Set of Continuous Functions, Functionals, and Equicontinuity Define the subset of $C^0[0,1]$ to be:
$P = \{F(x) = \int_0^x f(t) \, dt : f \in C^0[0,1], \|f\|_\infty \le 1\}$
1) Show that $P$ is not closed.
2) Show that $P$ is bounded and equicontinuous (using the infinity norm)
3) Show that the functional $J: C^0[0,1] \to R$ given by $J(F) = \int_0^1 F(x)\, dx$ achieves its max value on $P$.
Thanks in advance for your help and explanations. Thus far I have been working on the first two parts and have an intuition for why they are true, but I am struggling to construct a formal proof. I see that the functions are all differentiable and have been thinking about pointwise convergence. For part 2, I have been considering using the mean value theorem. Any help would be greatly appreciated.
 A: 1) Observe that every function in $P$ is differentiable. To show that $P$ is not closed, it suffices to construct a sequence in $P$ which converges uniformly to a non differentiable function $F$. For every $n\geq 1$, consider the continuous function $f_n$ defined by $1$ on $[0,1/2-1/n]$, $-1$ on $[1/2+1/n,1]$, and by an affine piece connecting these two portions. Then $\|f_n\|_\infty$, so $F_n(x):=\int_0^xf_n(t)dt$ belongs to $P$. I claim that $F_n$ converges to $F(x)=x$ on $[0,1/2]$ and $F(x)=1-x$ on $[1/2,1]$. This is straighforward by the dominated convergence theorem. Since $F_n$ is easily seen to be uniformly Cauchy, uniform convergence to $F$ follows. Of course, $F$ is not differentiable at $1/2$. So $P$ is not closed.
2) This is easier. First
$$
|F(x)|\leq \int_0^x|f(t)|dt\leq \int_0^x\|f\|_\infty dt=\|f\|_\infty x\leq \|f\|_\infty\leq 1\qquad\forall x\in[0,1]
$$
for all $F$ in $P$. Taking the sup, this yields $\|F\|_\infty\leq 1$ for all $F$ in $P$, so $P$ is bounded.
Second take $F$ in $P$ and check
$$
|F(x)-F(x_0)|=\lvert\int_{x_0}^x f(t)dt\rvert\leq\lvert \int_{x_0}^x |f(t)|dt\rvert\leq |x-x_0|\|f\|_\infty\leq |x-x_0|.
$$
It follows that $P$ is (uniformly) equicontinuous.
3) Note that for every $F$ in $P$
$$
|J(F)|\leq\int_0^1\left(\int_0^x|f(t)|dt\right)dx\leq\|f\|_\infty\int_0^1xdx\leq \frac{1}{2}.
$$
And this is achieved for $f(t)=1$ and $F(x)=\int_0^x1dt=x$ in $P$.
