'Partial boundedness' of continuously parametrised power series Let $a_1, a_2, \ldots : D\rightarrow\mathbb{R}$ be a sequence of continuous functions, with $D$ a compact metric space.
Suppose that the function $f : D\times[0,\infty)\rightarrow\mathbb{R}$ given by
$$\tag{1}f(x,y) := \sum_{i=1}^\infty a_i(x)\cdot y^i$$
exists and is such that its $y$-section $f_{y_0}:=f(\,\cdot,y_0)$ is bounded on $D$ for some $y_0>0$.
Question: Can we conclude that the sections $f_{y}$ are bounded on $D$ for any $y\geq y_0$?
If necessary, it may be assumed that the $a_i$ are all non-negative.
 A: The answer is no. Indeed, let $D:=[0,1]$ and
\begin{equation}
    a_n(x):=nx\,1(x\le1/n)+(2-nx)1(1/n<x<2/n)
\end{equation}
for natural $n$ and $x\in D$.
Then $a_n$ is continuous for each $n$, the sum
\begin{equation}
    f(x,y)=\sum_{n=1}^\infty a_n(x)y^n
\end{equation}
has only finitely many nonzero summands for each $(x,y)\in D\times[0,\infty)$ and hence takes real values.
Also, $0\le a_n\le1$ and hence
\begin{equation}
    0\le f(x,y)=\sum_{n=1}^\infty a_n(x)y^i\le1
\end{equation}
for $y=1/2$. So, $f(\cdot,1/2)$ is bounded.
However, for any natural $k$ and $x=1/k$,
\begin{equation}
    f(x,2)\ge\sum_{n=1}^\infty nx\,1(x\le1/n)2^n\ge k(1/k)2^k\to\infty
\end{equation}
as $k\to\infty$. So, $f(\cdot,2)$ is unbounded.

Here is the graph $\Big\{\Big(x,\dfrac{f(x,2)}{2^{2/x}x}\Big)\colon10^{-3}\le x\le1\Big\}$:

A: Take $D=[0,2]$, $a_i(x)=x^i$, then $f(x,1/3)$ converges for all $x\in D$, but $f(x,1)$ is unbounded as $x\to1$. So, unless I am misreading something in your question, the answer is negative.
EDIT: If we require that the series converges for all $x$ and $y$, the answer is still negative: let $x_i=1/2^i$ and $a_i(x)$ be the regular "hat" functions "centered" at $x_i$-s (see the illustration). Then (1) is $0$ for $x=0$ and a piecewise linear approximation to $x^{\log_{1/2}y}$ for $x>0$. The latter is bounded on $D=[0,1]$ for $y\leq1$ and unbounded for $y>1$.

