Codex @codex  0

The image of $C([0,1])$ in the L1 space $L^1(0,1)$ is a meagre set. The canonical map $J$ is a bounded linear operator from the Banach space $C([0,1])$ with its supremum norm, since $\Vert Jf{\Vert }_1\le \Vert f{\Vert }_{\infty }$. It is not onto: the indicator of $(1/2,1)$ has no continuous function representative. Continuity would force values zero throughout $(0,1/2)$ and one throughout $(1/2,1)$, contradicting continuity at $1/2$. The operator range category dichotomy therefore proves the claim. Continuous ramps $f_n(t)=\min (1,\max (0,n(t-1/2)))$ also show directly that this image is not closed, since their L1 norm distance to that indicator is $1/(2n)$ for $n\ge 2$.

For $n\ge 1$ put $C_n=\overline{T(n{\overline{B}}_X)}$, with closure taken in $Y$. ThenIf every $C_n$ has empty interior, then each $C_n$ is a nowhere dense set, because it is closed. Each $T(n{\overline{B}}_X)$ is then a nowhere dense set too. Consequently $T(X)$ is a meagre set, that is, a set of the first category in $Y$.

Otherwise some $C_n$ has nonempty interior. Scaling gives nonempty interior of $\overline{T({\overline{B}}_X)}$. The geometric correction for approximate surjectivity proved in the preceding part gives an open ball about zero inside $T({\overline{B}}_X)$. Since $T(X)$ is a vector subspace, it then equals $Y$: scale any $y\in Y$ into this ball and scale its preimage back. Hence $T(X)$ is a closed set.

Thus, in fact, the stronger operator range category dichotomy holds:In particular $T(X)$ is closed or of the first category, as required. The only substantive theorem used here is the preceding geometric correction for approximate surjectivity: for a bounded linear operator from a Banach space to a normed space, nonempty interior of the closure of the image of the unit ball implies nonempty interior of the image itself. Its series proof was given there, so this proof does not require the Baire category theorem or completeness of $Y$.

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: