Positivity of a continous Linear Functional This is one of the homework problems.
Show that a continuous linear functional $F$ on $C ([0, 1])$ is positive if and only if $F (1) = ∥F ∥$
I am not sure how to approach in either implication, I feel the reverse implication should follow easily, since the norm is positive but how do you go from the identity to an arbitrary positive function?
I am not sure how to even get started on the other direction.
Any help will be appreciated 
 A: I'll first mention  an argument specific to this context, and later a more general argument. 
Since $F$ is a bounded functional on $C[0,1]$, there exists a measure $\mu$ such that $$F(f)=\int f\,d\mu$$ for all $f\in C[0,1]$. So now one can discuss the assertion in terms of $\mu$. 

The general argument I know works on any C$^*$-algebra. 
If $F$ if positive, it satisfies Cauchy-Schwarz, in the sense that $$|F(\bar g f)|\leq F(|f|^2)\,F(|g|^2).$$
Then 
$$
|F(f)|^2\leq F(|f|^2)\,F(1)
\leq \|F\|\,\|f\|^2\,\|F\|\,\|1\|=\|F\|^2\|f\|^2.
$$
It follows that, since all positive functions with norm one can be written as $|f|^2$ with $\|f\|=1$, 
$$
\|F\|^2=\sup\{F(|f|^2)\,F(1):\ f\in C[0,1]\}=F(1)\,\sup\{F(g):\ 0\leq g\leq1\}\leq F(1)^2.
$$
So $\|F\|\leq F(1)\leq \|F\|$, so $\|F\|=F(1)$. 
For the converse, assume that $\|F\|=F(1)$. Changing to $F/\|F\|$ we may assume that $\|F\|=F(1)=1$. For real $f$ with $\|f\|=1$, and $n\in\mathbb Z$, 
\begin{align}\tag{1}
|F(f)+in|&=|F(f+in )|\leq\|f+in\|=\|(f-in)(f+in)\|^{1/2}\\ \ \\
&=\|f^2+n^2\|^{1/2}=(1+n^2)^{1/2}
\end{align}
(the last equality because $f^2\geq0$). In particular 
$$
|n+\text{Im}\,F(f)|\leq(1+n)^{1/2}.
$$
Thus
$$
n-(1+n^2)^{1/2}\leq\text{Im}\,F(f)\leq -n+(1+n^2)^{1/2}
$$
for all $n$, which implies that $\text{Im}\,F(f)=0$. So $F$ is real, and taking $n=0$ in $(1)$ we get $F(f)^2=|F(f)|^2\leq1$. So $-1\leq F(f)\leq 1$. 
Now for $f$ with $0\leq f\leq 1$, we have $2f-1$ real with $-1\leq 2f-1\leq 1$, so by the above $-1\leq 2F(f)-1\leq1$, which is equivalent to $0\leq F(f)\leq1$. So $F$ is positive. 
