[proofplan] We treat the Hermitian case first and reduce the general normal case to it. For Hermitian $x$ (i.e.\ $x = x^*$), the $C^*$-identity $\|y^*y\| = \|y\|^2$ applied with $y = x$ gives $\|x^2\| = \|x\|^2$; iterating $n$ times yields $\|x^{2^n}\| = \|x\|^{2^n}$, and the [Beurling–Gelfand Spectral Radius Formula](/theorems/2672) $r_A(x) = \lim_n \|x^n\|^{1/n}$ then forces $r_A(x) = \|x\|$. For general normal $x$, the element $x^*x$ is Hermitian, so the previous case gives $r_A(x^*x) = \|x^*x\| = \|x\|^2$ (the second equality is the $C^*$-identity applied to $x$). Submultiplicativity of the spectral radius for commuting elements ([Spectral Radius is Subadditive and Submultiplicative for Commuting Elements](/theorems/2679)) — applicable here because $x$ and $x^*$ commute (definition of normal) — combined with the always-true bound $r_A(\cdot) \le \|\cdot\|$ and the (always-true) identity $r_A(x^*) = r_A(x)$, then sandwiches $\|x\| \le r_A(x) \le \|x\|$. [/proofplan] [step:Hermitian case: prove $\|x^{2^n}\| = \|x\|^{2^n}$ by induction] Assume $x = x^*$ (Hermitian). The $C^*$-identity in $A$ states that for every $y \in A$, \begin{align*} \|y^*y\|_A = \|y\|_A^2. \end{align*} Apply this with $y = x$: \begin{align*} \|x^2\|_A = \|x^* x\|_A = \|x\|_A^2. \end{align*} Note that $x^2$ is again Hermitian: $(x^2)^* = (x^*)^2 = x^2$ (using $x^* = x$). So we may iterate. We claim $\|x^{2^n}\|_A = \|x\|_A^{2^n}$ for every $n \in \mathbb{N}$. *Base case $n = 0$:* $\|x^{2^0}\| = \|x^1\| = \|x\| = \|x\|^{2^0}$. *Inductive step:* Suppose $\|x^{2^n}\|_A = \|x\|_A^{2^n}$. Set $y_n := x^{2^n}$. Then $y_n^* = (x^{2^n})^* = (x^*)^{2^n} = x^{2^n} = y_n$, so $y_n$ is Hermitian. The $C^*$-identity applied to $y_n$ gives \begin{align*} \|y_n^2\|_A = \|y_n^* y_n\|_A = \|y_n\|_A^2. \end{align*} But $y_n^2 = (x^{2^n})^2 = x^{2^{n+1}}$, so \begin{align*} \|x^{2^{n+1}}\|_A = \|y_n^2\|_A = \|y_n\|_A^2 = (\|x\|_A^{2^n})^2 = \|x\|_A^{2^{n+1}}, \end{align*} completing the induction. [/step] [step:Hermitian case: conclude $r_A(x) = \|x\|$ via the spectral radius formula] By the [Beurling–Gelfand Spectral Radius Formula](/theorems/2672), for every $x$ in a unital Banach algebra $A$, \begin{align*} r_A(x) = \lim_{n \to \infty} \|x^n\|_A^{1/n}. \end{align*} Theorem 2672 requires only that $A$ be a unital Banach algebra and $x \in A$. A $C^*$-algebra is in particular a Banach algebra, and $x \in A$ is given, so the hypothesis is met. Specialise to the subsequence $n = 2^k$, $k = 0, 1, 2, \dots$ (a subsequence of a convergent sequence has the same limit): \begin{align*} r_A(x) = \lim_{k \to \infty} \|x^{2^k}\|_A^{1/2^k}. \end{align*} By the previous step, $\|x^{2^k}\|_A = \|x\|_A^{2^k}$, so $\|x^{2^k}\|_A^{1/2^k} = \|x\|_A$ for every $k$. The limit of a constant sequence is the constant: \begin{align*} r_A(x) = \|x\|_A. \end{align*} This proves the Hermitian case. [guided] The strategy is to extract from the $C^*$-identity a doubling phenomenon $\|x^2\| = \|x\|^2$, iterate it to $\|x^{2^n}\| = \|x\|^{2^n}$, and feed the result into the [Beurling–Gelfand Spectral Radius Formula](/theorems/2672), which expresses the spectral radius as the geometric-mean limit $r_A(x) = \lim_n \|x^n\|^{1/n}$. **Why does the doubling iterate?