[proofplan] We first prove the estimate when both algebras are unital, without assuming that the homomorphism preserves units. The image of the unit is a projection $p$, so the range of $\varphi$ lies in the corner $pBp$, where $\varphi$ becomes a unital $*$-homomorphism. Spectral inclusion in this corner gives the norm estimate for self-adjoint elements, and the $C^*$-identity upgrades it to all elements. The general case follows by applying the unital case to the induced homomorphism between unitizations. [/proofplan] [step:Pass a nonunital homomorphism between unital algebras to a corner where it is unital] Assume first that $A$ and $B$ are unital $C^*$-algebras, with units $1_A$ and $1_B$. Define \begin{align*} p:=\varphi(1_A)\in B. \end{align*} Since $\varphi$ is multiplicative and preserves the involution, \begin{align*} p^2=\varphi(1_A)^2=\varphi(1_A^2)=\varphi(1_A)=p \end{align*} and \begin{align*} p^*=\varphi(1_A)^*=\varphi(1_A^*)=\varphi(1_A)=p. \end{align*} Thus $p$ is a projection in $B$. Let $pBp$ denote the corner \begin{align*} pBp:=\{pbp:b\in B\}\subset B, \end{align*} equipped with the inherited norm, multiplication, and involution. This is a unital $C^*$-algebra with unit $p$. For every $a\in A$, \begin{align*} p\varphi(a)p=\varphi(1_A)\varphi(a)\varphi(1_A)=\varphi(1_Aa1_A)=\varphi(a), \end{align*} so $\varphi(A)\subset pBp$. Hence the same map, now viewed as \begin{align*} \varphi:A\to pBp, \end{align*} is a unital $*$-homomorphism, because $\varphi(1_A)=p$, the unit of $pBp$. [/step] [step:Prove spectral inclusion for self-adjoint elements in the corner] Let $x\in A$ be self-adjoint. We prove \begin{align*} \sigma_{pBp}(\varphi(x))\subseteq \sigma_A(x), \end{align*} where $\sigma_A(x)$ is the spectrum of $x$ in $A$ and $\sigma_{pBp}(\varphi(x))$ is the spectrum of $\varphi(x)$ in the unital algebra $pBp$. Let $\lambda\in \mathbb C\setminus \sigma_A(x)$. Then $\lambda 1_A-x$ is invertible in $A$. Define \begin{align*} y:=(\lambda 1_A-x)^{-1}\in A. \end{align*} Since $\varphi:A\to pBp$ is unital, multiplicative, and complex-linear, \begin{align*} (\lambda p-\varphi(x))\varphi(y)=\varphi(\lambda 1_A-x)\varphi(y)=\varphi((\lambda 1_A-x)y)=\varphi(1_A)=p. \end{align*} Similarly, \begin{align*} \varphi(y)(\lambda p-\varphi(x))=\varphi(y)\varphi(\lambda 1_A-x)=\varphi(y(\lambda 1_A-x))=\varphi(1_A)=p. \end{align*} Therefore $\lambda p-\varphi(x)$ is invertible in $pBp$, with inverse $\varphi(y)$. Hence $\lambda\notin \sigma_{pBp}(\varphi(x))$, which proves the spectral inclusion. [guided] The point of passing to $pBp$ is that spectra are defined relative to a unit. The element $\varphi(x)$ may not be tested against $1_B$ in the right way if $\varphi$ is not unital, but it is tested against the unit $p$ of the corner $pBp$. Let $x\in A$ be self-adjoint. We want to show that every resolvent point of $x$ remains a resolvent point of $\varphi(x)$ in the corner. Thus take \begin{align*} \lambda\in \mathbb C\setminus \sigma_A(x). \end{align*} By the definition of the spectrum in the unital algebra $A$, the element $\lambda 1_A-x$ is invertible in $A$. Define its inverse by \begin{align*} y:=(\lambda 1_A-x)^{-1}\in A. \end{align*} Then \begin{align*} (\lambda 1_A-x)y=1_A \end{align*} and \begin{align*} y(\lambda 1_A-x)=1_A. \end{align*} Because $\varphi:A\to pBp$ is complex-linear, multiplicative, and sends $1_A$ to the unit $p$ of $pBp$, it sends the element $\lambda 1_A-x$ to \begin{align*} \varphi(\lambda 1_A-x)=\lambda p-\varphi(x). \end{align*} Applying $\varphi$ to the first inverse identity gives \begin{align*} (\lambda p-\varphi(x))\varphi(y)=\varphi((\lambda 1_A-x)y)=\varphi(1_A)=p. \end{align*} Applying $\varphi$ to the second inverse identity gives \begin{align*} \varphi(y)(\lambda p-\varphi(x))=\varphi(y(\lambda 1_A-x))=\varphi(1_A)=p. \end{align*} Thus $\varphi(y)$ is a two-sided inverse for $\lambda p-\varphi(x)$ in the unital algebra $pBp$. Therefore $\lambda\notin \sigma_{pBp}(\varphi(x))$. Since every $\lambda\notin \sigma_A(x)$ lies outside $\sigma_{pBp}(\varphi(x))$, we have \begin{align*} \sigma_{pBp}(\varphi(x))\subseteq \sigma_A(x). \end{align*} [/guided] [/step] [step:Convert spectral inclusion into a norm estimate for self-adjoint elements] Let $x\in A$ be self-adjoint. Since $\varphi$ preserves the involution, \begin{align*} \varphi(x)^*=\varphi(x^*)=\varphi(x), \end{align*} so $\varphi(x)$ is self-adjoint in $pBp$. For a self-adjoint element $z$ in a $C^*$-algebra, the norm is equal to its spectral radius: \begin{align*} \|z\|=r(z):=\sup\{|\lambda|:\lambda\in\sigma(z)\}. \end{align*} Applying this formula in $pBp$, using the spectral inclusion from the previous step, and then applying the same formula in $A$, we obtain \begin{align*} \|\varphi(x)\|_{pBp}=r_{pBp}(\varphi(x))\leq r_A(x)=\|x\|_A. \end{align*} The norm on $pBp$ is inherited from $B$, so \begin{align*} \|\varphi(x)\|_B=\|\varphi(x)\|_{pBp}\leq \|x\|_A. \end{align*} Thus $\varphi$ is contractive on self-adjoint elements of $A$. [/step] [step:Apply the self-adjoint estimate to $a^*a$ and use the $C^*$-identity] Let $a\in A$. The element $a^*a$ is self-adjoint because \begin{align*} (a^*a)^*=a^*a. \end{align*} By the self-adjoint estimate, \begin{align*} \|\varphi(a^*a)\|_B\leq \|a^*a\|_A. \end{align*} Using preservation of multiplication and involution, and then the $C^*$-identity in $B$ and $A$, we get \begin{align*} \|\varphi(a)\|_B^2=\|\varphi(a)^*\varphi(a)\|_B. \end{align*} Also, \begin{align*} \varphi(a)^*\varphi(a)=\varphi(a^*)\varphi(a)=\varphi(a^*a). \end{align*} Therefore \begin{align*} \|\varphi(a)\|_B^2=\|\varphi(a^*a)\|_B\leq \|a^*a\|_A=\|a\|_A^2. \end{align*} Taking nonnegative square roots gives \begin{align*} \|\varphi(a)\|_B\leq \|a\|_A. \end{align*} This proves the theorem when $A$ and $B$ are unital. [/step] [step:Reduce arbitrary $C^*$-algebras to the unital case by unitization] Now let $A$ and $B$ be arbitrary complex $C^*$-algebras. Let $A^+$ and $B^+$ denote their standard $C^*$-unitizations, and regard $A$ and $B$ as closed two-sided ideals in $A^+$ and $B^+$ by the canonical isometric embeddings \begin{align*} a\mapsto (a,0) \end{align*} and \begin{align*} b\mapsto (b,0). \end{align*} By the universal property of unitization, [citetheorem:8573], the $*$-homomorphism $\varphi:A\to B\subset B^+$ extends to a unital $*$-homomorphism \begin{align*} \varphi^+:A^+\to B^+ \end{align*} defined by \begin{align*} \varphi^+(a,\lambda)=(\varphi(a),\lambda) \end{align*} for every $a\in A$ and every $\lambda\in\mathbb C$. Applying the unital case to $\varphi^+$ gives, for every $a\in A$, \begin{align*} \|\varphi^+(a,0)\|_{B^+}\leq \|(a,0)\|_{A^+}. \end{align*} Because the canonical embeddings of $A$ and $B$ into their unitizations are isometric, \begin{align*} \|\varphi(a)\|_B=\|\varphi^+(a,0)\|_{B^+} \end{align*} and \begin{align*} \|(a,0)\|_{A^+}=\|a\|_A. \end{align*} Hence \begin{align*} \|\varphi(a)\|_B\leq \|a\|_A. \end{align*} Since this holds for every $a\in A$, $\varphi$ is contractive. A contractive [linear map](/page/Linear%20Map) between normed spaces is continuous, so $\varphi$ is continuous. [/step]