Let $\overline{u}(r)$ denote the spherical average of *u* on the ball of radius *r*, that is,

$\overline{u}(r)={\u2a0d}_{\partial {B}_{r}}u\mathit{d}\sigma .$

The following result is a slight modification of [1, Theorem 1.1].

#### Theorem 2.1.

Let *u* be a nonnegative solution of

$-\mathrm{\Delta}u=g(u)\mathit{\hspace{1em}}\text{in}{B}_{1}\setminus \{0\},$

with an isolated singularity at the origin. Suppose that $g(t)$ is a locally Lipschitz function, which in a neighborhood of infinity satisfies the conditions below:

(i)

$g(t)$ is nondecreasing in *t*,

(ii)

${t}^{-\frac{n+2}{n-2}}g(t)$ is nonincreasing,

(iii)

$g(t)\ge c{t}^{p}$ for some $p\ge \frac{n}{n-2}$ and $c>0$.

Then

$u(x)=(1+O(|x|))\overline{u}(|x|)\hspace{1em}\text{as}x\to 0.$(2.1)

The original result in [1, Theorem 1.1] requires condition (i) above to be satisfied for all $t>0$, but a careful analysis of its proof shows that this condition is enough to hold in a neighborhood of infinity.

It is not hard to see that $g(t)={t}^{\alpha}|\mathrm{log}t{|}^{\beta}$ fulfills conditions (i)–(iii) in Theorem 2.1.
Moreover, it follows from [1, Lemma 2.1] that ${u}^{\alpha}|\mathrm{log}u{|}^{\beta}\in {L}^{1}({B}_{1})$ and *u* is a distribution solution of (1.1) in ${B}_{1}$, i.e., for any $\eta \in {C}_{c}^{\mathrm{\infty}}({B}_{1})$, we have

$-{\displaystyle {\int}_{{B}_{1}}}u\mathrm{\Delta}\eta dx={\displaystyle {\int}_{{B}_{1}}}{u}^{\alpha}|\mathrm{log}u{|}^{\beta}\eta dx.$

The next lemma provides an asymptotic upper bound for $\overline{u}$.

#### Lemma 2.2.

We have

$\overline{u}(r)=O\left({r}^{-\frac{2}{\alpha -1}}{\left(\mathrm{log}{\displaystyle \frac{1}{r}}\right)}^{-\frac{\beta}{\alpha -1}}\right)$(2.2)

and

${\overline{u}}^{\prime}(r)=O\left({r}^{-\frac{\alpha +1}{\alpha -1}}{\left(\mathrm{log}{\displaystyle \frac{1}{r}}\right)}^{-\frac{\beta}{\alpha -1}}\right)$(2.3)

as $r\to 0$.

#### Proof.

Throughout this proof, $c>0$ depends at most on *n*, α and β, and may differ from one line to another. As mentioned earlier, we have ${u}^{\alpha}|\mathrm{log}u{|}^{\beta}\in {L}^{1}({B}_{1})$, and thus from the divergence theorem and (1.1), we deduce that

$-{\overline{u}}^{\prime}(r)={\displaystyle \frac{c}{{r}^{n-1}}}{\displaystyle {\int}_{{B}_{r}}}{u}^{\alpha}|\mathrm{log}u{|}^{\beta}dx.$(2.4)

In particular, $\overline{u}(r)$ is monotone decreasing in *r*. Moreover, if (2.2) holds, then one may easily derive (2.3) from (2.4) and (2.1).

Henceforth, we shall prove (2.2). Especially, we shall assume that $\overline{u}(r)\ne O(1)$ as $r\to 0$, since the case $\overline{u}(r)=O(1)$ already satisfies (2.2). Under this assumption, we have $\overline{u}({r}_{k})\to \mathrm{\infty}$ for some ${r}_{k}\to 0$. Then the monotonicity of $\overline{u}$ implies that $\overline{u}(r)\to \mathrm{\infty}$ as $r\to 0$.

