# Finding sequence of continuously differentiable functions with bounded derivative that converge to non-differentiable function

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.

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

## 1 Answer

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 $$\left |x-\frac{1}{2}\right|$$