Find a sequence that converges uniformly with g is differentiable

Question

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

Answer 1

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?

Answer 2

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.$