Taking *r* small enough, and using (2.1) and the fact that $s\mapsto {s}^{\alpha}{(\mathrm{log}s)}^{\beta}$ is increasing for large *s*, we deduce that

$-{\overline{u}}^{\prime}(r)\ge cr{\overline{u}}^{\alpha}(r){(\mathrm{log}\overline{u}(r))}^{\beta}.$

Hence, from the assumption $\overline{u}(r)\to \mathrm{\infty}$ as $r\to 0$ and the fact ${\overline{u}}^{\prime}(r)<0$, it follows that

${\int}_{\overline{u}(r)}^{\mathrm{\infty}}\frac{ds}{{s}^{\alpha}{(\mathrm{log}s)}^{\beta}}=-{\int}_{0}^{r}\frac{{\overline{u}}^{\prime}(r)dr}{{\overline{u}}^{\alpha}(r){(\mathrm{log}\overline{u}(r))}^{\beta}}\ge c{r}^{2}.$

Note that for any sufficiently large *s* satisfying $2|\beta |\le (\alpha -1)\mathrm{log}s$, we have

$-\frac{1}{\alpha -1}\frac{d}{ds}\left(\frac{1}{{s}^{\alpha -1}{(\mathrm{log}s)}^{\beta}}\right)=\left(1-\frac{\beta}{(\alpha -1)\mathrm{log}s}\right)\frac{1}{{s}^{\alpha}{(\mathrm{log}s)}^{\beta}}\ge \frac{1}{2{s}^{\alpha}{(\mathrm{log}s)}^{\beta}},$

whence we may proceed from the integral above as

$\frac{1}{{\overline{u}}^{\alpha -1}(r){(\mathrm{log}\overline{u}(r))}^{\beta}}\ge c{r}^{2}$

for sufficiently small $r>0$. Thus, we arrive at

${\overline{u}}^{\alpha -1}(r){(\mathrm{log}\overline{u}(r))}^{\beta}=O({r}^{-2})\mathit{\hspace{1em}}\text{as}r\to 0.$(2.5)

Setting $w(s)$ to be the inverse function^{1} of $s{e}^{s}$, we know that $\frac{s}{w(s)}$ is the inverse function of $s\mathrm{log}s$.
Since ${t}^{\alpha -1}{(\mathrm{log}t)}^{\beta}={(cs\mathrm{log}s)}^{\beta}$, with $s={t}^{\frac{\alpha -1}{\beta}}$, we deduce from (2.5) and the choice of *w* that

$\overline{u}(r)=O\left({r}^{-\frac{2}{\alpha -1}}w{({r}^{-\frac{2}{\beta}})}^{-\frac{\beta}{\alpha -1}}\right)\mathit{\hspace{1em}}\text{as}r\to 0.$

However, since $\mathrm{log}s-\mathrm{log}\mathrm{log}s\le w(s)\le \mathrm{log}s$ for sufficiently large *s*, we arrive at (2.2).
∎

Let us next define

$\psi (t,\theta )={r}^{\frac{2}{\alpha -1}}{\left(\mathrm{log}\frac{1}{r}\right)}^{\frac{\beta}{\alpha -1}}u(r,\theta ),$(2.6)

with $t=-\mathrm{log}r$ and $\theta \in {\mathbb{S}}^{n-1}$.

#### Lemma 2.3.

We have

${\psi}_{tt}+{\mathrm{\Delta}}_{\theta}\psi +a{\psi}_{t}-b\psi +{\zeta}^{\beta}{\psi}^{\alpha}=0$(2.7)

for large $t>1$ and $\theta \in {\mathbb{S}}^{n-1}$, where

$a(t)={\displaystyle \frac{4}{\alpha -1}}-n+2-{\displaystyle \frac{2\beta}{(\alpha -1)t}},$(2.8)$b(t)=\left((n-2)-{\displaystyle \frac{2}{\alpha -1}}+{\displaystyle \frac{\beta}{(\alpha -1)t}}\right)\left({\displaystyle \frac{2}{\alpha -1}}-{\displaystyle \frac{\beta}{(\alpha -1)t}}\right)-{\displaystyle \frac{\beta}{(\alpha -1){t}^{2}}}$(2.9)

and

$\zeta (t,\theta )=\frac{2}{\alpha -1}-\frac{\beta}{\alpha -1}\frac{\mathrm{log}t}{t}+\frac{\mathrm{log}\psi (t,\theta )}{t}.$(2.10)

