Show that $\mathcal{F} = \{F(x) = \int_{0}^{x}f(t)dt : f\in C[0,1], \|f\|_{\infty} \leq 1\}$ is not closed as a subset of $C[0,1]$
To do this I was going to show that a sequence of functions in $\mathcal{F}$ converges to a function that is not in $\mathcal{F}$. In particular I have a candidate function:
$$f_{n}(x) =\Bigg\{ \begin{array}{11} 0, & x\in [0, \frac{1}{2} - \frac{1}{n}] \\ 1 + n(x - \frac{1}{2}), & x \in [\frac{1}{2} - \frac{1}{n}, \frac{1}{2}] \\ 1, & x \in [\frac{1}{2}, 1] \end{array} $$
What I wanted to do was show that this function converges to the characteristic function:
$$\chi_{[\frac{1}{2}, 1]} = \Bigg\{ \begin{array}{11} 0, & x \in [0,\frac{1}{2}]\\ 1,& x \in [\frac{1}{2}, 1] \end{array}$$
Now here is where I am having some trouble. I've established that $f_{n}(x)$ belongs to $\mathcal{F}$, so that is fine. But my idea was to show that $f_{n}(x)$ converges uniformly to $\chi_{[\frac{1}{2}, 1]}$. But IF it did converge uniformly that would mean that $\chi_{[\frac{1}{2}, 1]}$ was continuous. Which it is not. o I'm thinking I have to show pointwise convergence. To do this as I write it out now I'm realizing what could be done, I would have to split my question into two cases:
Case 1:
$x_{0} \in [0, \frac{1}{2}]$, where $x_{0}$ is a fixed term.
Then $$\|f_{n}(x) - \chi_{[\frac{1}{2}, 1]}\| = \|1 + n(x - \frac{1}{2}) - 0\| = \|1 + n(x - \frac{1}{2})\|$$
Case 2:
$x_{0} \in [\frac{1}{2}, 1]$
Then $$\|f_{n}(x) - \chi_{[\frac{1}{2}, 1]}\| = \|1 - 1\| = 0$$
And then as I mentioned previously, this is converging to a discontinuous function so my set $\mathcal{F}$ is not closed.
Is this the right way to think and approach this problem?
Edit: Should I use single absolute value bars to distinguish from unifrom convergence?
Every function in $\mathcal F$ is differentiable on $[0,1]$ by the FTC. But your $f_n$'s are not differentiable at $1/2,$ so they can't be in $\mathcal F.$
I did the problem this way: The functions $((x-1/2)^2+1/n)^{1/2}$ $\to |x-1/2|$ uniformly on $[0,1].$ This implies the functions
$$F_n(x)=\int_0^x\frac{t}{(t^2+1/n)^{1/2}}\,dt$$
converge uniformly to $|x-1/2|-1/2$ on $[0,1].$ Note that each $F_n$ is in $\mathcal F.$ Since $|x-1/2|-1/2$ is not differentiable at $1/2,$ it is not in $\mathcal F.$ It follows that $\mathcal F$ is not closed in $C[0,1].$