Let $C[0,1]$ be the space of all continuous functions on $[0,1]$. Then for any $g\in C[0,1]$ define the norm on $C[0,1]$ as $\|g\|=\max_{t\in[0,1]}|g(t)|$.
Then let $f:C[0,1]\to\mathbb{R}$ as $$ f(g)=\int_0^1 g(t)h(t)\,dt $$ for some function $h\in L^1[0,1]$. Show that $f$ is a bounded linear functional and compute the norm of $f$.
MY ATTEMPT: Holder's inequality quickly establishes the fact that $f$ is a bounded linear functional since the max norm on $C[0,1]$ coincides with the $L^\infty$ norm on $C[0,1]$. If we let $\|f\|_*$ denote the norm of $f$ then we also get that $\|f\|_*\le \|h\|_1$ for all $g\in C[0,1]$.
I am certain that we actually have $\|f\|_*=\|h\|_1$ but I don't know how to turn the inequality around.
The usual trick is to find a function $g^*\in C[0,1]$ such that $f(g^*)=\|h\|_1$ which demonstrates that the upper bound is attainable for some function in $C[0,1]$. Unfortunately, the usual suspect for $g^*=\operatorname{sgn}(h)$ is not continuous. Any ideas out there?
Let $g^*(x) = \operatorname{sgn} h(x)$. We see that $|g^*(x)| \le 1$ for all $x$, and $\int h g^* = \|h\|_1$.
Since $h \in L^1[0,1]$, for any $\epsilon>0$ we can find a $\delta>0$ such that if $m A < \delta$, then $\int_A |h| < \epsilon$.
Using Lusin's theorem (See Theorem 2.24 in Rudin's "Real & Complex Analysis") for all $\delta>0$ we can find a continuous $c$ such that $m \{x | c(x) \neq g^*(x) \} < \delta$ and $\sup_x |c(x)| \le \sup_x |g^*(x)|$.
Choose the $c$ above and let $A = \{x | c(x) \neq g^*(x) \}$, then $|\int h g^* - \int h c| = |\int h (g^*-c)| = |\int_A h (g^*-c)| \le 2 \int_A |h| < 2 \epsilon$, which gives $|f(c) - \|h\|_1| \le 2 \epsilon$. Since $\sup_x |c(x)| \le 1$, we see that $\|c\| \le 1$ and $|f(c)| \ge \|h\|_1- 2 \epsilon$.
Since $\epsilon>0$ was arbitrary, we have $\|f\|_* \ge \|h\|_1$.
As an aside, the bound is not necessarily achieved. A standard example is to use $h = 1_{[0,{1 \over2})} - 1_{({1 \over2}, 1])}$. Then the corresponding $f$ satisfies $\|f\|_* = 1$, but $f(c) <1$ for any $c$ with $\|c\| \le 1$.