On what subintervals of $[0,\infty)$ is $f_n(x)=\frac{x^2}{1+x^n}$ uniformly convergent?

Question

I have calculated the pointwise convergence which is $f(x)=x^2$ for $x<1\qquad$ and for $x=1$ we get $f(x)=\frac{1}{2} \quad$ and lastely for $f(x)=0$ for $x>1$. Now, I know that this sequence is not uniformly convergent for the set $[0,\infty)$ but I know that it IS uniformly convergent for $[0,1)$ since we can do the following: $|f_n(x)-f(x)| =...= |\frac{x^n}{1+x^n}|<|x^n|$ and since we know that $x<1, \forall \epsilon >0,$ we can find a $N$ such that $x^n < \epsilon \forall n\ge N$. The problem I'm having now is how can I find all subintervals of $[0,\infty)$ where this sequence is uniformly convergent?

Answer 1

• The sequence is not uniformly convergent on $(a,1)$ for $0 <a < 1$. Indeed, we have $\frac1{\sqrt[n]{2}} \to 1$ from the left so for large enough $n\in\Bbb{N}$ we have $\frac1{\sqrt[n]{2}} \in (a,1)$ and
  $$\left|f\left(\frac1{\sqrt[n]{2}}\right) - f_n\left(\frac1{\sqrt[n]{2}}\right)\right| = \frac1{\sqrt[n]{4}} -\frac{\frac1{\sqrt[n]{4}}}{1+\frac12} = \frac1{3\sqrt[n]4} \xrightarrow{n\to\infty} \frac13 \ne 0.$$

• The sequence is uniformly convergent on $[0,b]$ for $0 < b < 1$. Namely, for $x \in [0,b]$ we have $$|f(x)-f_n(x)| = x^2-\frac{x^2}{1+x^n} = {x^2}\frac{x^n}{1+x^n} \le x^n \le b^n \xrightarrow{n\to\infty} 0$$ uniformly in $x$ since the bound $b^n$ is independent of $x$.

• The sequence is not uniformly convergent on $(1,b)$ for $b > 1$. Namely, we have $\sqrt[n]{2} \to 1$ from the right so for large enough $n\in\Bbb{N}$ we have $\sqrt[n]{2} \in (1,b)$ and $$\left|f_n(\sqrt[n]{2}) - f(\sqrt[n]{2})\right| = \frac{\sqrt[n]{2^2}}{1+2}-0 = \frac{\sqrt[n]{4}}{3} \xrightarrow{n\to\infty} \frac13 \ne 0.$$

• The sequence is uniformly convergent on $[a,b]$ for $1 < a < b$. Namely, for any $x \in [a,b]$ we have $$|f_n(x)-f(x)| = \frac{x^2}{1+x^n} \le \frac{b^2}{1+a^n} \xrightarrow{n\to\infty} 0$$ uniformly in $x$ since the bound $\frac{b^2}{1+a^n}$ is independent of $x$.

• The sequence is furthermore uniformly convergent on $[a,+\infty)$ for $a>1$. Namely, by calculating the derivative $$f_n'(x) = \frac{x(2-x^n(n-2))}{(1+x^n)^2}$$we see that $f_n$ is strictly decreasing for $x \ge \sqrt[n]{\frac{2}{n-2}}$. Since this bound converges to $1$, for large enough $n\in\Bbb{N}$ it will be less than $a$ so we will have that $f_n$ is strictly decreasing on $[a,+\infty)$. Therefore for $x \in [a,+\infty)$ we have $$|f_n(x)-f(x)| = f_n(x) \le f_n(a) = \frac{a^2}{1+a^n} \xrightarrow{n\to\infty} 0$$ uniformly in $x$ since the bound $\frac{a^2}{1+a^n}$ is independent of $x$.

Therefore we conclude that $f_n \to f$ uniformly on $[a,b]$ where $0 \le a < b$ if and only if $$0\le a < b < 1 \quad \text{ or } \quad 1 < a \le b$$ and on the ray $[a,+\infty)$ for $1 < a$.

Answer 2

It is not uniformly convergent on $[0, 1)$, you have to be cautious here. First, you made a little mistake: $$ \lvert f_n(x) - f(x) \rvert = \left \lvert \frac{x^{n+2}}{1+x^n} \right \rvert $$ Then: $$ \sup_{x \in [0, 1)} \lvert f_n(x) - f(x) \rvert = \sup_{x \in [0, 1)} \left \lvert \frac{x^{n+2}}{1+x^n} \right \rvert = \frac{1^{n+2}}{1+1^n} = \frac{1}{2} $$ This does not tend to zero, so we do not have uniform convergence on $[0, 1)$ to $f(x) = x^2$. But we do have uniform convergence on every interval $[a, b]$ (or $(a, b]$ or $[a, b)$ or $(a, b)$, it does not matter) where $a, b \in (0, 1)$, because: $$ \sup_{x \in [a, b]} \lvert f_n(x) - f(x) \rvert = \sup_{x \in [a, b]} \left \lvert \frac{x^{n+2}}{1+x^n} \right \rvert = \frac{a^{n+2}}{1+{a^n}} \overset{n \rightarrow \infty}{\longrightarrow} 0 $$

Obviously, for $x = 1$, $f_n(x) = \frac{1}{2}$ so the pointwise convergence for $x=1$ is to the constant function $\frac{1}{2}$.

For $x>1$, the pointwise limit is $\displaystyle \lim_{n \rightarrow \infty} f_n(x) = 0$. So uniform convergence is not possible on intervals $J \subseteq \mathbb{R}$ where $1 \in \overline{J}$, because $f$ would be discontinuous which is not allowed since all the $f_n$ are in fact continuous. Now consider intervals of the form $[c, d]$ (or $(c, d)$ or $[c, d)$ or $(c, d]$) where $c, d>1$. Then: $$ \sup_{x \in [c, d]} \lvert f_n(x) - 0\rvert = \sup_{x \in [c, d]} \left \lvert \frac{x^2}{1+x^n} \right \rvert \leq \left \lvert \frac{d^2}{1+c^n} \right \rvert \overset{n \rightarrow \infty}{\longrightarrow} 0 $$

For the case $[a, \infty)$ where $a>1$ consider mechanodroid's answer.