Let $\mathcal{C}^0([0,1];\mathbb{C})$ be the set of all continuous complex-valued functions on $[0,1]$. Prove that :
$a)$ $E:=\{f\in\mathcal{C}^0([0,1];\mathbb{C}) : f(0)=0\}$ is a Banach space.
$b)$ $F=\{f\in E: \int_0^1f(x)\mathrm{d}x=0\}$ is a proper closed subspace of E.
$c)$ Riesz's lemma is not true for $r=1$.
I tried to prove a) saying that you can use a Cauchy sequence to prove that the limit belongs to $E$. Instead for b) I could demonstrate only that $F(1)=F(0)$ (one primitive of $f$) but I don't know if it's useful. Can you help me?
b) Convergence in $C^0[0,1]$ is uniform convergence, and uniform convergence on $[0,1]$ implies convergence of integrals. (And it preserves the property $f(0)=0$ too.) Hence $F$ is closed.
c) There is no function $f$ in the closed unit ball of $E$ that is at distance $1$ from $F$. Indeed, fix any $f$ in the closed unit ball. From $\sup_{[0,1]} |f|\le 1$ we have $$\left|\int_0^1 f\right|\le 1\tag{1}$$ Furthermore, condition $f(0)=0$ implies that strict inequality holds in $(1)$: say, $$\left|\int_0^1 f\right| = c<1$$ There exists a function $g$ in the unit ball of $E$ such that $\int_0^1 g>c$: namely, $g(x)=\min(1, |x|/\epsilon)$ works for $\epsilon$ small enough. Hence, letting $$h=f- \frac{c}{\int_0^1 g}g$$ yields $h\in F$ and $\|h-f\|<1$.