Let $f_n(x) =\sqrt[n] {1+x^n}$, defined for $x\geq0. $

pointwise; $f_n(x) \rightarrow f(x)=\begin{cases} 1,x\in[0,1)\\ x , x\geq 1 \end{cases} $

I was wondering about uniform convergence. $f_n$ are monotone (relative to $n$) for any $x\in[a,b]$ for any $a,b\geq0$, (Decreasing for $x\leq1$ and increasing for $x>1)$, and $f$ is continous, so we can use Dini`s theorem to show uniform convergence,in the closed interval $[a,b]$. However, does this mean $f_n$ converge uniformly in $(0,\infty$)?

The example above is just an excercise I stumbled upon that lead me to this question, I'd much rather have a general answer if we know that $f_n$ are monotone (relative to $n$) for any $x\in[a,b]$ for any $a,b\geq0$, does this imply $f_n$ converge uniformly in $(0,\infty$)?


1 Answer 1


Dini's theorem states that if $(g_n)_n$ is a monotone sequence of continuous functions $g_n : [a,b] \to \mathbb{R}$ converging pointwise to $g : [a,b] \to \mathbb{R}$, then the convergence is uniform.

In your example using Dini you can only conclude that $f_n|_{[0,1]} \to 1$ uniformly on $[0,1]$ and that $f_n|_{[a,b]} \to x$ uniformly on all segments $[a,b] \subseteq [1, +\infty)$.

However, the convergence indeed is uniform on $[0, +\infty)$. To show this, note that for $x \ge 1$ we have

$$1 = (1+x^n) - x^n = \left(\sqrt[n]{1+x^n}\right)^n - x^n = \left(\sqrt[n]{1+x^n} - x\right)\sum_{k=0}^{n-1} \left(\sqrt[n]{1+x^n}\right)^{n-k}x^k \ge n \left(\sqrt[n]{1+x^n} - x\right)$$

so $\sqrt[n]{1+x^n} - x \le \frac1n \xrightarrow{n\to\infty} 0$ uniformly on $[1, +\infty)$. This together with the uniform convergence $f_n|_{[0,1]} \to 1$ gives $f_n \to f$ uniformly on $[0, +\infty)$.

In general, if $g_n \to g$ pointwise on $(0, +\infty)$ and uniformly on all segments $[a,b] \subseteq (0, +\infty)$, it still doesn't follow that the convergence is uniform on $(0, +\infty)$.

An example is the sequence of partials sums of the exponential function $g_n(x) = \sum_{k=1}^n \frac{x^k}{k!}$. Then $g_n \to \exp$ uniformly on all segments $[a,b] \subseteq (0, +\infty)$: $$\exp(x) - g_n(x) = \sum_{k=n+1}^\infty \frac{x^k}{k!} \le \sum_{k=n+1}^\infty \frac{b^k}{k!} \xrightarrow{n\to\infty} 0$$ but the convergence is not uniform on $(0, +\infty)$. I discussed this example here.

  • $\begingroup$ Thank you for the answer, However , I fail to understand why does $1=1+x^n-x$ ? $\endgroup$
    – Sar
    Commented Jul 22, 2018 at 23:05
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    $\begingroup$ @Sar It should be $x^n$, thanks for the correction. $\endgroup$ Commented Jul 22, 2018 at 23:06
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    $\begingroup$ Perhaps more simply, if $x\geq1$ and $n\geq 1$ then $0\leq(1+x^n)^{1/n}-x=\;$ $\;x((1+x^{-n})^{1/n}-1)\;$ $\leq\;x(1+n^{-1}x^{-n}-1)\;=\;$ $(nx^{n-1} )^{-1} \;\leq 1/n$. $\endgroup$ Commented Jul 23, 2018 at 5:12
  • $\begingroup$ I'm reviewing this question and I realized I can't prove the last example you provided doesn't converge uniformly. Can you give me a hint on to how to prove it without using measure theory ? (I haven't learned this topic yet.) $\endgroup$
    – Sar
    Commented Jul 24, 2018 at 11:35
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    $\begingroup$ @Sar For $n\in\mathbb{N}$ consider $x_n = \sqrt[n+1]{(n+1)!}$ $$\sup_{x\in (0,+\infty)}\left|e^x - \sum_{k=1}^n \frac{x^k}{k!}\right| = \sup_{x\in (0,+\infty)}\sum_{k=n+1}^\infty \frac{x^k}{k!} \ge \sup_{x\in (0,+\infty)}\frac{x^{n+1}}{(n+1)!} \ge \frac{x_n^{n+1}}{(n+1)!} = 1$$ so $\sup_{x\in (0,+\infty)}\left|e^x - \sum_{k=1}^n \frac{x^k}{k!}\right|$ doesn't converge to $0$ when $n\to\infty$. (In fact, the supremum is $+\infty$) $\endgroup$ Commented Jul 24, 2018 at 11:56

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