1
$\begingroup$

I am having some trouble with the following problem, which states the following: Suppose $\{f_n\}, n=0,1,2,...$ is a sequence of continuously differentiable functions $f_n :[0,1] \rightarrow \mathbb{R}$ converging pointwise to $f$, and that there is a constant $M>0$ such that $|f_n'(x) < M|$ holds for all $x \in [0,1]$ and all $n\geq 1$. Give an example of a sequence satisfying all of the above hypothesis for which the limit function is, however, not differentiable. Sketch graphs of a few of the $f_n$'s and the limit function $f$.

Now, I have already established that under the conditions above, $f_n$ actually converges uniformly to $f$. And as a result, $f$ is continuous. I ran across a helpful related question here: Sequence of differentiable functions converging to non-differentiable function and the sequence $f_n := \sqrt{ 1/n + (x-1/2)^2}$ is close to what I want, but unfortunately it does not meet the requirement of having some $M>0$ such that$|f_n'(x) < M|$ holds for all $x \in [0,1]$ and all $n\geq 1$. Is there a way to modify this sequence to get the desired result/is there a better example?

I think I can visualize more or less what the solution will require: We want some sequence of functions that are smooth everywhere but for which there is some smooth portion which gets "sharper" as $n$ increases, until it becomes a "corner" in the limit. But I'm having trouble making that happen while satisfying the bounded derivative part of the hypothesis.

$\endgroup$
2
  • 1
    $\begingroup$ But $$\frac{x - 1/2}{\sqrt{1/n + (x - 1/2)^2}}$$ is bounded by $1$ in absolute value. $\endgroup$ – Daniel Fischer Aug 20 '20 at 20:16
  • $\begingroup$ Oh, I made a complete novice mistake and completely forgot about the $x- 1/2$ factor in the numerator. $\endgroup$ – French Toast Crunch Aug 20 '20 at 20:23
1
$\begingroup$

How about $f_n(x) = (x-\frac{1}{2})^{\large {1+\frac{1}{2n-1}}}=\left((x-\frac{1}{2})^2\right)^{n/(2n-1)}$ which converges to $\left |x-\frac{1}{2}\right| $

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.