Showing that the sequence $\{f'_{n}(x)\}$ converges pointwise but not uniformly on $(-1,1)$.

Question

Consider the sequence $\{f_{n} \}$ where for each $n \in \mathbb{N}$ the function $f_{n}\colon \mathbb{R} \rightarrow \mathbb{R}$ is given by: $$f_{n}(x)= \begin{cases} \lvert x \rvert & \text{if } \frac{1}{n} \leq \lvert x \rvert \leq 1 \\ \frac{n}{2}x^{2}+\frac{1}{2n} & \text{if $ \lvert x \rvert \leq \frac{1}{n}$} \end{cases}$$ I want to show that the sequence $\{f'_{n}(x)\}$ converges pointwise but not uniformly on $(-1,1)$.

For starters, I think

$$f'_{n}(x)= \begin{cases} 1 & \text{if } \frac{1}{n} \leq x \leq 1 \\ nx & \text{if $ \lvert x \rvert \leq \frac{1}{n}$} \\ -1 & \text{if } -1 \leq x \leq \frac{-1}{n} \\ \end{cases}$$

So I think this implies that this sequence of functions converges to a discontinuous function (the piecewise function which is $-1$ on $[1,0)$ and $1$ on $(0,1]$. This should imply that the convergence is not uniform because otherwise the limit function would be continuous.

Is this reasoning okay? If not or more generally is there a better way to do this?

Answer 1

It's almost entirely correct. You just forgot to add that $\lim_{n\to\infty}f_n'(0)=0$. So, yes, $(f_n')_{n\in\mathbb N}$ converges pointwise to a non-continuous functions and, since each $f_n'$ is continuous, the convergence cannot be uniform.

Answer 2

Yes, it is correct.

To be entirely precise, $(f_n')_n$ converges to the function $\operatorname{sgn} : [-1,1] \to \mathbb{R}$ defined as $$\operatorname{sgn} x = \begin{cases} -1, & \text{if $x < 0$} \\ 0, & \text{if $x = 0$} \\ 1, & \text{if $x > 0$} \end{cases}$$ which is not continuous at $0$.

You can also notice that

$$\left|\operatorname{sgn}\left(\frac1{2n}\right) - f_n'\left(\frac1{2n}\right)\right| = 1 - \frac12 = \frac12, \forall n\in\mathbb{N}$$

which shows that the convergence is not uniform.