#### Proof.

We take ${r}_{0}>0$ small enough such that $\mathrm{log}u>0$ in ${B}_{{r}_{0}}$, and set ${t}_{0}=-\mathrm{log}{r}_{0}$.
In what follows, we take $t\ge {t}_{0}$ and $0<r\le {r}_{0}$, unless stated otherwise. For notational convenience, let us write

$\varphi (r)={r}^{\frac{2}{\alpha -1}}{\left(\mathrm{log}\frac{1}{r}\right)}^{\frac{\beta}{\alpha -1}},$

so that $\psi (t,\theta )=\varphi (r)u(r,\theta )$. Since ${\partial}_{t}=-r{\partial}_{r}$ and ${\partial}_{tt}=r{\partial}_{r}+{r}^{2}{\partial}_{rr}$, we have

${\psi}_{tt}+{\mathrm{\Delta}}_{\theta}\psi ={r}^{2}\varphi \mathrm{\Delta}u+\left(2r{\varphi}^{\prime}-(n-2)\varphi \right)r{u}_{r}+(r{\varphi}^{\prime}+{r}^{2}{\varphi}^{\prime \prime})u,$

where the left and right side are evaluated in $(t,\theta )$ and, respectively, in $(r,\theta )$, and by ${\varphi}^{\prime}$ and ${\varphi}^{\prime \prime}$, we denoted $\frac{d\varphi}{dr}$ and, respectively, $\frac{{d}^{2}\varphi}{d{r}^{2}}$.
Setting

$\eta (r)=\frac{2}{\alpha -1}+\frac{\beta}{(\alpha -1)\mathrm{log}r},$

we observe that $r{\varphi}^{\prime}=\eta \varphi $ and ${r}^{2}{\varphi}^{\prime}=({\eta}^{2}-\eta +r{\eta}^{\prime})\varphi $, and therefore

${\psi}_{tt}+{\mathrm{\Delta}}_{\theta}\psi =-{r}^{2}\varphi {u}^{\alpha}{(\mathrm{log}u)}^{\beta}+(2\eta -n+2)r{u}_{r}\varphi +({\eta}^{2}+r{\eta}^{\prime})\varphi u$$=-{r}^{2}\varphi {u}^{\alpha}{(\mathrm{log}u)}^{\beta}-(2\eta -n+2){\psi}_{t}+\left((n-2)\eta -{\eta}^{2}+r{\eta}^{\prime}\right)\psi ,$

where we used the fact that ${\psi}_{t}=-r{\varphi}^{\prime}u-r\varphi {u}_{r}=-\eta \psi -r\varphi {u}_{r}$ and $\psi =\varphi u$ in deriving the second identity.

In view of (2.8) and (2.9), it is not hard to check that

$a(t)=2\eta (r)-n+2$(2.11)

and

$b(t)=(n-2)\eta (r)-{\eta}^{2}(r)+r{\eta}^{\prime}(r).$(2.12)

On the other hand, we know from (2.6) that

$\mathrm{log}u(r,\theta )=\frac{2t}{\alpha -1}-\frac{\beta}{\alpha -1}\mathrm{log}t+\mathrm{log}\psi (t,\theta ),$

from which we may also deduce that

$\zeta (t,\theta )=\frac{\mathrm{log}u(r,\theta )}{\mathrm{log}\frac{1}{r}}.$(2.13)

One may also notice from (2.6) that

${r}^{2}\varphi (r)u(r,\theta )={t}^{-\beta}{\psi}^{\alpha}(t,\theta ).$(2.14)

Hence, inserting (2.11)–(2.14), we arrive at equation (2.7), which completes the proof.
∎

Let us define

$\overline{\psi}(t)={\u2a0d}_{{\mathbb{S}}^{n-1}}\psi (t,\theta )\mathit{d}\theta ={r}^{\frac{2}{\alpha -1}}{\left(\mathrm{log}\frac{1}{r}\right)}^{\frac{\beta}{\alpha -1}}\overline{u}(r)$(2.15)

and

$\overline{\zeta}(t)=\frac{2}{\alpha -1}-\frac{\beta}{\alpha -1}\frac{\mathrm{log}t}{t}+\frac{\mathrm{log}\overline{\psi}(t)}{t}.$(2.16)

Averaging (2.7) over ${\mathbb{S}}^{n-1}$, we obtain

${\overline{\psi}}^{\prime \prime}+a{\overline{\psi}}^{\prime}-b\overline{\psi}+{\overline{\zeta}}^{\beta}{\overline{\psi}}^{\alpha}+{\u2a0d}_{{\mathbb{S}}^{n-1}}({\psi}^{\alpha}{\zeta}^{\beta}-{\overline{\psi}}^{\alpha}{\overline{\zeta}}^{\beta})\mathit{d}\theta =0$(2.17)

for large *t*.

