Operator norm equality

Question

I came across this problem and am getting stuck on how to prove it. Any help would be appreciated.

Suppose $L:C(\textbf{T}) \rightarrow \mathbb{C}$, where $L(f)=\int_0^1 {f(x)g(x)}dx$ for all $f \in C(\textbf{T})$ (the v.s. of continuous, 1-periodic functions on $\mathbb{R}$). Then $\|L\|=\|g\|_1$, where $\|L\|$ is the operator norm of $L$ and $\|g\|_1=\int_0^1|g(x)|dx$.

My approach is to show $\|L\|\leq\|g\|_1$ and $\|L\|\geq\|g\|_1$. I know that L is a linear functional on $C(\textbf{T})$ and I know that $\|L(f)\|=\|\int_0^1f(x)g(x)dx\| \leq \int_0^1 \|f(x)g(x)\|dx$ but am not sure where to go from there.

Answer 1

Hints:

1. $|f(x)| \le \|f\|$.
2. Approximate $|g|$ by a linear combination of indicator functions for disjoint closed sets.