Thinking of sequence where $f_n'$ does not converge to $f'$

Question

everyone, we see in Rudin this example.

I was trying to think of another example that satisfies this property, but I could not. I could think of sequences of functions that were similar in form with a manipulation of $\sin(nx)$ on the top but could not think of anything else that satisfies the property/example above. Is there any examples satisfiying the property above not of the form $\sin(nx)$? Thank you.

Answer 1

Define$$f_n(x)=\frac{nx}{1+n^4x^2}.$$Then, again, $\lim_{n\to\infty}f_n'(0)=+\infty$, whereas $(\forall x\in\mathbb{R}):\lim_{n\to\infty}f_n(x)=0$.

Answer 2

Consider the sequence of functions $f_n(x) = \frac{x}{1+nx^2}$. We have that $f'_n(x) = \frac{1-nx^2}{(1+nx^2)^2}$. Now obviously we have that $f(x) = 0$ and $f'(x) = 0$. On the other side we have that

$$\lim_{n \to \infty} f'_n(x) = \begin{cases} 0, & x \not = 0 \\ 1, & x = 0 \end{cases} $$

So the two "derivatives" doesn't coincide at $0$. This has something to do with the fact that $f'_n(x)$ doesn't converge uniformly to $f'(x)$ on any interval contaning $0$.

On the other side it's possible for $g'_n(x)$ not to converge uniformly on an interval and yet $\lim_{n \to \infty} g_n'(x) = g'(x)$ on it. Such an example is $g_n(x) = \frac{e^{-n^2x^2}}{n}$, with $g'_n(x)$ not converging uniformly on an interval containing the origin, yet $\lim_{n \to \infty} g_n'(x) = 0 = g'(x)$ on $\mathbb{R}$

Answer 3

Let $g : \mathbb{R} \rightarrow \mathbb{R}$ be bounded, so that $|g(x)| < \alpha$ $\forall x \in \mathbb{R}$; and also $g'(x) \neq 0$ $\forall x\in \mathbb{R}$. Let the function, $$ \begin{align} f_n : \mathbb{R} \rightarrow& \mathbb{R} \\ x \mapsto& \frac{g(nx)}{n} \end{align} $$ It follows that $f_n \to 0$ as $n \to \infty$; but $f'_n(x) = g'(nx)$ and $g'(0) \neq 0$. We also have that $f_n' \nrightarrow f'$ as $n \to \infty$.

The sigmoidal functions have this property, for example, $$ \begin{align} g : \mathbb{R} \rightarrow& \mathbb{R} \\ x \to& \frac{1}{1 + e^{-x}} \end{align} $$

Answer 4

Another example is the sequence of functions $$\begin{array}{l|rcl} f_n : & [-1,1] & \longrightarrow & \mathbb{R} \\ & x & \longmapsto & \sqrt{\frac{1}{n^2}+x^2} \end{array}$$ These functions are differentiable according to the chain rule as $g_n(x): x \longmapsto \frac{1}{n^2}+x^2$ is strictly positive on $[-1,1]$. We also have $$x^2 \le f_n^2(x) = \frac{1}{n^2}+x^2 \le (\vert x \vert +\frac{1}{n})^2$$ hence $$\vert x \vert \le f_n(x) \le \vert x \vert +\frac{1}{n}$$ as $f_n$ is positive. So by the squeeze test, $(f_n)$ converges uniformly to the absolute value function $f(x) = \vert x \vert$. But this function is not differentiable at $0$. Thus, the uniform limit of differentiable functions need not be differentiable.

The picture below helps you to understand graphically what is happening:

And for more details, you can go to my counterexamples website.