[proofplan] The statement reduces to the Hermitian case: $x = x^*$ implies $\varphi(x) \in \mathbb{R}$. To prove this, we use the trick of perturbing $x$ along the imaginary axis: for $t \in \mathbb{R}$ the element $z_t = x + it \cdot 1$ has known norm via the $C^*$-identity, namely $\|z_t\|^2 = \|x^2 + t^2 \cdot 1\| \leq \|x\|^2 + t^2$. Comparing this with $|\varphi(z_t)|^2 = |\varphi(x) + it|^2$ produces a polynomial inequality in $t$ that forces the imaginary part of $\varphi(x)$ to vanish. The general case follows from the Hermitian decomposition $x = h + ik$ with $h, k$ Hermitian. The unitary case is then a one-line computation using $x^*x = 1$. [/proofplan] [step:Recall that characters on a unital Banach algebra are continuous of norm one] By [Characters are Continuous](/theorems/2675), a character $\varphi$ on a unital Banach algebra $A$ satisfies $\|\varphi\| = 1$; in particular $\varphi(1) = 1$ and $|\varphi(x)| \leq \|x\|$ for all $x \in A$. Theorem 2675 requires $A$ to be a unital Banach algebra; a $C^*$-algebra is in particular a unital Banach algebra, so the hypothesis is met. We use only $\varphi(1) = 1$ and the bound $|\varphi(z)| \leq \|z\|$ for all $z \in A$. We also note that $\varphi$ is multiplicative ($\varphi(xy) = \varphi(x)\varphi(y)$) and linear by the definition of a character. [/step] [step:Reduce to the Hermitian case by linear decomposition] Every $x \in A$ admits the decomposition $x = h + ik$ with \begin{align*} h &:= \tfrac{1}{2}(x + x^*), & k &:= \tfrac{1}{2i}(x - x^*), \end{align*} both Hermitian: $h^* = \tfrac{1}{2}(x^* + x) = h$ and $k^* = \tfrac{-1}{2i}(x^* - x) = \tfrac{1}{2i}(x - x^*) = k$. Then $x^* = h - ik$. If we know $\varphi(h), \varphi(k) \in \mathbb{R}$ (the Hermitian case, proved in the next step), then by linearity of $\varphi$: \begin{align*} \varphi(x^*) = \varphi(h - ik) = \varphi(h) - i\varphi(k) = \overline{\varphi(h) + i\varphi(k)} = \overline{\varphi(x)}, \end{align*} where the second equality uses $\overline{a + ib} = a - ib$ for $a, b \in \mathbb{R}$. Hence it suffices to prove that $\varphi(h) \in \mathbb{R}$ for every Hermitian $h \in A$. [/step] [step:Prove $\varphi(x) \in \mathbb{R}$ for Hermitian $x$ via an imaginary-axis perturbation] Fix $x = x^* \in A$ Hermitian. Write $\varphi(x) = \alpha + i\beta$ with $\alpha, \beta \in \mathbb{R}$; we shall prove $\beta = 0$. For $t \in \mathbb{R}$ define \begin{align*} z_t := x + it \cdot 1 \in A. \end{align*} By linearity and $\varphi(1) = 1$, \begin{align*} \varphi(z_t) = \varphi(x) + it = \alpha + i(\beta + t), \end{align*} so \begin{align*} |\varphi(z_t)|^2 = \alpha^2 + (\beta + t)^2. \end{align*} We bound $|\varphi(z_t)|^2$ from above using the contractivity $|\varphi(z_t)| \leq \|z_t\|$ and the $C^*$-identity. The $C^*$-identity in $A$ states that $\|y^* y\| = \|y\|^2$ for every $y \in A$; applied with $y = z_t$: \begin{align*} \|z_t\|^2 = \|z_t^* z_t\|. \end{align*} Compute $z_t^* z_t$. Since $x^* = x$ and $(it \cdot 1)^* = -it \cdot 1$ (the involution is conjugate-linear), $z_t^* = x - it \cdot 1$, and \begin{align*} z_t^* z_t = (x - it \cdot 1)(x + it \cdot 1) = x^2 + itx - itx + t^2 \cdot 1 = x^2 + t^2 \cdot 1, \end{align*} where the cross terms cancel because $x$ commutes with $1$. Therefore \begin{align*} \|z_t\|^2 = \|x^2 + t^2 \cdot 1\| \leq \|x^2\| + t^2 \|1\| = \|x^2\| + t^2 \leq \|x\|^2 + t^2, \end{align*} using the triangle inequality, $\|1\| = 1$ in a unital $C^*$-algebra, and submultiplicativity $\|x^2\| \leq \|x\|^2$. Combining, \begin{align*} \alpha^2 + (\beta + t)^2 = |\varphi(z_t)|^2 \leq \|z_t\|^2 \leq \|x\|^2 + t^2. \end{align*} Expanding the left-hand side: $\alpha^2 + \beta^2 + 2\beta t + t^2 \leq \|x\|^2 + t^2$, equivalently \begin{align*} \alpha^2 + \beta^2 + 2\beta t \leq \|x\|^2 \qquad \text{for all } t \in \mathbb{R}. \end{align*} The left-hand side is an affine function of $t$ with slope $2\beta$. If $\beta \neq 0$ the left-hand side is unbounded above as $t \to +\infty$ (if $\beta > 0$) or as $t \to -\infty$ (if $\beta < 0$), contradicting the constant upper bound $\|x\|^2$. Hence $\beta = 0$, i.e.