I'm having some problems in series of this kind: $$\sum_{k=0}^{+\infty} a_kf^k(x) \ \ \ , x\in \text{Dom}(f)$$ ($f^k$ is k-th power, not k-th derivative, or iterated compositin). If I make the substitution $t=f(x)$ the series becomes: $$\sum_{k=0}^{+\infty} a_kt^k \ \ \ , t\in \text{Im}(f)$$ It's pretty obvious that if this serie converges pointwise in $t \in A \subseteq \text{Im}(f) $ then also the starting series converges pointwise in $x \in f^{-1}(A)$. I was wondering if this is the case also for uniform convergence. I think yes, and this is my attempt of proof. Let's suppose that the "t-series" converges uniformly in $A$, we want to prove that the "x-series" converges uniformly in $f^{-1}(A)$. Let's suppose it's not, than Cauchy-uniforme convergence criterion is not satisfied: $$\exists \varepsilon'>0:\forall N\in \mathbb{N} \exists p>N \exists q>0 \exists x'\in f^{-1}(A):|\sum_{k=p}^{p+q} a_kf^k(x')|\geq \varepsilon' $$ Since $f(x') \in A$: $$\exists \varepsilon'>0:\forall N\in \mathbb{N} \exists p>N \exists q>0 \exists t'\in A:|\sum_{k=p}^{p+q} a_kt'^k|\geq \varepsilon' $$ This means that the "t-series" doesn't satisfy Cauchy-uniform convergence criterion in $A$, therefore it doesn't converge uniformly in $A$ and this is absurd. Is this correct? Moreover can this be generalized similarly to any function series: $$\sum_{k=0}^{+\infty} g_k(f(x)) \ \ \ , x\in \bigcap_{k\in\mathbb{N}}\text{Dom}(g_k \circ f)$$
Thank you in advance
Why not just use a direct proof rather than contradiction? Let $S_n(t) = \sum_{k=0}^n a_k t^k$ be the partial sums. If $\sum_k a_k t^k$ converges uniformly to $L(t)$ for $t \in A$, it means for every $\epsilon > 0$ there exists $N$ such that for all $n > N$, $|S_n(t) - L(t)| < \epsilon$ for all $t \in A$. This includes the case $t = f(x)$ for $x \in f^{-1}(A)$, i.e. for all $n > N$, $|S_n(f(x)) - L(f(x))| < \epsilon$ for all $x \in f^{-1}(A)$, which means $\sum_{k=0}^\infty a_k f(x)^k$ converges uniformly to $L(f(x))$ for $x \in f^{-1}(A)$.