# Find a sequence that converges uniformly with g is differentiable

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

• Don't vandalize your questions (or, more generally, edit them in a substantial way) after you've received answers. – Gae. S. Nov 25 '20 at 11:52

For $$0 we have $$0<\sqrt{b}-\sqrt{a}<\sqrt{b-a}$$ (which can be easily shown by squaring the inequality and using $$0). 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?

• So take N > 1/$\varepsilon^2$, and do the proof? – user840729 Nov 24 '20 at 13:40
• 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. – ym94 Nov 24 '20 at 13:49
• 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? – user840729 Nov 24 '20 at 13:58
• 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. – ym94 Nov 24 '20 at 14:41
• I see, thank you very much – 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 $$\mathbb R.$$

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