$f_n(x) = [x_n + g (x)]^2$ is not uniformly convergent, for every $x\in X$.

Question

The statement is:

Let $(x_n)_{n\in\mathbb{N}}$ be a sequence of non-zero real numbers such that $x_n → 0$ when $n → + ∞$. Show that if g: X → R is unlimited, then the sequence of functions $f_n$: X → R defined by

$$f_n(x) = [x_n + g (x)]^2$$ for every $x \in X$, it will not converge uniformly.

I am not sure that what I did is correct. Let's go:

As $f_n=x_n^2+2x_ng(x)+g(x)^2$, is that true that if one term of this sum is not uniformly convergent, then $f_n$ won't be? See:

$x_n$ is uniformly convergent due to the fact that $x_n → 0$ when $n → + ∞$. And $g(x)$ is uniformly convergent due to the fact that for every $\varepsilon>0, \exists n_0\in\mathbb{N}$ such that $$n>n_0\Rightarrow |g(x)^2-g(x)^2|=0<\varepsilon, \forall x\in X.$$

However, $x_ng(x)$ is not uniformly convergent (I will hide the procedure because it is a routine calculation, it just uses the fact that g is unlimited). So, $2x_ng(x)$ isn't uniformly convergent as well.

So, can I conclude that $f_n$ is not uniformly convergent based on the fact that one of the terms isn't uniformly convergent?

Answer 1

If $f_n$ were unif. conv, then $x_ng(x)=f_n(x)-x_n^2-g(x)^2$ would be unif conv, and you have checked it is not possible.

So, can I conclude that fn is not uniformly convergent based on the fact that one of the terms isn't uniformly convergent?

NO, BUT, you can conclude that it is not unif. conv, since one term is NOT unif conv and the other terms are.