The question asks to find sequence of differentiable continuous functions $g_n:\mathbb{R}\rightarrow\mathbb{R}$ such that:

$g_n\rightarrow g$ uniform on $\mathbb{R}$, but $g$ is not differentiable on $\mathbb{R}$

I have the that $g_k(x)=\sqrt{x^2+1/k} \rightarrow |x|$, but I need to be able to prove this is true.

I have $| \sqrt{x^2+1/k} - (|x|)|<\varepsilon$,

  • 3
    $\begingroup$ Don't vandalize your questions (or, more generally, edit them in a substantial way) after you've received answers. $\endgroup$ – Gae. S. Nov 25 '20 at 11:52
  • $\begingroup$ Please don't change your question after you have received and accepted an answer. (Adding typos seems particularly useless.) $\endgroup$ – tripleee Nov 25 '20 at 12:04

For $0<a<b$ we have $0<\sqrt{b}-\sqrt{a}<\sqrt{b-a}$ (which can be easily shown by squaring the inequality and using $0<x<y \iff 0<x^2<y^2$). Therefore, using this and $|x|=\sqrt{x^2}$ for all $x\in \mathbb{R}$, we have

$| g_n (x)-|x|| =\sqrt{x^2+\frac{1}{n}}-\sqrt{x^2}\leq \sqrt{x^2+\frac{1}{n}-x^2}=\frac{1}{\sqrt{n}}$ for all $x\in \mathbb{R}$ and $n\in \mathbb{N}$.

Can you take it from here?

  • $\begingroup$ So take N > 1/$\varepsilon^2$, and do the proof? $\endgroup$ – user840729 Nov 24 '20 at 13:40
  • $\begingroup$ Exactly. Then, given $\varepsilon>0$, we can choose $N>\frac{1}{\varepsilon^2}$. Then $|g_n(x)-g(x)|<\varepsilon$ for all $n\geq N$ and all $x\in \mathbb{R}$ proving uniform convergence. $\endgroup$ – ym94 Nov 24 '20 at 13:49
  • $\begingroup$ Thank you, so for pointwise convergence of $g'_n$, we just have to find the limit of this, which I have found to be $x/\sqrt{x^2}$, do I need to do anything else or because the limit exists it is pointwise convergent? $\endgroup$ – user840729 Nov 24 '20 at 13:58
  • $\begingroup$ First of all, you need to compute $g_n'$, which is $g_n'(x)=\frac{x}{g_n(x)}$. As, we already know $g_n\to g$ uniformly (for the following argument pointwise convergence would suffice, too), we therefore conclude that for any fixed $x\in \mathbb{R}$ we have $g_n'(x)\to \frac{x}{g(x)}=\frac{x}{|x|}$ which proves pointwise convergence. $\endgroup$ – ym94 Nov 24 '20 at 14:41
  • $\begingroup$ I see, thank you very much $\endgroup$ – user840729 Nov 24 '20 at 15:27

Let $g(x)=|x|$. Then

$$|g_n(x)-g(x)|= \frac{|(g_n(x)-g(x))(g_n(x)+g(x))|}{g_n(x)+g(x)}=\frac{1/n}{g_n(x)+g(x)} \le \frac{1}{\sqrt{n}}.$$

Hence, $(g_n)$ converges uniformly to $g$ on $ \mathbb R.$

  • $\begingroup$ Does this alone prove uniform convergence, or do I use this to take N > $1/\varepsilon^2$ and do the epsilon delta proof? $\endgroup$ – user840729 Nov 24 '20 at 13:42

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