So I was having a discussion with a friend about this problem and we have conflicting views. Here it is
We let $f_n: E \rightarrow \mathbb{R}$ be continuous functions for $1 \leq n \leq N$ and we let $a_k^{(n)}$ be $N$ convergent sequences of numbers. Assume lim$_{k\rightarrow \infty}a_k^{(n)} = a_n$. Let $f = \sum_{n=1}^N a_n f_n$.
He thinks that $\sum_{n=1}^{N} a_k^{\{n\}} f_n$ converges uniformly to $f$.
But I do not think so. However, I can't think of any counterexample. I am fairly certain that this is not true. However, I do think that it converges uniformly to $f$ if $E$ is compact. This just makes sense to me, but I cannot figure out how to rigorously prove this.
So, (1) I think that $\sum_{n=1}^{N} a_k^{\{n\}} f_n$ does not converge uniformly to $f$. Edit: SOLVED
(2) $\sum_{n=1}^{N} a_k^{\{n\}} f_n$ does converge uniformly to $f$ if $E$ is compact.
(3)Edit: SOLVED $\sum_{n=1}^{N} a_k^{\{n\}} f_n$ converges point-wise to $f$, my friend said this is true but can't prove. This is what I have so far. I can fix $x$ in $E$ and then look at |$\sum_{n=1}^{N} a_k^{\{n\}}f_n(x)$ - $\sum_{n=1}^{N} a_nf_n(x)$|but from here I am stuck on how to show that it is indeed pointwise convergent.
Any help showing these would be great!
Consider the continuous function $f:(0,\infty)\to \mathbb{R},$ $f(x)=1/x.$ Consider $a_k=\frac{k}{k+1}\to 1.$ Then,
$$\left|\frac{k}{k+1}f(x)-f(x)\right|=\frac{1}{k+1}\frac{1}{x}.$$ That is, $\frac{k}{k+1}f$ doesn't converge uniformly to $f.$ Indeed, assume that given $\epsilon >0$ there exists $N\in\mathbb{N}$ such that
$$n\ge N \implies \left|\frac{n}{n+1}f(x)-f(x)\right|=\frac{1}{n+1}\frac{1}{x}<\epsilon.$$ Now, this is equivalent to say $$x> \frac{1}{(n+1)\epsilon},$$ and thus the inequality is not satisfied for all $x\in (0,1/(n+1)\epsilon),$ what gives us a contradiction. Thus, we have shown that the convergence is no uniform.
If $E$ is bounded the answer is yes and can be obtained from the inequality
$$\left\|\sum_{n=1}^N a^n_kf_n-\sum_{n=1}^n a_nf_n\right\|_{\infty}\le \sum_{n=1}^N|a_k^n-a_n|\|f_n\|_{\infty}.$$
Since $f_n$ is continuous and $E$ is compact we have that $$\|f_n\|_\infty=\max_{x\in E} |f_n(x)|=M_ n<\infty.$$ Consider $M=\max\{M_ 1,\ldots,M_N\}.$ So, we have
$$\left\|\sum_{n=1}^N a^n_kf_n-\sum_{n=1}^n a_nf_n\right\|_{\infty}\le M\sum_{n=1}^N|a_k^n-a_n|.$$ That is, for any $x\in E$ it is
$$\left|\sum_{n=1}^N a^n_kf_n(x)-\sum_{n=1}^n a_nf_n(x)\right| \le M\sum_{n=1}^N|a_k^n-a_n|.$$ Since
$$\sum_{n=1}^N|a_k^n-a_n|\to 0$$ we have shown that the convergence is uniform.