What is the norm of the functional $F: C[0,1] \to \mathbb C, F(f) = \int_0^1f(t)g(t) \, dt$ with $g \in L^1[0,1]$?

Let $g \in L^1[0,1]$ be a (complex-valued) integrable function on $[0,1]$. Consider the map $$ F: C[0,1] \to \mathbb C, \ \ F(f) = \int_0^1f(t)g(t) \, dt. $$ This map is a bounded linear functional on the space $C[0,1]$ w.r.t. the supremum norm. Is it true that the norm of $F$ (as a linear functional) is equal to the $L^1$-norm of $g$, i.e. $$ \| F \| = \| g \|_{L^1[0,1]}. $$ My attempt would be to use the Riesz representation theorem: $F$ is represented by a unique complex regular Borel measure $\mu$ on $[0,1]$ and the norm of $F$ is the total variation of $\mu$. Is it true that in the above case this measure is simply $g dt$ and the total variation of $g dt$ is the $L^1$-norm of $g$?

Solution 1:

One inequality is fairly straightforward: $$ |F(f)| \le \int_0^1|f(t)||g(t)|dt \le \|g\|_{L^1[0,1]}\|f\|_{C[0,1]}. $$ This gives $$ \|F\|\le \|g\|_{L^1[0,1]}. $$ To show the opposite inequality, let $0 < a < b < 1$, and choose $h > 0$ small enough that $h < a$ and $h < 1-b$. Let $f_{a,b,h}$ be the piecewise linear and continuous function that is given by $$ f_{a,b,h}(t)= \int_0^x\frac{1}{h}\chi_{[a-h,a]}(t)-\frac{1}{h}\chi_{[b,b+h]}(t) dt. $$ This function is piecewise linear and continuous. It is $0$ on $[0,a-h]$, rises linearly to $1$ at $a$, remains $1$ on $[a,b]$, declines linearly to $0$ at $b+h$ and is $0$ on $[b+h,1]$. Then, $$ F(f_{a,b,h})=\int_0^1f_{a,b,h}(t)g(t)dt \\ \lim_{h\downarrow 0}F(f_{a,b,h})=\int_a^b g(t)dt $$ Therefore, for all $0 \le a < b \le 1$, $$ \left|\int_a^bg(t)dt\right| \le \lim_{h\downarrow 0}|F(f_{a,b,h})|\le \|F\|\lim_{h\downarrow 0}\|f_{a,b,h}\|=\|F\|(b-a) \\ \left|\frac{1}{b-a}\int_a^b g(t)dt\right|\le \|F\|. $$ Therefore, $$ |g(x)| \le \|F\|,\;\; a.e. x\in [0,1] \\ \implies \|g\|_{L^1[0,1]} \le \|F\| $$ Therefore, $\|F\|=\|g\|_{L^1[0,1]}$.