Does there exists a non-zero function $$f\in C_0([0,1]):=\{f:[0,1]\to \mathbb R:\ f\text{ is continuous and } f(0)=f(1)=0\},$$ such that $(-\Delta)^{\frac\alpha 2}f\in C_0([0,1]) $, where $(-\Delta)^{\frac\alpha 2}$ is the Dirichlet fractional Laplacian defined by $$ (-\Delta)^{\frac\alpha 2}f(x):=\int_0^1(f(x)-f(y))\frac{dy}{|x-y|^{1+\alpha}}+f(x)\int_{\mathbb R\backslash (0,1)}\frac{dy}{|x-y|^{1+\alpha}},\ \ x\in(0,1), $$ and $(-\Delta)^{\frac\alpha 2}f(0):=\lim_{x\to 0^+}(-\Delta)^{\frac\alpha 2}f(x)$, $(-\Delta)^{\frac\alpha 2}f(1):=\lim_{x\to 1^-}(-\Delta)^{\frac\alpha 2}f(x)$, where $\alpha \in(0,2)$ (principal value definition of the first integral).
My question is motivated by the last part of Theorem 2.7, as the existence of the function I am looking for would allow me to apply that theorem.
Note that the above definition of $(-\Delta)^{\frac \alpha 2}$ agrees with the definition of the restricted fractional Laplacian in formula (3.1) here, which differs from definition of the spectral fractional Laplacian in formula (3.4) here.
CLAIM. If $f\in C^\infty(\mathbb R)$ and supported in $[a, b]\subset (0, 1)$ then $(-\Delta)^{\frac a2}f\in C^\infty(0,1)$ and it satisfies the Dirichlet boundary conditions on $[0, 1]$.
Proof. (Sketchy) Let $(\phi_n)_{n\ge 1}$ be a sequence of orthonormal Dirichlet eigenfunctions. We decompose $$ f(x)=\sum_{n=1}^\infty \hat{f}(n)\phi_n(x).$$ By the smoothness assumption, the sequence $\hat{f}(n)$ decays faster than any polynomial, and so $$ (-\Delta)^{\frac a 2} f = \sum_{n=1}^\infty n^a\hat{f}(n)\phi_n(x)$$ is an absolutely convergent series.
Remark. In this post, the Dirichlet fractional Laplacian is defined by the formula $$(-\Delta)^{\frac{a}{2}}f(x) : = \sum_{n=1}^\infty n^a \hat f(n) \phi_n(x).$$