Clarifying the meaning of uniform convergence

Question

I just want to clarify the notion of uniform convergence.

Suppose, I have a sequence of function like this,

$$f_n(x) = \frac{n}{e+n^2x^2}$$

Clearly for $|x| > 0$, $f_n(x) \to 0$ if $n \to \infty$. I was tempted to say that for $|x| > 0$ the function converges uniformly to $0$. But I realize, wouldn't we have variable value for $N$ if $x < 1$?

Here's the rough argument that leads me to that conclusion: (valid only for $|x| \ge 1$)

$$\frac{n}{e+n^2x^2} < \frac{1}{nx^2} < \frac{1}{n} < \epsilon$$

Should the uniform convergence interval then only be for $|x| \ge 1$?

Does this corresponds to the definition of uniform convergence? Am I misinterpreting it? Or instead, as long as the sequence of function converges to the same function, whatever the value of $x$ is, then it is uniformly convergent?

Answer 1

The sequence converges uniformly on any interval $[a,+\infty)$, no matter how small $a > 0$ is, but not on $(0,+\infty)$.

To show uniform convergence on $[a,+\infty)$, note that $f_n(x) \leq n/(e + n^2a^2)$ on this interval. Thus if $b_n = \sup_{x \geq a} |f_n(x)|$, we have $b_n \leq n/(e + n^2a^2)$, hence $b_n \to 0$. This proves that $f_n$ converges uniformly to $0$ on $[a,+\infty)$.

However, for $(0,+\infty)$, note that $f_n(1/n) = n/(e + 1)$. Thus if we let $c_n = \sup_{x > 0} |f_n(x)|$, we have $c_n \geq n/(e + 1)$, so $c_n$ does not tend to zero. Therefore $f_n$ does not converge uniformly to zero on $(0,+\infty)$. (And since the sequence does converge pointwise to $0$, it can't converge uniformly to any other function.)