#### Lemma 2.4.

We have

$\psi (t,\theta )-\overline{\psi}(t)=\overline{\psi}(t)O({e}^{-t}),$(2.18)$\left|{\displaystyle \frac{\partial}{\partial t}}(\psi (t,\theta )-\overline{\psi}(t))\right|+\left|{\nabla}_{\theta}(\psi (t,\theta )-\overline{\psi}(t,\theta ))\right|=\overline{\psi}(t)O({e}^{-t}),$(2.19)$\overline{\psi}(t)=O(1)\hspace{1em}\text{and}\hspace{1em}{\overline{\psi}}^{\prime}(t)=O(1)$(2.20)

as $t\to \mathrm{\infty}$.

#### Proof.

In this proof, $C>0$ will depend on *n* only and may differ from one line to another.
The estimates in (2.20) follow immediately from (2.15), (2.2) and (2.3).
Moreover, since

$\psi (t,\theta )-\overline{\psi}(t)={r}^{-\frac{2}{\alpha -1}}{\left(\mathrm{log}\frac{1}{r}\right)}^{-\frac{\beta}{\alpha -1}}(u(r,\theta )-\overline{u}(r)),$

(2.18) can be easily deduced from (2.1). Thus, we are only left with proving (2.19).

For notational convenience, let $\overline{u}(x)$ denote $\overline{u}(|x|)$.
Also let us denote by ${A}_{r}$ the annulus ${B}_{2r}\setminus {\overline{B}}_{r/2}$.
From (2.1), we have

$-\mathrm{\Delta}(u-\overline{u})={\overline{u}}^{\alpha}|\mathrm{log}\overline{u}{|}^{\beta}O(r)\mathit{\hspace{1em}}\text{in}{A}_{r}\text{as}r\to 0.$

Therefore, it follows from the interior gradient estimates that

$|\nabla (u-\overline{u})|\le C({\displaystyle \frac{1}{r}}{\parallel u-\overline{u}\parallel}_{{L}^{\mathrm{\infty}}({A}_{r})}+{r}^{2}\parallel {\overline{u}}^{\alpha}{|\mathrm{log}\overline{u}{|}^{\beta}\parallel}_{{L}^{\mathrm{\infty}}({A}_{r})})$$\le C({\parallel \overline{u}\parallel}_{{L}^{\mathrm{\infty}}({A}_{r})}+{r}^{2}\parallel {\overline{u}}^{\alpha}{|\mathrm{log}\overline{u}{|}^{\beta}\parallel}_{{L}^{\mathrm{\infty}}({A}_{r})})\hspace{1em}\text{on}\partial {B}_{r}.$

We regard (1.1) as $-\mathrm{\Delta}u=m(x)u$ in ${B}_{1}\setminus \{0\}$, where $m={u}^{\alpha -1}|\mathrm{log}u{|}^{\beta}$.
In view of (2.5) and (2.1), we have that $m(x)=O({|x|}^{-2})$, so the Harnack inequality implies

$\underset{{A}_{r}}{sup}u\le C\underset{{A}_{r}}{inf}u.$

Using this observation along with (2.1), (2.5) and the above gradient estimate, we find

$|\nabla (u-\overline{u})|\le C(\overline{u}+{r}^{2}{\overline{u}}^{\alpha}|\mathrm{log}\overline{u}{|}^{\beta})\le C\overline{u}\mathit{\hspace{1em}}\text{on}\partial {B}_{r}.$(2.21)

Since

$\psi (t,\theta )-\overline{\psi}(t)={r}^{\frac{2}{\alpha -1}}{\left(\mathrm{log}\frac{1}{r}\right)}^{\frac{\beta}{\alpha -1}}(u(r,\theta )-\overline{u}(r)),$