** The $C^*$-identity $\|y^* y\| = \|y\|^2$ applied to $y = x$ gives $\|x^* x\| = \|x\|^2$, and Hermiticity $x^* = x$ converts this into $\|x^2\| = \|x\|^2$. The crucial closure property is that $x^2$ is again Hermitian: $(x^2)^* = (x^*)^2 = x^2$. So we can apply the $C^*$-identity again with $y = x^2$, obtaining $\|x^4\| = \|x^2\|^2 = \|x\|^4$. Inductively, $y_n := x^{2^n}$ is Hermitian for every $n$, and the $C^*$-identity applied to $y_n$ doubles the exponent: \begin{align*} \|x^{2^{n+1}}\| = \|y_n^2\| = \|y_n^* y_n\| = \|y_n\|^2 = (\|x\|^{2^n})^2 = \|x\|^{2^{n+1}}. \end{align*} **Why use the $2^n$ subsequence?** The Beurling-Gelfand formula gives the limit along the **whole** sequence $\|x^n\|^{1/n}$. Computing $\|x^n\|$ for arbitrary $n$ in a Banach algebra is hard. But the $C^*$-doubling controls $\|x^{2^n}\|$ exactly. Convergence of a sequence $a_n$ to $L$ implies convergence of any subsequence to $L$, so $\|x^{2^n}\|^{1/2^n} \to r_A(x)$. We have just computed $\|x^{2^n}\|^{1/2^n} = \|x\|$ for every $n$, so $r_A(x) = \|x\|$. **What if $x$ were not Hermitian?** Then $x^2$ might not equal $x^* x$, and the doubling fails. Concretely, in a $C^*$-algebra one can have nilpotent (so $x^2 = 0$ but $x \neq 0$, hence $\|x^2\| = 0 \neq \|x\|^2$) elements that are not Hermitian. So Hermiticity is essential at this stage; we will need a separate argument to bootstrap from Hermitian to general normal. [/guided] [/step] [step:General normal case: reduce to the Hermitian case via $x^*x$] Assume $x \in A$ is normal: $x^* x = x x^*$. The element $x^* x$ is Hermitian: $(x^* x)^* = x^* (x^*)^* = x^* x$. Hence the Hermitian case (Steps 1–2) applies to $x^* x$: \begin{align*} r_A(x^* x) = \|x^* x\|_A. \tag{$\heartsuit$} \end{align*} Apply the $C^*$-identity to $x$: \begin{align*} \|x^* x\|_A = \|x\|_A^2, \end{align*} so $(\heartsuit)$ becomes \begin{align*} r_A(x^* x) = \|x\|_A^2. \tag{$\spadesuit$} \end{align*} Now we bound $r_A(x^* x)$ from above using submultiplicativity of the spectral radius for commuting elements. Apply [Spectral Radius is Subadditive and Submultiplicative for Commuting Elements](/theorems/2679) to the pair $(x^*, x)$. Theorem 2679 requires $A$ a unital Banach algebra (verified — $C^*$-algebra) and two **commuting** elements (verified — $x$ is normal, so $x^* x = x x^*$). Theorem 2679 gives \begin{align*} r_A(x^* x) \le r_A(x^*)\, r_A(x). \tag{$\clubsuit$} \end{align*} We now use two further facts: 1. $r_A(x^*) = r_A(x)$. *Proof:* The spectrum satisfies $\sigma_A(x^*) = \overline{\sigma_A(x)} = \{\overline{\lambda} : \lambda \in \sigma_A(x)\}$ (this is a standard fact: $z \cdot 1 - x^* = (\overline{z}\cdot 1 - x)^*$, and the involution preserves invertibility because $(y^{-1})^* = (y^*)^{-1}$ in a $*$-algebra; so $z \cdot 1 - x^*$ is invertible iff $\overline{z}\cdot 1 - x$ is invertible, i.e.