Norm of linear operator that sends continuous functions to their integral

Question

Let $A=\{\phi\in C[0,1]:\phi(0)=0\}$ with the supremum norm.
Further let $F:A\rightarrow\mathbb{F}$ with $F(\phi)=\int^1_0 \phi(x)dx$.

How do I show that

1. $F$ is bounded?
2. $||F||=1$?
3. Is there an $\phi\in A$ such that $||\phi||=1$ and $|F(\phi)|=||F||$?

What I know:
1 Bounded means that there exists a constant $C$ such that $||F(\phi)||\leq C||\phi||$ for all $\phi\in A$. Further the supremum norm is $||\phi||=\sup\{|\phi(x)|:x\in[0,1]\}$.
Thus I want $\sup\{|\int_0^1\phi(x)dx|\}\leq C \sup\{|\phi(x)|\}$. What is the constant $C$?

2 The norm of $F$ is $||F||=\sup\{|F(\phi)|:||\phi||\leq1\}$. How can I show that this equals $1$?

3 I think there must, but I can't think of a concrete example.

Answer 1

1. We have $$ |F(\phi)| = \left| \int_0^1 \phi(x) \, dx \right| \leq \int_0^1 |\phi(x)| \, dx \leq \int_0^1 \|\phi\|_{\infty} \, dx = \|\phi\|_{\infty}$$ and so $F$ is bounded and $\| F \| \leq 1$.
2. If we could find a function $\phi$ such that $\| \phi \|_{\infty} = 1$ and $|F(\phi)| = 1$, the supremum in the definition of $\| F \|$ would actually be a maximum. The obvious candidate for such $\phi$ is the constant function $\phi \equiv 1$ but unfortunately $\phi$ does not belong to $A$. However, we can modify $\phi$ slightly and use it to show that $\| F \| = 1$. Let $\phi$ be a piecewise linear function that goes from $0$ to $1$ on $[0,\delta]$ and equals to $1$ on $[\delta, 1]$. Then $\|\phi\|_{\infty} = 1$, $\phi \in A$ and $$|F(\phi)| = \frac{1}{2}\delta + (1 - \delta) = 1 - \frac{1}{2}\delta$$ which shows that $\|F\| \geq 1 - \frac12 \delta$ for all $\delta > 0$. Together with part $(1)$, we get $\| F \| = 1$.
3. Let $\phi \in A$ with $\| \phi \|_{\infty} = 1$ . Since $\phi$ is continuous and $\phi(0) = 0$, there exists $\delta > 0$ such that $|\phi(x)| < \frac{1}{2}$ on $[0,\delta]$. But then $$ -1 + \frac{\delta}{2} = -\frac{\delta}{2} - (1 - \delta) \leq \int_0^\delta \phi(x) \, dx + \int_{\delta}^1 \phi(x) \, dx = \int_0^1 \phi(x) \, dx,\\ \int_0^1 \phi(x) \, dx = \int_0^{\delta} \phi(x) \, dx + \int_{\delta}^1 \phi(x) \, dx \leq \frac{\delta}{2} + (1 - \delta) = 1 - \frac{1}{2}{\delta}$$ and so we cannot have $| F(\phi) | = 1$.