# pointwise convergence does not preserve differentiability

In my real analysis lecture, when the lecturer give an example which shows that pointwise convergence does not preserve differentiability, I am not quite understand. The example is as follow:

Suppose $$h_n(x)=x^{1+\frac{1}{2n-1}}$$ on the set $[-1,1]$. Notice that $$\lim_n{h_n(x)}=\lim_{n}{x^{1+\frac{1}{2n-1}}}=x\lim_n{x^{\frac{1}{2n-1}}}$$ This limit, taking the odd roots of $x$, will tend to $1$ if $x>0$ , $-1$ if $x<0$ and $0$ if $x=0$ , which is eauivalent to $|x|$. Hence, we have $\lim_n{x^{1+\frac{1}{2n-1}}}=|x|$, which is not differentiable

Here is my doubt. How do we know that the limit tends to these values? Is there any way to see it?

## 3 Answers

You should consider that pointwise convergence means that you take a (stable) $x\in [-1.1]$ and regard it as a number, rather than as a variable. Thus, if $x\neq0$ then $x^{\frac{1}{2n-1}}\longrightarrow 0$ , while if $x=0$ then the limit is again $0$ obviously.

This is not an answer, but is another example that might be easier to follow:

Let $\epsilon > 0$ and $f_\epsilon(x) = \begin{cases} -x & x < \epsilon \\ \frac{1}{2 \epsilon} x^2 + \frac{1}{2\epsilon} & |x| \le \epsilon \\ x & x > \epsilon \end{cases}$

This is an approximation to $x \mapsto |x|$ that is continuously differentiable everywhere (the value and derivative of the separate functions match at $x = \pm \epsilon$). It is easy to see that $f_{\frac{1}{n}}(x) \to |x|$ everywhere, $f_{\frac{1}{n}}$ is $C^1$, but $x \mapsto |x|$ is not differentiable at $x=0$.

You must know the following results, which is rather easy to prove: If $f:I\rightarrow\mathbb{R}$ is a continuous function defined in some interval $I\subseteq\mathbb{R}$ and $\left\{x_n\right\}_{n\in\mathbb{N}}$ is a real sequence which converges for some number $x\in I$, then the sequence $\left\{f(x_n)\right\}_{n\in\mathbb{N}}$ converges to $f(x)$, that is, $\lim f(x_n)=f(\lim x_n)$.

For a fixed $x>0$, the function $t\geq 0\mapsto x^t=\exp(t\cdot\log x)\in\mathbb{R}$ is continuous, Hence

$$\lim_{n\rightarrow\infty}x^\frac{1}{2n-1}=x^{\lim_{n\rightarrow\infty}(1/2n-1)}=x^0=1$$

The case $x<0$ is analogous, since $x^\frac{1}{2n-1}=-|x|^\frac{1}{2n-1}=-exp\left(\frac{1}{2n-1}\log|x|\right)$