Past exam of the functional analysis II course of the University of Oxford / 2023 / 2 / b / iv

Let $J:C([0,1])\to L^1(0,1)$ send a continuous function to its almost everywhere equivalence class. The domain with the supremum norm and the target with the L1 norm are Banach spaces, andThus $J$ is a bounded linear operator, and the requested vector subspace is $J(C([0,1]))$.

To show this image of a linear map is not closed, consider the continuous functionsand $g=1_{(1/2,1)}$. Their L1 norm difference is the area of one triangle:But $g$ has no continuous function representative. Indeed, a continuous function equal to zero almost everywhere on $(0,1/2)$ must be identically zero there: any nonzero value would persist on an open interval of positive Lebesgue measure. Likewise any continuous representative would be identically one on $(1/2,1)$. Continuity at $1/2$ makes these requirements incompatible. Therefore $g\notin J(C([0,1]))$, although it lies in its closure.

The preceding operator range category dichotomy proves continuous functions are meagre in L1: