Let u be a positive solution of $s\in (0, 1) $ \begin{equation} \left\{\begin{aligned} (\Delta )^{s} u &= 0 \text{ in } (2T, 2T)\\ u &=g\quad\text{in}\quad \mathbb R\setminus(2T, 2T). \end{aligned} \right. \end{equation} If $u \in C^2(T, T)\cap C^1 [T, T]$ and $g$ is a bounded positive function on $\mathbb R$, is there a Harnack inequality in one dimension for the above kind of equation.
Harnack's inequality for the fractional Laplace operator was proved already by M. Riesz in 1938. For an exact statement and further information, see Lemma 2.1 in R.F. Bass and M. Cranston.
By the way, in your case $$\begin{aligned} u(x) & = \frac{1}{\Gamma(1 + s) \lvert\Gamma(s)\rvert} \int_{\mathbb{R} \setminus (2T,2T)} \biggl( \frac{4 T^2  x^2}{y^2  4 T^2}\biggr)^{\!s} \frac{1}{x  y} \, g(y) dy \\ & \qquad + (4 T^2  1)^s \biggl(\frac{c_1}{2 T  x} + \frac{c_2}{2 T + x}\biggr) \end{aligned}$$ for $x \in (2T, 2T)$ for some constants $c_1, c_2 \geqslant 0$ by the result of K. Bogdan or Z.Q. Chen and R. Song (see also [Hmissi, Fonctions harmoniques pour les potentiels de Riesz sur la boule unite, Exposition. Math. 12(3) (1994): 281–288]), without assuming any smoothness of $u$.

$\begingroup$ I have changed the question a bit. Through google, I checked there seems to be many types of Harnack inequality. But I am not sure if it is applicable in this case. $\endgroup$ – sadiaz Aug 4 '18 at 10:27

$\begingroup$ I updated the answer and added a reference to Bass–Cranston that might be helpful for you. I am not sure what do you mean by "many types of Harnack inequality", though (parabolic? boundary? for other operators?). $\endgroup$ – Mateusz Kwaśnicki Aug 4 '18 at 12:01

$\begingroup$ I am looking for the interior Harnack kind of inequality. $\endgroup$ – sadiaz Aug 4 '18 at 12:12

$\begingroup$ @sadiaz: Does the Bass–Cranston reference work for you then? $\endgroup$ – Mateusz Kwaśnicki Aug 4 '18 at 12:17