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Does $(f_n)$ converge pointwise/uniformly on $I$ if $$f_n(x) = \frac{x^n}{1+x^n} ~~~~~~ I=[0,1]$$

My attempt:

If $x \in [0,1): \displaystyle \lim_{n \to \infty}f_n(x) = 0$

If $x=1: \displaystyle \lim_{n \to \infty}f_n(x) = \frac{1}{2}$

Hence $(f_n)$ converges pointwise to $$f(x) = \begin{cases}0 & \text{if } 0\leq x < 1 \\ \frac{1}{2} & \text{if } x=1\end{cases}$$ on $I=[0,1]$.

Now, for each $n \in \mathbb{N}$ $$\big|~f_n(x) - f(x)~\big| = \displaystyle \begin{cases}\displaystyle\frac{x^n}{1+x^n} & \text{if }~ 0 \leq x <1\\ 0 &\text{if } x=1\end{cases}$$

Hence $$\| f_n(x) - f(x) \| = \sup\bigg\{ \bigg|\displaystyle \frac{x^n}{1+x^n} \bigg| : x \in [0,1) \bigg\} = \frac{1}{2}$$ which does not converge to $0$.

Hence $(f_n)$ does not converge uniformly to $f$ on $I=[0,1]$.

Is this correct?

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    $\begingroup$ This looks good. You can also use the fact that the $f_n$ converges to a discontinuous function to conclude that the $f_n$ do not converge uniformly, but I suspect that the point of the exercise was to apply basic principles. $\endgroup$ – Tim Raczkowski May 21 '15 at 23:17
  • $\begingroup$ If $f_n$ converged uniformly, then it would converge to a continuous function. Imasmuch as $f$ is discontinuous at $1$, $f_n$ cannot converge uniformly. And you're done. $\endgroup$ – Mark Viola May 21 '15 at 23:22
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If $f_{n}$ converged uniformly on $[0,1]$, the limit function would necessarily be continuous, but as you have shown, it is not. Therefore, the sequence does not converge uniformly.

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It converges pointwise but not uniformly. Take the sequence $x_n=1-\frac{1}{n}$. The we get that the $\sup f_n(x)\geq 1-\frac{1}{1+e}$. The sup is not neccesarilly equal to 1/2. However, it is positive.

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