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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.

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    $\begingroup$ 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. $\endgroup$ – User8128 Dec 19 '18 at 7:13
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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 $\mathbb R$.

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  • $\begingroup$ 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. $\endgroup$ – ramanujan Dec 19 '18 at 7:21
  • $\begingroup$ This example also serves as a sequence of continuously differentiable functions converging to a function which is not continuously differentiable (not even differentiable). $\endgroup$ – ramanujan Dec 20 '18 at 13:36
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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.

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  • $\begingroup$ Am I correct about$f_n$? $\endgroup$ – ramanujan Dec 19 '18 at 6:41
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    $\begingroup$ @KaviRamaMurthy The answer is actually incorrect for $f_n$. See my answer below. $\endgroup$ – User8128 Dec 19 '18 at 7:09

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