# Dirichlet fractional Laplacian and zero boundary conditions

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.

(I asked the same question on Math.SE but it received low attention. I then though it might be better to post it here, let me know if it's not ok)

Yes, there are many functions with this property. In fact, these functions are dense in $C_0([0,1])$.

For every $g \in C_0([0,1])$ there is a unique $f \in C_0([0,1])$ such that $(-\Delta)^{\alpha/2} f(x) = g(x)$ for $x \in [0,1]$. This $f$ is given more-or-less explicitly: $$f(x) = \int_0^1 G_{(0,1)}(x, y) g(y) dy,$$ where $G_{(0,1)}(x,y)$ is the Green function for $(-\Delta)^{\alpha/2}$ in $(0, 1)$: $$G_{(0,1)}(x,y) = \frac{|x - y|^{\alpha - 1}}{2^\alpha (\Gamma(\alpha/2))^2} \int_0^{R(x,y)} \frac{s^{\alpha/2 - 1}}{\sqrt{1 + s}} \, ds$$ with $$R(x,y) = \frac{4 x (1 - x) y (1 - y)}{|x - y|^2} \, .$$

(Edited). In fact, the class of functions $f$ defined above constitute the domain of $(-\Delta)^{\alpha/2}$ on $(0, 1)$, with zero exterior condition, and therefore it is dense in $C_0([0,1])$. A rigorous proof of this claim is somewhat complicated, though.

The fractional Laplacian $-(-\Delta)^{\alpha/2}$ is the Feller generator of (the transition semigroup of) the symmetric $\alpha$-stable Lévy process $X_t$. Let $D = (0, 1)$ and let $X_t^D$ be the process $X_t$ killed when it first exits $D$. Since every boundary point of $D$ is regular, $X_t^D$ is a Feller process, and hence its Feller generator is a densely defined operator on $C_0(D)$. Denote this Feller generator by $L$.

By the general theory of strongly continuous semigroups, $f$ is in the domain of $L$ if and only if $f(x) = \int_D G_D(x, y) g(y) dy$ for some $g \in C_0(D)$, and in this case $g = -L f$. What remains to be proved is that if $f$ belongs to the domain of $L$, then $L f(x) = -(-\Delta)^{\alpha/2} f(x)$.

It was proved by Dynkin (in his brilliant book Markov processes) that $f \in C_0(D)$ is in the domain of $L$ if and only if the limit in $$\tilde{L} f(x) = \lim_{r \to 0^+} \frac{\mathbb{E}^x f(X(\tau_{B(x, r)})) - f(x)}{\mathbb{E}^x \tau_{B(x, r)}}$$ exists for every $x \in D$ and it defines a function $\tilde{L} f$ in $C_0(D)$; in this case $\tilde{L} f(x) = L f(x)$ for all $x \in D$. (The operator $\tilde{L}$ is the Dynkin characteristic operator).

Now the important fact is that the definition of $\tilde{L} f(x)$ does not depend on $D$! (It does not matter whether we use the killed process $X_t^D$ or the free process $X_t$ in the definition, as long as $f = 0$ in the complement of $D$).

Furthermore, convergence in the definition of $\tilde{L} f(x)$ implies convergence to the same limit in the definition of $-(-\Delta)^{\alpha/2} f(x)$ as a singular integral (for any fixed point $x$). This phenomenon seems to be specific to the fractional Laplacian, and it appears that it has been first observed here.

Final remark: Regarding the converse claim: "if $f \in C_0(D)$, $(-\Delta)^{\alpha/2} f(x)$ is well-defined as a singular integral for each $x \in D$ and it defines a $C_0(D)$ function", I believe it is not stated explicitly in the literature unless $D$ is full space.

• 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) – Rgkpdx Mar 5 '18 at 18:52
• 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 $f$. – Rgkpdx Mar 6 '18 at 10:44
• 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?). – Rgkpdx Mar 6 '18 at 10:44
• 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. – Rgkpdx Mar 6 '18 at 15:35