Continuity of $\limsup$ related to Cauchy-Hadamard formula I'm investigating to what extent the Cauchy-Hadamard theorem can be generalized to Banach algebras. Let $B$ be a real Banach algebra, i.e. a $\mathbb{R}$-algebra with unity and equipped with a norm $\lVert . \rVert : B \to [0, +\infty[$ which has the additional properties that
$$
\lVert 1 \rVert = 1 \text{ and} \\
\forall x, y \in B : \lVert x y \rVert \leq \lVert x \rVert \lVert y \rVert.
$$
Given a sequence $(c_n)_{n \in \mathbb{N}} \in B^{\mathbb{N}}$ for which
$$
L = \limsup_{n \to +\infty} \lVert c_n \rVert^{1/n} < +\infty
$$
we can define the map
$$
\tau : B \to [0, +\infty[ : z \mapsto \limsup_{n \to +\infty} \lVert c_n z^n \rVert^{1/n} .
$$
Indeed, we have
$$
\limsup_{n \to +\infty} \lVert c_n z^n \rVert^{1/n} \leq \limsup_{n \to +\infty} \lVert c_n \rVert^{1/n} \lVert z \rVert = L\lVert z \rVert < +\infty
$$
for each $z \in B$. The above inequality also proves that $\tau$ is continuous in $0$, but I'm wondering whether $\tau$ is also continuous elsewhere. If $B = \mathbb{R}$ or $B = \mathbb{C}$, then the first inequality above is an equality and this is obvious; this makes for the easy Cauchy-Hadamard formula in these cases. Also note that $L < +\infty$ is a necessary condition: otherwise $\tau$ takes the value $0$ at $z = 0$, but $+\infty$ for $z \in \mathbb{R} \setminus \lbrace 0 \rbrace$.
Both proofs or counterexamples for the general case are welcome; my gut feeling says that $\tau$ should be continuous. If needed, the extra assumption that $B$ is commutative can be made for the proof. EXTRA: (and of minor importance) Is the map $\tau$ open if $L > 0$?
 A: $\lVert x^n \rVert^{\frac{1}{n}}$ always converges to the spectral radius $\rho(x)$. So your $\tau$ is just a multiple of that in the case that the $c_n$ are multiples of $1$. In general the spectral radius is not continuous but only upper semi continuous.
Counterexample (due to Kakutani): Let $H$ be a real, separable, infinite-dimensional Hilbert space and $B=B(H)$ the Banach algebra of bounded operators on $H$. Let $x=(x_n)$ be a sequence of positive numbers with $x_n \to 0$ that we will choose later. Use $x$ to define $y=(y_n)$ in the following way: Set $y_n := x_k$ where $k$ is the number of consecutive $1$'s at the end of the binary representation of $n$. I.e.,
$$ y = (x_0, x_1, x_0, x_2, x_0, x_1, x_0, x_3, ...).$$
Now, fix an orthonormal basis (e_n) on $H$. Subject to this basis, define $A \in B(H)$ to be the diagonal multiplication operator with multiplier $y$ followed by a (right-)shift. If $h \in H$ with $h = \sum_n h_n e_n$, then $Ah = \sum_n y_n h_n e_{n+1}$. In the same spirit define $A_l$ with $y^l$ instead of $y$, where $y^l$ is defined be exchanging all the $x_l$ with zeros.
Clearly, $(A_l)^{2^{l+1}} = 0$ so that $\rho(A_l)=0$. At the same time $\lVert A_l -A \rVert = x_l \to 0$. All that is left now is to show that $x$ can be chosen such that $\rho(A) > 0$.
We can calculate
\begin{align*} \rho(A) &= \lim_n \lVert A^n \rVert^{\frac{1}{n}}\\
&= \lim_m \lVert A^{2^m-1} \rVert^{\frac{1}{2^m-1}}\\
&\geq \limsup_m \Big(\prod_{i=0}^{2^m-2} y_i\Big)^{\frac{1}{2^m-1}}\\
&= \limsup_m \Big(\prod_{l=0}^{m-1} x_l^{2^{-1-l}} \Big)^{\frac{2^m}{2^m-1}}\\
&= \prod_{l=0}^{\infty} x_l^{2^{-1-l}}
 \end{align*}
The simple choice of $x_l= \frac{1}{l+1}$ (or $x_l = \frac{1}{2^{l+1}}$) makes the last infinite product absolutely convergent. In particular, we then have $\rho(A)>0$. 
