How to justify, $\sum_{n=1}^{\infty} a_{n} x^{n} - \sum_{n=1}^{\infty}a_{n}y^{n}=\sum_{n=1}^{\infty} a_{n} (x^{n}-y^{n})$?

Question

Let $\{a_{n}\}_{n\in \mathbb N} \subset \mathbb C$ so that the series, $\sum_{n=1}^{\infty} a_{n} x^{n},$ converges absolutely for all $x\in \mathbb R$ and we let $K_{1}$ be a compact subset of $\mathbb R$ and define $f:K_{1}\subset \mathbb R \to \mathbb C$ such that $$f(x):= \sum_{n=1}^{\infty} a_{n} x^{n}; x\in K_{1}$$

Let $x, y\in K_{1}$ with $x\neq y$; and I wants to compute,

$f(x)- f(x)= \sum_{n=1}^{\infty} a_{n} x^{n} - \sum_{n=1}^{\infty}a_{n}y^{n};$

My question: (1) Is it true that, $f(x)-f(y)= \sum_{n=1}^{\infty} a_{n} (x^{n}-y^{n})$; if yes, how to justify it ? (2) Is $\sum_{n=1}^{\infty} a_{n} (x^{n}- y^{n})$ converges absolutely ?

Thanks,

Answer 1

If $(u_N)_N$, $(v_N)_N$ are both convergent sequences, then also the sequence $(u_N- v_N)_N$ converges and the limit is

$$\lim_{N\rightarrow \infty} u_N - \lim_{N\rightarrow \infty} v_N = \lim_{N\rightarrow \infty} (u_N - v_N)$$

Now set $u_N=\sum_{n=1}^N a_n x^n$ and $v_N = \sum_{n=1}^N a_n y^n$. This justifies $(1)$.

Also $(2)$ is true. If $f(x)$ and $f(y)$ converge absolutely, then also $\sum_{n=1}^\infty a_n (x^n - y^n)$ does:

$$\sum_{n=1}^\infty |a_n (x^n - y^n)|\le \sum_{n=1}^\infty (|a_n x^n| + |a_n y^n|)=\sum_{n=1}^\infty |a_n x^n| + \sum_{n=1}^\infty |b_n x^n| <\infty$$