Uniform convergence of $f_n(x)=\frac{\sqrt{1+{(nx)}^2}}{n}$ and ${f_n}^\prime$

Question

Discuss the uniform convergence of $f_n(x)=\frac{\sqrt{1+{(nx)}^2}}{n}$ and its first derivative on real line.

I think both $f_n$ and ${f_n}^\prime$ do not converge uniformly. If we put $x=0,1$ in $f_n$ then we get point wise limit 0 and 1 respectively. So $f_n$ doesn't converge uniformly.

Now, ${f_n}^\prime(x)=\frac{nx}{\sqrt{1+(nx)^2}}.$ If we put $x=1/n$ then it becomes $1/\sqrt{2}$ . So sup norm never becomes zero. But point wise limit is zero. So it is not uniformly convergent. Correct? Thanks.

Answer 1

You're answer for $f_n$ is NOT correct actually. The functions $f_n$ do converge uniformly to the function $f(x) = \lvert x \rvert $. Indeed, we see that $$f_n(x) = \sqrt{\frac{1}{n^2} + x^2}$$ and thus using the inequality $$\lvert b \rvert \le \sqrt{a^2 + b^2} \le \lvert a \rvert + \lvert b \rvert,$$ which holds for all $a,b\in \mathbb R$, we have $$\lvert x \rvert \le f_n(x) \le \frac 1 n + \lvert x \rvert, \,\,\,\,\,\,\, \forall x \in\mathbb R.$$ Sending $n \to \infty$ clearly shows that $f_n$ converges uniformly to $f(x) = \lvert x \rvert$ on $\mathbb R$.

Answer 2

Pointwise limit of $f_n'(x)$ is $0$ for $x=0$, $1$ for $x >0$ and $-1$ for $x<0$. The limit function is not continuous and hence the convergence of $(f_n')$ is not uniform.