Showing a set of functions in the collection of continuous functions on $\left[0,1\right]$ is closed

Question

I have a set $$A=\left\{f\in\mathcal{C}\left(\left[0,1\right]\right)\restriction-1\le f^{\prime}\left(x\right)\le2,f\left(0\right)=0\right\}$$ where $\mathcal{C}\left(\left[0,1\right]\right)$ is endowed with the supremum norm. I need to show that this set is closed. I have done everything except show that a sequence of functions $\left(f_{n}\right)$ that converges to $f_{*}$ has a sequences of derivatives $\left(f_{n}^{\prime}\right)$ that converge uniformly to $g=f_{*}^{\prime}$. This way I can show that $$-1\le f_{*}^{\prime}\left(x\right)\le 2$$ I want to say that since $$-1\le\lim_{n\rightarrow\infty}f_{n}^{\prime}\left(x\right)\le2$$ then $$-1\le g\left(x\right)\le2$$ But, I am not sure how to show that this sequence of derivatives converges uniformly so that I can say $g=f_{*}^{\prime}$. Any help would be greatly appreciated.

Answer 1

Fix any point $x$.

Given you have uniform convergence, and given for close enough $x,y$, the term $h_n(x)=\frac{f_n(x)−f_n(y)}{x-y}$ is absolutely bounded for all $n$ by the bound on slopes, for large enough $n$, you could make $h(x)=\frac{f(x)−f(y)}{x-y}$ converge to $h_n(x)$ uniformly for that interval.

Given the uniform convergence above, you should be able to take limits ($n \to \infty $ and $y \to x$) in the order that you want, to show what you want to show.

Answer 2

COUNTER-EXAMPLE. Consider a Koch snowflake curve: https://en.wikipedia.org/wiki/Koch_snowflake

We will take a "small" snowflake: $f:[0,1]\to [0,1/6]^2$, with $f(0)=(0,0)$, is the uniform limit of a sequence $f_n:[0,1]\to [0,1/6]$ of piecewise-linear continuous functions such that $\|f_n(x)-f_n(y)\|\leq |x-y|/2$ for all $n$ and for all $x,y\in [0,1],$ and with $f_n(0)=(0,0)$.

We can "smooth" the "corners" of each $f_n$ to get a sequence $g_n$ that also converges uniformly to $f,$ retaining the property that $\|g_n(x)-g_n(y)\|\leq |x-y|/2$, with the important additional property that, with $g_n(x)=(A_n(x),B_n(x)),$ the function $A_n(x)$ is differentiable, with $|A'_n(x)|\leq 1/2$ for all $x\in [0,1]$. And also $g_n(0)=(0,0).$

Now with $f(x)=(A(x),B(x)),$ the function $A(x)$ is the uniform limit of $(A_n)_n.$ But $A(x)$ is nowhere-differentiable.

So the function $A(x)$ is in the closure of the set $A$ but not in the set $A.$