\ $z \in \sigma_A(x^*)$ iff $\overline{z} \in \sigma_A(x)$). Conjugation preserves moduli, so $\sup\{|z| : z \in \sigma_A(x^*)\} = \sup\{|\overline{\lambda}| : \lambda \in \sigma_A(x)\} = r_A(x)$. 2. $r_A(x) \le \|x\|_A$. *Proof:* Always true in a Banach algebra: if $|\lambda| > \|x\|_A$ then $\lambda \cdot 1 - x = \lambda(1 - x/\lambda)$ is invertible by the Neumann series (since $\|x/\lambda\| < 1$), so $\lambda \notin \sigma_A(x)$. Hence $\sigma_A(x) \subseteq \{|\lambda| \le \|x\|_A\}$. Combining ($\clubsuit$) with these two facts: \begin{align*} r_A(x^* x) \le r_A(x^*)\, r_A(x) = r_A(x) \cdot r_A(x) = r_A(x)^2 \le r_A(x) \cdot \|x\|_A. \end{align*} With ($\spadesuit$), this gives \begin{align*} \|x\|_A^2 = r_A(x^* x) \le r_A(x) \cdot \|x\|_A. \end{align*} If $\|x\|_A = 0$ then $x = 0$ and $r_A(0) = 0 = \|0\|_A$, so the conclusion holds. Otherwise, $\|x\|_A > 0$ and we may divide both sides by $\|x\|_A$: \begin{align*} \|x\|_A \le r_A(x). \end{align*} The reverse inequality $r_A(x) \le \|x\|_A$ is fact 2 above. Hence $r_A(x) = \|x\|_A$, completing the proof of the general normal case. [guided] The general normal case reduces to the Hermitian case by the **standard $C^*$ trick**: control $\|x\|^2$ via the $C^*$-identity and the Hermitian element $x^*x$. We have: - The $C^*$-identity $\|x^*x\| = \|x\|^2$ — converts the norm-squared of $x$ to the norm of the Hermitian $x^*x$. - The Hermitian case applied to $x^*x$ — converts $\|x^*x\|$ to $r_A(x^*x)$. - Submultiplicativity for commuting elements applied to the pair $(x^*, x)$ — converts $r_A(x^*x)$ to $r_A(x^*) \cdot r_A(x)$. - The fact $r_A(x^*) = r_A(x)$ — symmetry of the spectrum under conjugation in a $*$-algebra. **Verification that submultiplicativity applies.** Theorem 2679 requires the two factors to commute. Here the factors are $x^*$ and $x$, and they commute exactly because $x$ is normal. *This is the only place where normality enters.* If $x$ failed to be normal, $x^*$ and $x$ would not commute, theorem 2679 would not apply, and the argument would collapse — and indeed the conclusion $r_A(x) = \|x\|$ fails for non-normal elements (e.g.\ a strict triangular nilpotent has $r_A = 0$ but positive norm). **Why $r_A(x^*) = r_A(x)$ in a $C^*$-algebra?** Because the involution preserves invertibility: $(y^*)^{-1} = (y^{-1})^*$. So $\lambda \cdot 1 - x^* = (\overline{\lambda}\cdot 1 - x)^*$ is invertible iff $\overline{\lambda}\cdot 1 - x$ is invertible. Thus $\sigma_A(x^*) = \overline{\sigma_A(x)}$, and complex conjugation preserves absolute values, giving the same supremum. **Putting the chain together.** We have \begin{align*} \|x\|^2 = \|x^*x\| \overset{\text{Hermitian case}}{=} r_A(x^*x) \overset{\text{2679}}{\le} r_A(x^*)\, r_A(x) = r_A(x)^2 \le r_A(x)\,\|x\|. \end{align*} The last step uses the universal Banach-algebra bound $r_A(x) \le \|x\|$. Cancelling $\|x\|$ (when non-zero), $\|x\| \le r_A(x)$. Combined with $r_A(x) \le \|x\|$, equality holds. The case $\|x\| = 0$ is also fine because then $x = 0$ and both sides vanish. [/guided] [/step]