Correcting the calculation of a norm and continuity of a sequence of functions. Here is the question:
Let $h \in L^{1}([0,1], m)$ and let $F$ be a linear functional on the normed space $C[0,1]$ (with the maximum norm), defined by $F(g) = \int_{0}^{1} hg.$ Prove that $F$ is a bounded linear functional on $C[0,1]$ and determine its norm.
Here is a trial:
To find its norm.
Assume that $0=x_0<x_1<...<x_n=1$ such that $h=\sum_{k=1}^n c_k \chi_{[x_{k-1},x_k)}$ for some constants $c_k$. Let $g_n$ be some continuous function such that $g_n\equiv \frac{c_k}{|c_k|}$ on $[\frac{n+1}{n}x_{k-1}, \frac{n-1}{n}x_k]$ for every $k$ such that $c_k\neq 0$ and $0$ otherwise. then we know that $|g_n|\leq 1$. So, $\|g_n\|_{\max}=1$ for all $n.$ So,$\|h\|_{1} = \int |h| \textrm{d}m =  \int \|g_n\|_{\max} |h|\textrm{d}m$ implying that $\|F\| = \|h\|_{1}.$
Comment  received on this solution:
How are $g_n$ continuous and the part of the last sentence "implying that $\|F\| = \|h\|_{1}.$" is not clear, so could anyone help me improve this solution, either by giving me another more clear solution or by improving it?
 A: You're on the right track, but you need take into account what happens if $h$ is not a step function and explain why you may assume $g_n$'s are continuous. Here is my suggestion. First of all, $F$ is well defined since $g$ is continuous, so bounded in any $L^p$-norm on $[0,1]$ and we get for free that  $$|F(g)| \leq ||h||_1||g||_\infty$$
from which it follows that $||F||_{sup} \leq ||h||_1$. 
Conversely, let us show $||F||_{sup} \geq ||h||_1$. Let $sgn(h)$ be the sing of $h$, which takes the value $\frac{\overline{h}(x)}{|h(x)|}$ if $h(x)\neq 0$ and $0$ otherwise. This function is integrable. Since $C([0,1])\subset L^1([0,1])$ is dense we may approximate $sgn(h)$ by a  sequence of continuous functions $f_n$ in $L^1$-norm with $||f_n||_{sup}\leq 1$ (this may require an argument though). Now$$|\int sgn(h)h - F(f_n)| = |\int(sgn(h)-f_n)h| \to 0.$$
Lastly since $ \int sgn(h)h = ||h||_1$ and $||f_n||_{sup} \leq 1$ for all $f_n$, it follows that $||F|| \geq ||h||_1$
A: The fact that $F$ is linear is clear. Boundness follows form the fact that if $||g||_\infty =1$ than $|F(g)| \le \int |h| = ||h||_1$ so we get $||F|| \le ||h||_1 \lt \infty$.
So  we get $F$ is a bounded linear operator and $||F||\le ||h||_1$.
Lets show there is an equality.
$sgn(h) \in L^1[0,1]$. Since $C[0,1]$ is dense in $L^1[0,1]$ we can find $g_n \in C[0,1]$ s.t. $g_n \to sgn(h)$ in the $L^1[0,1]$ norm. We can have such $g_n$'s with $||g_n||_\infty \le1$. 
Then we will get $||F|| =sup_{||g||_\infty \le 1} |F(g)|\ge |F(g_n)| = |\int_0^1g_n h|\to \int_0^1|h| =||h||_1 $
So, $||F|| \ge ||h||_1 $ and we are done. 
