[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.[/step]

[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: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]