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Actually, looking at Theorem 2.3 in arxiv.org/pdf/1604.06421.pdf does it follow that f∉Dom(L(0,1))? (where f is the f defined in the comment of mine about the counter example.). Also your Green function $G_{(0,1)}$ does not seem to allow for $f$ to be smooth.
Also, I do not see how the characteristic operator $\tilde Lf$ can equal $-(-\Delta)^{\frac \alpha 2}f$ as values of $f$ outside any ball are ignored in the limiting definition ($\tilde Lf(x)=\tilde L g(x)$ for any $f,g$ such that $f=g$ a neighborhood of $x$, no?).
Sorry, but is the following a counter-example? Let $f$ be smooth and positive inside its compact support in $(0,1)$. Then $(-\Delta)^{\frac \alpha 2}f(0)\neq 0=- L_{(0,1)}f(0)$,where $(L_{(0,1)},Dom(L_{(0,1)}))$ is the generator of the killed-$\alpha$-stable process (working on $C_0([0,1])$) and $f\in Dom(L_{(0,1)})$. To conclude note that zero-extension of $f$ belongs the domain of the $\alpha$-stable process and its the generator agrees with $-(-\Delta)^{\frac \alpha 2}$ on $f$.
I am indeed interested in the pointwise equality of the generator of the killed process with the fractional Laplacian applied to the function extended to zero outside of $(0,1)$. Are there such functions? Is $(-\Delta)^{\frac \alpha 2}$ in your first line the generator? (sorry but I'm not sure you answered this part)
