Correction: Wang, M. and Yin, S. Some Liouville Theorems on Finsler Manifolds. Mathematics, 2019, 7, 351

The authors are sorry to report that the proof of case III of Theorem 1.2 in their recently published paper [1] was incorrect. Upon revising the manuscript, they mistakenly thought that $u_k$ is in $H^2$ and $\Delta u_k=0$ holds on some particular subset of the manifold M. Consequently, the authors wish to make the following corrections to the paper at this time:

We first remark that the proof of case III is different from the one in [2] because there is a mistake there. For every $k\in {\mathbb{R}}^{+}$, set

$$u_k=\left\{\begin{array}{cc}k,\hfill & u\ge k;\hfill \\ u,\hfill & u<k.\hfill \end{array}\right.$$

In what follows, we will follow the arguments in [3] (p. 178) with some modifications. Let $\beta$ be a symmetric, convex, and bounded smooth function with $|{\beta }^{\prime }|<1$ and $\left|s\right|<\beta <ϵ+\left|s\right|$, where $0<ϵ<1$ is such that $u-ϵ>0$. Define

$${\tilde{u}}_k=\frac{u+k}{2}-\frac{\beta (u-k)}{2}.$$

Then, for any positive integer k, it holds that ${\tilde{u}}_k>\frac{u+k}{2}-\frac{ϵ+|u-k|}{2}>0$. Moreover, ${\tilde{u}}_k$ is a superharmonic function in a weak sense. Indeed, by definition, we have $d{\tilde{u}}_k=\frac{1}{2}(1-{\beta }^{\prime })du$, which yields $\nabla {\tilde{u}}_k=\frac{1}{2}(1-{\beta }^{\prime })\nabla u$ by Legendre transformation. As ${\tilde{u}}_k\in H_{loc}^2$ and thus $\Delta {\tilde{u}}_k=0$ a.e. on $M\setminus M_u$, for $\psi$ defined in Case I, we have

$$\begin{array}{cc}\hfill 2{\int }_{B_{x_0}^{-}\left(R\right)}\psi \Delta {\tilde{u}}_{k}d\mu & =-2{\int }_{B_{x_0}^{-}\left(R\right)}d\psi (\nabla {\tilde{u}}_k)d\mu =-{\int }_{B_{x_0}^{-}\left(R\right)}(1-{\beta }^{\prime })d\psi (\nabla u)d\mu \hfill \\ & =-{\int }_{B_{x_0}^{-}\left(R\right)}d\left[(1-{\beta }^{\prime })\psi \right](\nabla u)d\mu -{\int }_{B_{x_0}^{-}\left(R\right)}{\beta }^{\prime \prime }\psi F{(\nabla u)}^{2}d\mu \hfill \\ & \le -{\int }_{B_{x_0}^{-}\left(R\right)}d\left[(1-{\beta }^{\prime })\psi \right](\nabla u)d\mu ={\int }_{B_{x_0}^{-}\left(R\right)}(1-{\beta }^{\prime })\psi \Delta ud\mu \hfill \\ & \le 0.\hfill \end{array}$$

The last step holds because $(1-{\beta }^{\prime })\psi$ is differentiable almost everywhere on $B_{x_0}^{-}\left(R\right)$ with bounded differential, and u is superharmonic. Moreover, ${\tilde{u}}_k$ is smooth on the open subset $M_u$ and is also superharmonic, in the classical sense, on $M_u$. Notice that $\psi$ is differentiable almost everywhere on $B_{x_0}^{-}\left(R\right)$ with bounded differential. Hence, by similar arguments, we can also obtain (4) (see [1], p. 6) for ${\tilde{u}}_k$ on $B_{x_0}^{-}\left(R\right)$ as in case I. Set $v_k={\tilde{u}}_k^{\frac{q}{2}}$ for any $q\in (0,1)$. Then we have (5) (see [1], p. 7) as follows:

$$\begin{array}{cc}& {(1-\frac{1}{q})}^2{\int }_{B_{x_0}^{-}\left(R\right)}{\psi }^{2}F{(\nabla v_k)}^{2}d\mu \hfill \\ \hfill \le & \widehat{C}{\left({\int }_{B_{x_0}^{-}\left(R\right)\setminus {\overline{B}}_{x_0}^{-}\left(r_0\right)}{v}_k^2\right)}^{\frac{1}{2}}{\left({(1-\frac{1}{q})}^2{\int }_{B_{x_0}^{-}\left(R\right)\setminus {\overline{B}}_{x_0}^{-}\left(r_0\right)}{\psi }^{2}F{(\nabla v_k)}^{2}d\mu \right)}^{\frac{1}{2}}\hfill \\ \hfill =& \widehat{C}{(V_q\left(R\right)-V_q\left(r_0\right))}^{\frac{1}{2}}{\left({(1-\frac{1}{q})}^2{\int }_{B_{x_0}^{-}\left(R\right)\setminus {\overline{B}}_{x_0}^{-}\left(r_0\right)}{\psi }^{2}F{(\nabla v_k)}^{2}d\mu \right)}^{\frac{1}{2}}.\hfill \end{array}$$

On the other hand, note that ${\tilde{u}}_k\le k$, and thus

$${\int }_{B_{x_0}^{-}\left(R\right)}{\tilde{u}}_k^{q}d\mu \le {\int }_{B_{x_0}^{-}\left(R\right)}{k}^{q}d\mu =k^{q}V\left(R\right),$$

which implies that

$${\int }_1^{\infty }\frac{r}{V_q\left(r\right)}dr=\infty .$$

Then by the same discussion in the proof of (2) (see [1], pp. 5–7) and Case I of (1) (see [1], p. 7), we show that this ${\tilde{u}}_k$ is constant. Take then a sequence ${\beta }_n$ (such that each ${\beta }_n$ satisfies the same properties as $\beta$) uniformly converging to the absolute value function. Every ${\tilde{u}}_{k,n}$ is then constant. These constants are bounded (they are in $(0,k)$). Thus, up to pass to a subsequence ${\tilde{u}}_{k,n}$ converges uniformly to $u_k$ and to a constant at the same time. Hence, $u_k$ must be constant. k being arbitrary, u is also constant.