The continuity of linear functionals with respect to uniform convergence of entire functions on balls

Question

Let $X$ be a Banach space and $H_b (X)$ be the algebra of complex-valued entire functions on $X$ which are bounded on bounded sets, with the topology of uniform convergence on bounded sets.

Let $\varphi \in H_b^* (X)$ (which is the dual space of $H_b (X)$).

Each $\varphi \in H_b^* (X)$ is continuous with respect to the norm of uniform convergence on some ball in $X$.

Define the radius function $R$ on $H_b^* (X)$ by declaring $R(\varphi) $ to be the infimum of all $r >0$ such that $\varphi$ is continuous with respect to the norm of uniform convergence on the ball $r B$.

I don't understand the above bold statement. I am a bit confused because I think $\varphi \in H_b^* (X)$ is continuous on every ball in $X$ (so, the radius of $\varphi$ is always $0$).

I want to know what I missed.

Answer 1

For example, consider $X=\mathbb{C}$. The map $\varphi(f) = f(1)$ is a continuous linear functional on $H_b(\mathbb{C})$.

Consider the sequence $f_n(z)=z^n$ of elements of $H_b(\mathbb{C})$. On the disk $\frac12B = \{z:|z|<1/2\}$ this sequence converges uniformly to $0$. However, $\varphi(f_n)=1$ does not converge to $0$. So, $\varphi$ is not continuous with respect to the uniform convergence on $\frac12B$.

Based on the above, one can see that $R(\varphi)=1$: uniform convergence on the unit disk is sufficient for convergence of the values of $\varphi$, while uniform convergence on smaller disks is not.