For $\|f\|_\infty\leq1$, $|Tf(x)|\leq1-e^{-x}$, with equality for f = 1; hence $\|T\|=1-e^{-1}<1$. This makes I − T invertible by its Neumann series. T is not onto since every Tf vanishes at zero. The triangle inequality gives $\|I+T\|\leq1+\|T\|$, and f = 1 attains equality at x = 1. Thus D.
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