\ $\varphi(x) = \alpha \in \mathbb{R}$. [guided] We want to show $\varphi(x) \in \mathbb{R}$ when $x$ is Hermitian. The strategy is to **isolate the imaginary part** of $\varphi(x)$ and show it must vanish, by exhibiting a one-parameter family of inequalities that constrain it. Write $\varphi(x) = \alpha + i\beta$ with $\alpha, \beta \in \mathbb{R}$. We shall prove $\beta = 0$. **The trick.** Perturb $x$ along the imaginary axis: for $t \in \mathbb{R}$ set $z_t := x + it \cdot 1$. Two things will happen as $t$ varies: 1. $\varphi(z_t)$ varies linearly in $t$, and we can read off how $\beta$ enters. 2. $\|z_t\|^2$ depends on $t^2$ but is **bounded above** by something quadratic in $t$ (via the $C^*$-identity). Comparing $|\varphi(z_t)|^2 \leq \|z_t\|^2$ for all $t \in \mathbb{R}$ then forces $\beta = 0$. **Computing $\varphi(z_t)$.** Since $\varphi$ is linear and $\varphi(1) = 1$ (we recalled this in Step 1 from theorem 2670), $\varphi(z_t) = \varphi(x) + it\varphi(1) = \alpha + i(\beta + t)$. Hence \begin{align*} |\varphi(z_t)|^2 = \alpha^2 + (\beta + t)^2 = \alpha^2 + \beta^2 + 2\beta t + t^2. \end{align*} Notice the cross term $2\beta t$ — that is the term we will exploit. **Computing $\|z_t\|^2$ via the $C^*$-identity.** The $C^*$-identity says $\|y^*y\| = \|y\|^2$ for every $y \in A$. Why use it here? Because it converts a norm of $z_t$ into a norm of the more tractable Hermitian object $z_t^* z_t$. We need $z_t^*$: since $x = x^*$ (Hermiticity, the hypothesis) and the involution is conjugate-linear ($(it \cdot 1)^* = -it \cdot 1$), \begin{align*} z_t^* = (x + it \cdot 1)^* = x^* - it \cdot 1 = x - it \cdot 1. \end{align*} *Without Hermiticity we would have $x^*$ instead of $x$ here, and the next computation would not collapse.* Now \begin{align*} z_t^* z_t = (x - it)(x + it) = x^2 + itx - itx + t^2 = x^2 + t^2 \cdot 1, \end{align*} the imaginary cross-terms cancelling because $x$ commutes with the scalar multiple $it \cdot 1$. The $C^*$-identity gives \begin{align*} \|z_t\|^2 = \|z_t^*z_t\| = \|x^2 + t^2 \cdot 1\|. \end{align*} By the triangle inequality and submultiplicativity: \begin{align*} \|x^2 + t^2 \cdot 1\| \leq \|x^2\| + t^2 \|1\| \leq \|x\|^2 + t^2, \end{align*} using $\|1\| = 1$ in a unital $C^*$-algebra and $\|x^2\| \leq \|x\| \cdot \|x\| = \|x\|^2$. **The forcing inequality.** We use that characters on a unital Banach algebra are contractive (theorem 2675), so $|\varphi(z_t)| \leq \|z_t\|$. Squaring and combining the two computations: \begin{align*} \alpha^2 + \beta^2 + 2\beta t + t^2 \leq \|x\|^2 + t^2 \qquad \text{for all } t \in \mathbb{R}. \end{align*} The $t^2$'s cancel: \begin{align*} \alpha^2 + \beta^2 + 2\beta t \leq \|x\|^2 \qquad \text{for all } t \in \mathbb{R}. \end{align*} **Why this forces $\beta = 0$.** The left-hand side is an affine function of $t$. An affine function $a + bt$ is bounded above on all of $\mathbb{R}$ if and only if its slope $b$ vanishes. Here the slope is $2\beta$, so $\beta = 0$. Hence $\varphi(x) = \alpha \in \mathbb{R}$. **Failure mode.** Why didn't we just say "$\varphi(x) \in \overline{\sigma_A(x)} \subseteq \mathbb{R}$ since the spectrum of a Hermitian element is real"? Because that result is **what we are trying to prove** — the next theorem **Spectral Inclusions and Permanence** uses the present theorem in its proof. We need a direct argument from the $C^*$-identity, and the perturbation $z_t$ is the standard one. [/guided] [/step] [step:Conclude the general formula and the Hermitian/unitary corollaries] Combining the Hermitian case (Step 3) with the decomposition (Step 2): for every $x \in A$, write $x = h + ik$ with $h, k$ Hermitian; then $\varphi(h), \varphi(k) \in \mathbb{R}$ by Step 3, so \begin{align*} \varphi(x^*) = \varphi(h - ik) = \varphi(h) - i\varphi(k) = \overline{\varphi(h) + i\varphi(k)} = \overline{\varphi(x)}. \end{align*} *Hermitian corollary.* If $x = x^*$, then $x = h + ik$ with $k = 0$, so $\varphi(x) = \varphi(h) \in \mathbb{R}$. *Unitary corollary.* If $x$ is unitary, i.e.\ $x^* x = x x^* = 1$, then by multiplicativity of $\varphi$ and the formula just proved: \begin{align*} |\varphi(x)|^2 = \varphi(x) \overline{\varphi(x)} = \varphi(x) \varphi(x^*) = \varphi(x x^*) = \varphi(1) = 1, \end{align*} where $\varphi(1) = 1$ by Step 1. Hence $|\varphi(x)| = 1$, i.e.\ $\varphi(x) \in S^1$. [/step]