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If an entire Banach algebra-valued function has bounded spectral radius, that spectral radius is constant. The spectral-radius three-circle inequality on circles about an arbitrary point, with the outer radius tending to infinity, bounds every interior value by the maximum on any smaller circle. Upper semicontinuity of the spectral radius then lets the smaller radius tend to zero. Every value is bounded by the value at the arbitrary center; exchanging centers gives equality. The algebra-valued function itself can be nonconstant, for example $zN$ for a fixed nonzero square-zero matrix $N$.

For a Banach algebra-valued holomorphic function, set $p_n(w)=\Vert f(w{)}^{2^n}{\Vert }^{1/2^n}$. These continuous functions decrease to the spectral radius by the spectral radius formula. If $M={\sup }_{\partial K}r(f(w))$, the continuous functions $\max \{p_n,M\}$ on $\partial K$ decrease to the continuous constant $M$. The Dini theorem gives uniform convergence. Apply the norm maximum principle for a Banach-space-valued holomorphic function to each power, and then let $n\to \infty$. This proof does not assume continuity of spectral radius.

For a holomorphic Banach algebra-valued function, put $M_r(f)={\sup }_{|z|=r}r(f(z))$. Apply the Hadamard three-circle theorem to the holomorphic powers $f^{2^n}$. The continuous functions $\Vert f(z{)}^{2^n}{\Vert }^{1/2^n}$ decrease to the spectral radius. Supremum convergence for decreasing continuous functions passes the boundary suprema to the limit, proving the inequality with the same logarithmic interpolation parameter. In particular, at the geometric-mean radius, the square of the middle spectral radius is at most the product of the two boundary suprema.

For a commutative complex unital Banach algebra, a character of an algebra annihilates $a-\varphi (a)1$, so this element cannot be invertible. Conversely a noninvertible $a-\lambda 1$ generates a proper ideal, which lies in a maximal ideal. The corresponding character then has value $\lambda$ at $a$. Thus the Gelfand transform determines every element's spectrum even when the algebra is not semisimple.

The identity $(x+\alpha e)-\lambda e=x-(\lambda -\alpha )e$ proves the formula directly from invertibility. In a nonzero complex unital Banach algebra, the nonemptiness of the Banach-algebra spectrum implies that if $x$ is a quasinilpotent element, then $\rho (x+e)=1$. Consequently every nonscalar element cannot simultaneously be quasinilpotent: adding the identity preserves nonscalarity while translating its spectrum.

Truncated convolution makes $L^1(0,1)$ a commutative nonunital Banach algebra. The Tonelli theorem gives $\Vert f\ast g{\Vert }_1\le \Vert f{\Vert }_1\Vert g{\Vert }_1$, and the Fubini theorem gives associativity over the integration simplex. Its unitization is $\mathbb{C}\oplus L^1(0,1)$ with norm $|\alpha |+\Vert f{\Vert }_1$ and product $(\alpha ,f)(\beta ,g)=(\alpha \beta ,\alpha g+\beta f+f\ast g)$. The constant function one has convolution powers $t^{n-1}/(n-1)!$, of norm $1/n!$. Thus it is a nonnilpotent quasinilpotent element in this unitization.