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So, I'm asked to prove that the sequence of functions given by:

$$ f_{n}(x) = \frac{x^n}{1 + x^{2n}} $$

converges uniformly on $[a,b]$ iff neither 1 nor -1 belong to the interval.

Here are my proof attempts:

If both $1$, $-1$ don't belong to $[a,b]$, if $|x| < 1$, $f_{n}$ converges uniformly to the zero function, by a direct calculation. The same if $|x|>1$.

Now suppose $f_{n}$ converges uniformly and that $1 \in [a,b]$. Then, $f_{n}(1) = 1/2$ for every $n$. Therefore, given any positive $\varepsilon$, $| f(1) - 1/2 |$ can be made $< \varepsilon$ for a suitably large choice of $n$, so that $f(1) = 1/2$.

Now, on the other hand, if $x < 1$, $\lim_{n \to \infty} f_{n}(x) = 0$, so that we have:

$$ \left|f(x) - f(1)\right| = \left|\lim_{n \to \infty} f_{n}(x) - f(1) \right| = 1/2 $$

for every $x<1$. This way, the limit function $f$ is not continuous.

Then, I thought that since the uniform limit of continuous functions must be continuous, we'd have a contradiction. The case -1 would be similar, and we'd be done. But I don't feel very comfortable about having to use continuity, and not very sure if this answer is correct or missing some subtle point. Can you figure some solution based solely on the definition of uniform convergence?

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    $\begingroup$ Does $(f_n)$ even converge pointwise at $x=-1$? $\endgroup$ – David Mitra Jul 3 '13 at 22:33
  • $\begingroup$ Indeed, -1 is the easiest case after all, since it has no pointwise limit. $\endgroup$ – ulilaka Jul 3 '13 at 22:36
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    $\begingroup$ Towards a proof that doesn't appeal to continuity of the limit function (your argument looks correct, incidentally), consider the quantity $\lim_{x\rightarrow1^-}f_n(x)$. For each fixed $n$, its value is $1/2$. So... $\endgroup$ – David Mitra Jul 3 '13 at 22:43

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