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?

  • 1
    $\begingroup$ 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. $\endgroup$ May 21, 2015 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, 2015 at 23:22

2 Answers 2


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.


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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