Series of Functions and Continuity Let $a > 0$, and $(f_n)_{n=0}^{\infty}$ a sequence of continuous functions $f_n:[-a,a] \rightarrow \mathbb{R}$. Assume that the series
\begin{equation}
\sum_{n=0}^{\infty} x^n f_n(t)
\end{equation}
converges for all $(x,t) \in [-a,a] \times [-a,a]$. Set
\begin{equation}
f(x,t) = \sum_{n=0}^{\infty} x^n f_n(t),
\end{equation}
for all $(x,t) \in [-a,a] \times [-a,a]$. Is $f$ continuous in $(0,0)$?
What if you add the hypothesis that the convergence is uniform in $t$ for every fixed $x \in [-a,a]$? 
I think the answer is negative to both the questions, but I could not find a counterexample. Thank you very much in advance for your attention.
 A: Finally, I found the answers to my two questions. Let us start from the first. In this case, the answer is generally negative, as the following construction shows. Let us recursively build a sequence $(f_n)$ of continuous functions $f_n: \mathbb{R} \rightarrow \mathbb{R}$, $f_n \geq 0$, with the following properties:
(i) for every $n$, the support of $f_n$ is contained in $\left( \frac{1}{2n+3}, \frac{1}{2n+1} \right)$,
(ii) for every $n$, we have $\sum_{k=0}^{n} \frac{1}{(2n+2)^k} f_k(\frac{1}{2n+2}) \geq 1$.
Clearly the series
\begin{equation}
\sum_{n=0}^{\infty} x^n f_n(t)
\end{equation}
converges for every $x$ and every $t$, but its sum $f(x,t)$ is not continuous in $(0,0)$.
Surprisingly, the answer to my second question is positive. Indeed, assume that for every fixed $x \in [-a,a]$ the series 
\begin{equation}
\sum_{n=0}^{\infty} x^n f_n(t)
\end{equation}
is uniformly convergent in t on $[-a,a]$. Put $g_n(t) = a^n f_n(t)$. Then the series $\sum_{n=0}^{\infty} g_n(t)$ is uniformly convergent. Define
\begin{equation}
G_n(t) = \sum_{k=0}^{n} g_k(t),
\end{equation}
and $G_{-1}(t)=0$. Fix $b$ such that $0 < b < a$, let $x \in [-b,b]$, and put $y=x/a$, $r = b/a$. Then from Abel's partial summation formula we have for $0 \leq p \leq q$
\begin{equation}
\sum_{n=p}^{q} x^n f_n(t) = \sum_{n=p}^{q-1}G_n(t)(y^n - y^{n+1}) + G_q(t) y^q - G_{p-1} y^p.
\end{equation}
Now, since $(G_n)$ is uniformly convergent, it is uniformly bounded by some constant $M >0$. So we have
\begin{equation}
\left| \sum_{n=p}^{q-1}G_n(t)(y^n - y^{n+1}) + G_q(t) y^q - G_{p-1} y^p \right| \leq \left| \sum_{n=p}^{q-1}G_n(t)(y^n - y^{n+1}) \right|  + \left| G_q(t)  - G_{p-1} \right| |y|^{q} + \left| G_{p-1}(t)(y^{q} - y^{p}) \right| \leq 2M \sum_{n=p}^{q} |y|^n  + \left| G_q(t)  - G_{p-1} \right| |y|^{q}  \leq 2M \sum_{n=p}^{q} r^n  + \left| G_q(t)  - G_{p-1} \right| r^{q}.
\end{equation}
We conclude that the series
\begin{equation}
\sum_{n=0}^{\infty} x^n f_n(t)
\end{equation}
is uniformly convergent in $(x,t)$ for $(x,t) \in [-b,b] \times [-a,a]$. In particular, since every $h_n(x,t)=x^n f_n(t)$ is a continuous function, the sum
\begin{equation}
f(x,t)= \sum_{n=0}^{\infty} x^n f_n(t), \quad (x,t) \in [-b,b] \times [-a,a],
\end{equation}
is continuous in $(0,0)$.
QED
