Let $Tf(x)=\int_{[0,x]}fdm$ for $0 \leq x \leq 1$. Use the Open Mapping Theorem to prove that $T: L^1[0,1] \rightarrow C_0((0,1])$ is not onto.

Question

Let $Tf(x)=\int_{[0,x]}fdm$ for $0 \leq x \leq 1$.

Use the Open Mapping Theorem to prove that $T: L^1[0,1] \rightarrow C_0((0,1])$ is not onto.

Where $C_0((0,1])=\{f \in C[0,1] : f(0)=0\}$

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My attempt:

I am examining the action of $T$ on the functions $t \rightarrow e^{int}$. I am sort of running into a wall. If somebody could help me I'd greatly appreciate it.. I've been trying to figure this out for days now!

Answer 1

HINT.- (1) Choose an open ball $B(0;r)$ in $L_1$ and take $S=B(0;r)\setminus\{0\}$. This $S$ is still open in $L_1$.

(2) Choose a sequence $\{f_n\}$ in $S$ converging to $0$ (you do have a lot of possibilities!).

(3) In general your choice $\{f_n\}$ will be such that $\{T(f_n)\}$ converges to $0$ in your $C_0$(to you to avoid exceptions).

Since in this case the sequence $\{T(f_n)\}$ converges to an element in its border, the set $T(S)$ can not be open. Consequently $T$ is not onto by the Open Mapping Theorem.