[proofplan] We prove the three spectral restrictions separately. For a self-adjoint element, we use the standard C*-algebraic real-spectrum theorem for self-adjoint elements, whose proof rests on the C*-identity and the Banach-algebra resolvent argument. For a unitary, we prove directly that every complex number off the unit circle lies in the resolvent by factoring $u-\lambda 1_A$ and applying the [Neumann series theorem](/theorems/8545). For a projection, we avoid polynomial spectral mapping and explicitly write an inverse for $p-\lambda 1_A$ whenever $\lambda\notin\{0,1\}$. [/proofplan]

[step:Place the spectrum of a self-adjoint element on the real line]Let $a\in A$ be self-adjoint, so $a^*=a$. By the standard real-spectrum theorem for self-adjoint elements in a unital C*-algebra, every spectral value of $a$ is real. This theorem applies because $A$ is a unital complex C*-algebra and $a=a^*$. Hence \begin{align*} \sigma(a)\subset \mathbb R. \end{align*}[/step]

[guided]Let $a\in A$ be self-adjoint, meaning $a^*=a$. The relevant structural fact about C*-algebras is the real-spectrum theorem for self-adjoint elements: in a unital complex C*-algebra, if an element equals its adjoint, then its spectrum is contained in the real line. The hypotheses of that theorem are exactly the hypotheses in force here: $A$ is unital, $A$ is complex, $A$ is a C*-algebra, and $a=a^*$. Therefore every $\lambda\in\sigma(a)$ satisfies $\lambda\in\mathbb R$, which is precisely \begin{align*} \sigma(a)\subset \mathbb R. \end{align*} Conceptually, this is the C*-algebraic version of the Hilbert-space fact that [self-adjoint operators](/page/Self-Adjoint%20Operators) have real spectral values.[/guided]

[step:Invert $u-\lambda 1_A$ outside the unit circle] Let $u\in A$ be unitary, so $u^*u=uu^*=1_A$. In particular, $u^{-1}=u^*$. The C*-identity gives \begin{align*} \|u\|_A^2=\|u^*u\|_A=\|1_A\|_A=1, \end{align*} so $\|u\|_A=1$. Similarly, \begin{align*} \|u^{-1}\|_A=\|u^*\|_A=\|u\|_A=1. \end{align*} Let $\lambda\in\mathbb C$ with $|\lambda|>1$. Define $a_\lambda\in A$ by \begin{align*} a_\lambda=\lambda^{-1}u. \end{align*} Then \begin{align*} \|a_\lambda\|_A=|\lambda|^{-1}\|u\|_A=|\lambda|^{-1}<1. \end{align*} By the [Neumann Series Theorem][citetheorem:8545], $1_A-a_\lambda$ is invertible. Since \begin{align*} u-\lambda 1_A=-\lambda(1_A-\lambda^{-1}u), \end{align*} the element $u-\lambda 1_A$ is invertible. Now let $\lambda\in\mathbb C$ with $0<|\lambda|<1$. Define $b_\lambda\in A$ by \begin{align*} b_\lambda=\lambda u^*. \end{align*} Then \begin{align*} \|b_\lambda\|_A=|\lambda|\|u^*\|_A=|\lambda|<1. \end{align*} Again by the [Neumann Series Theorem][citetheorem:8545], $1_A-b_\lambda$ is invertible. Since \begin{align*} u-\lambda 1_A=u(1_A-\lambda u^*), \end{align*} and both $u$ and $1_A-\lambda u^*$ are invertible, $u-\lambda 1_A$ is invertible. Finally, $\lambda=0$ is not spectral because $u$ is invertible. Thus every $\lambda\in\sigma(u)$ satisfies $|\lambda|=1$, and therefore \begin{align*} \sigma(u)\subset \{z\in\mathbb C: |z|=1\}. \end{align*} [/step]

[step:Write an explicit resolvent for a projection] Let $p\in A$ be a projection, so $p^2=p$. Let $\lambda\in\mathbb C\setminus\{0,1\}$. Define $q_\lambda\in A$ by \begin{align*} q_\lambda=\frac{1}{1-\lambda}p-\frac{1}{\lambda}(1_A-p). \end{align*} This element is well-defined because $\lambda\ne 0$ and $\lambda\ne 1$. Using $p^2=p$, we have \begin{align*} p(1_A-p)=p-p^2=0 \end{align*} and \begin{align*} (1_A-p)p=p-p^2=0. \end{align*} Hence \begin{align*} (p-\lambda 1_A)q_\lambda = (p-\lambda p-\lambda(1_A-p)) \left(\frac{1}{1-\lambda}p-\frac{1}{\lambda}(1_A-p)\right). \end{align*} The mixed products vanish, so this product equals \begin{align*} (1-\lambda)\frac{1}{1-\lambda}p + (-\lambda)\left(-\frac{1}{\lambda}\right)(1_A-p) = p+(1_A-p) = 1_A. \end{align*} The same computation, with the factors reversed and using the same two vanishing mixed products, gives \begin{align*} q_\lambda(p-\lambda 1_A)=1_A. \end{align*} Therefore $p-\lambda 1_A$ is invertible whenever $\lambda\notin\{0,1\}$. Equivalently, \begin{align*} \sigma(p)\subset \{0,1\}. \end{align*} [/step]