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

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.

• Your answer for $f_n$ is incorrect. You have proved that there are two points $x,y$ such that $f_n(x)$ and $f_n(y)$ converge to different values, but this doesn't mean the convergence is non-uniform. It turns out that $f_n$ does converge uniformly. See my answer below. – User8128 Dec 19 '18 at 7:13

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 $$\mathbb R$$.
• Ohh thank you very much. Nice idea. First it seems like $f_n$ converges point wise to 1 except at $x=0$. But if we think deeply than we realize it is uniformly convergent. – ramanujan Dec 19 '18 at 7:21
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 is not uniform.
• Am I correct about$f_n$? – ramanujan Dec 19 '18 at 6:41
• @KaviRamaMurthy The answer is actually incorrect for $f_n$. See my answer below. – User8128 Dec 19 '18 at 7:09