
\end{align*}
Because multiplication in $A=C_0(X)$ is pointwise and commutative, $fh=hf$. Hence the two left-hand sides are the same element of $A$, so
\begin{align*}
L(f)h=L(h)f.
\end{align*}
Evaluating this equality of functions at the point $x$ gives
\begin{align*}
(Lf)(x)h(x)=(Lh)(x)f(x).
\end{align*}
Since both $f(x)$ and $h(x)$ are nonzero, division by $f(x)h(x)$ is legitimate and yields
\begin{align*}
\frac{(Lf)(x)}{f(x)}=\frac{(Lh)(x)}{h(x)}.
\end{align*}
Therefore the scalar assigned to $x$ is independent of the compactly supported test function used to compute it.
[/guided]
[/step]
[step:Prove the recovered function is continuous and bounded]
Fix $x_0\in X$. Choose $f\in C_c(X)$ with $f(x_0)\ne 0$. Since $f:X\to\mathbb C$ is continuous, the set
\begin{align*}
U_f:=\{x\in X: f(x)\ne 0\}
\end{align*}
is an open neighbourhood of $x_0$. By the definition of $g$ and the independence just proved,
\begin{align*}
g(x)=\frac{(Lf)(x)}{f(x)}
\end{align*}
for every $x\in U_f$. Since $C(X)$ has been defined as the [algebra of continuous functions](/theorems/197) $X\to\mathbb C$, the inclusion $Lf\in C_0(X)$ implies that $Lf:X\to\mathbb C$ is continuous. Since $f$ is nonzero on $U_f$, the quotient $(Lf)/f$ is continuous on $U_f$. Hence $g$ is continuous at $x_0$, and since $x_0$ was arbitrary, $g\in C(X)$.
To prove boundedness, fix $x\in X$. By the locally compact Hausdorff bump-function lemma, choose $u_x\in C_c(X)$ such that
\begin{align*}
0\le u_x\le 1
\end{align*}
and
\begin{align*}
u_x(x)=1.
\end{align*}
Then $\|u_x\|_\infty=1$ and the definition of $g$ gives
\begin{align*}
|g(x)|=|(Lu_x)(x)|.
\end{align*}
Evaluation at a point is dominated by the supremum norm, so
\begin{align*}
|(Lu_x)(x)|\le \|Lu_x\|_\infty.
\end{align*}
Define \begin{align*}\mathcal L(A):=\mathcal L(A,A)\end{align*} to be the [Banach space](/page/Banach%20Space) of bounded linear maps from $A$ to $A$, equipped with the operator norm. By boundedness of $L:A\to A$,
\begin{align*}
\|Lu_x\|_\infty\le \|L\|_{\mathcal L(A)}\|u_x\|_\infty=\|L\|_{\mathcal L(A)}.
\end{align*}
Thus $|g(x)|\le \|L\|_{\mathcal L(A)}$ for all $x\in X$, so $g\in C_b(X)$.
[/step]
[step:Show the recovered function gives the original left multiplier]
Let $h\in C_c(X)$. For each $x\in X$, if $h(x)=0$, then
\begin{align*}
(gh)(x)=0.
\end{align*}
Choose $u_x\in C_c(X)$ with $u_x(x)=1$. The centralizer identity $L(hu_x)=L(h)u_x$ and the equality $hu_x=u_xh$ also give
\begin{align*}
L(hu_x)=L(u_x)h.
\end{align*}
Evaluating at $x$ yields
\begin{align*}
(Lh)(x)=(Lu_x)(x)h(x)=0.
\end{align*}
If $h(x)\ne 0$, the definition of $g(x)$ with test function $h$ gives
\begin{align*}
(Lh)(x)=g(x)h(x).
\end{align*}
Therefore $Lh=gh$ for every $h\in C_c(X)$.
Now let $a\in C_0(X)$. By the density theorem for compactly supported continuous functions on a locally compact [Hausdorff space](/page/Hausdorff%20Space), $C_c(X)$ is dense in $C_0(X)$ in the supremum norm. Choose a sequence $(h_n)_{n=1}^{\infty}$ in $C_c(X)$ such that
\begin{align*}
\|h_n-a\|_\infty\to 0.
\end{align*}
Boundedness of $L$ gives
\begin{align*}
\|Lh_n-La\|_\infty\to 0,
\end{align*}
and boundedness of $g$ gives
\begin{align*}
\|gh_n-ga\|_\infty\le \|g\|_\infty\|h_n-a\|_\infty\to 0.
\end{align*}
Since $Lh_n=gh_n$ for every $n$, [uniqueness of limits](/theorems/625) in the normed space $C_0(X)$ gives
\begin{align*}
La=ga.
\end{align*}
Thus $L=L_g$ on all of $A$.
[/step]
[step:Identify the right multiplier and conclude surjectivity]
We next prove $R=R_g$. Let $h\in C_c(X)$ and $x\in X$. Choose $u_x\in C_c(X)$ with $u_x(x)=1$. The centralizer relation
\begin{align*}
hL(u_x)=R(h)u_x
\end{align*}
evaluated at $x$ gives
\begin{align*}
h(x)(Lu_x)(x)=(Rh)(x).
\end{align*}
Since $g(x)=(Lu_x)(x)$, this becomes
\begin{align*}
(Rh)(x)=h(x)g(x).
\end{align*}
Thus $Rh=hg$ for every $h\in C_c(X)$. The same density argument used for $L$ extends this equality to every $a\in C_0(X)$, so $R=R_g$.
Therefore every multiplier $(L,R)\in M(C_0(X))$ equals $\Theta(g)$ for some $g\in C_b(X)$. Hence $\Theta:C_b(X)\to M(C_0(X))$ is surjective. We already proved that $\Theta$ is an injective unital $*$-homomorphism. Thus $\Theta$ is a bijective unital $*$-homomorphism, and therefore a $*$-isomorphism, so
\begin{align*}
M(C_0(X))\cong C_b(X)
\end{align*}
as unital commutative $C^*$-algebras.
[/step]
