Recall that a topological ring $A$ is Tate if there is an open subring $A_0$ such that the induced topology on $A_0$ is t-adic for some $t \in A_0$ that becomes a unit in $A.$ One can, given a Tate ring define a notion of powerbounded elements, denote the ring of powerbounded elements by $A^\circ.$ One says that the Tate ring $A$ is uniform if $A^\circ$ is bounded.
Is there an easy example of a Tate ring $A$ which is complete and reduced but not uniform?
While I do not have an idea as to how easy or difficult the following is compared to other examples, here is an example:
Let $A_0$ be the subring of $\mathbb{Z}_p[[T]]$ consisting of all the power series $\sum_i a_iT^i$ satisfying the condition $$v_p(a_i)\geq \sqrt{i}$$ (here $v_p$ denotes the $p$-adic valuation written additively). Note that this is a ring - to check that $A_0$ is closed under multiplication, one uses the fact that the function $\sqrt{x}$ is concave, hence subadditive. It is easy to see that $A_0$ is $p$-complete (as an additive group, it is $\prod_ip^{\lceil{\sqrt{i}}\rceil}\mathbb{Z}_p$). The Tate ring in question is then $A=A_0[\frac{1}{p}]$, with the topology given by $(A_0, p)$. Note that $A$ contains $\mathbb{Q}_p[T]$ as a subring.
To demonstrate some power-bounded elements, note that $p\mathbb{Z}_p[T] \subseteq A^{\circ}$. This follows from the fact that for a polynomial $f=\sum_{i=0}^n p a_iT^i$ with $a_i \in \mathbb{Z}_p$, we have $f^N \in A_0$ for all large enough $N$. To see that, note that $$f^N=p^N\left(\sum_{i=0}^na_iT^i\right)^N,$$ and so for $f^N$ to be in $A_0,$ it is enough to have $N \geq \sqrt{nN},$ which is achieved by taking any $N \geq n.$
However, the set $p\mathbb{Z}_p[T]$ is not itself bounded, since it contains e.g. the unbounded set $$pT, pT^2, pT^3, \dots$$ Thus, $A$ is not uniform.
