Prove that $\{f_n\}$ converges uniformly on $\mathbb{R}$ but $\{f_n^\prime\}$ does not.

Question

Consider $\{f_n\}$ be a sequence of functions from $\mathbb{R}$ to $\mathbb{R}$ defined by $$f_n (x) = \frac{\sqrt{1 + (nx)^2}}{n}$$.

Prove that $\{f_n\}$ converges uniformly on $\mathbb{R}$ but $\{f_n^\prime\}$ does not.

I have shown that $f_n \rightarrow |x|$ pointwise. But how to proceed for uniform convergence?

I want to do it without using basic definition of uniform convergence. There are so many tests like Weiesrtrass M-test.

Please help me.

Answer 1

1. Let $f(x) = |x|$. By noting that

$$ f_n(x) - f(x) = \frac{\sqrt{1+(nx)^2} - n|x|}{n} = \frac{1}{n\left(\sqrt{1+(nx)^2} + n|x|\right)} $$

and that $\sqrt{1+(nx)^2} + n|x| \geq 1$ holds for all $x \in \mathbb{R}$, we have

$$ \left| f_n(x) - f(x) \right| \leq \frac{1}{n} $$

uniformly in $x \in \mathbb{R}$ and hence the uniform convergence of $(f_n)$ to $f$ on $\mathbb{R}$ is established.

2. On the other hand, if $(f_n')$ converges uniformly to some function on $g$ on $\mathbb{R}$, then

• Since each $f_n'$ is continuous on $\mathbb{R}$, we know that $g$ must be continuous on $\mathbb{R}$ as well.

• By the theorem about uniform convergence and differentiability, it must be the case that the uniform limit $f$ of $f_n$ is also differentiable and $f' = g$ holds.

Now, since $f(x) = |x|$ is not differentiable, we have a contradiction and therefore $(f_n')$ cannot converge uniformly on $\mathbb{R}$.

Answer 2

$|f_n(x)-|x||=\frac 1 {n^{2}{\sqrt {\frac 1 {n^{2}}+x^{2}}+|x|}}\leq \frac 1 n$. For non-uniformity of convergence of $f_n'$, show that $f_n'(x) \to 1$ if $x>0$ and $f_n'(x) \to -1$ if $x<0$. Since the limit is not continuous it follows that the convergence is not uniform.