(2.19) follows from (2.21), (2.2) and (2.1).
∎

#### Corollary 2.5.

We have

${\u2a0d}_{{\mathbb{S}}^{n-1}}\left({\psi}^{\alpha}(t,\theta ){\zeta}^{\beta}(t,\theta )-{\overline{\psi}}^{\alpha}(t){\overline{\zeta}}^{\beta}(t)\right)\mathit{d}\theta =O({e}^{-t})\mathit{\hspace{1em}}\text{as}t\to \mathrm{\infty}.$(2.22)

#### Proof.

From (2.10) and (2.16), we know that

$\zeta (t,\theta )-\overline{\zeta}(t)=\frac{1}{t}\mathrm{log}\frac{\psi (t,\theta )}{\overline{\psi}(t)}$

for any $t>0$ and $\theta \in {\mathbb{S}}^{n-1}$. Due to (2.18), we have

$\zeta (t,\theta )-\overline{\zeta}(t)=\frac{1}{t}\mathrm{log}(1+O({e}^{-t}))=O\left(\frac{{e}^{-t}}{t}\right)\mathit{\hspace{1em}}\text{as}t\to \mathrm{\infty}.$

Using the above estimate together with Lemma 2.4, we have

${\psi}^{\alpha}(t,\theta ){\zeta}^{\beta}(t,\theta )-{\overline{\psi}}^{\alpha}(t){\overline{\zeta}}^{\beta}(t)=({\psi}^{\alpha}(t,\theta )-{\overline{\psi}}^{\alpha}(t)){\zeta}^{\beta}(t,\theta )+{\overline{\psi}}^{\alpha}(t)({\zeta}^{\beta}(t,\theta )-{\overline{\zeta}}^{\beta}(t))$$=(\psi (t,\theta )-\overline{\psi}(t))O(1)+(\zeta (t,\theta )-\overline{\zeta}(t))O(1)$$=O({e}^{-t})\hspace{1em}\text{as}t\to \mathrm{\infty}.$

An integration over ${\mathbb{S}}^{n-1}$ in the above estimate, will lead us to (2.22).
∎

#### Lemma 2.6.

We have either

$\underset{t\to \mathrm{\infty}}{lim}\overline{\psi}(t)=0$

or

$\underset{t\to \mathrm{\infty}}{lim}\overline{\psi}(t)=A,$(2.23)

with *A* given by (1.4).

#### Proof.

Let ψ and $\overline{\psi}$ be defined by (2.6) and (2.15), respectively.
Multiplying (2.17) by ${\overline{\psi}}^{\prime}$ and integrating it over $[t,T]$, from (2.18)–(2.20) and (2.22), we find

$\frac{1}{2}{[{\overline{\psi}}^{\prime 2}]}_{t}^{T}+{\int}_{t}^{T}a{\overline{\psi}}^{\prime 2}\mathit{d}s-\frac{1}{2}{\int}_{t}^{T}b{({\overline{\psi}}^{2})}^{\prime}\mathit{d}s+\frac{1}{\alpha +1}{\int}_{t}^{T}{\overline{\zeta}}^{\beta}{({\overline{\psi}}^{\alpha +1})}^{\prime}\mathit{d}s+O({e}^{-t})=0.$

By (2.9) and (2.20), we have $b(t)=O(1)$ and ${b}^{\prime}(t)=O({t}^{-2})$, which leads us to

${\int}_{t}^{T}b{({\overline{\psi}}^{2})}^{\prime}\mathit{d}s={\left[b{\overline{\psi}}^{2}\right]}_{t}^{T}-{\int}_{t}^{T}{b}^{\prime}{\overline{\psi}}^{2}\mathit{d}s=O\left(1+{\int}_{t}^{T}\frac{ds}{{s}^{2}}\right)=O(1).$(2.24)

Similarly, from (2.10) and (2.20), we find $\overline{\zeta}(t)=O(1)$ and ${\overline{\zeta}}^{\prime}(t)=O({t}^{-2}\mathrm{log}t)$, from which it follows that

${\int}_{t}^{T}{\overline{\zeta}}^{\beta}{({\overline{\psi}}^{\alpha +1})}^{\prime}\mathit{d}s=O\left(1+{\int}_{t}^{T}\frac{\mathrm{log}s}{{s}^{2}}\mathit{d}s\right)=O(1).$(2.25)

Since α is chosen as in (1.2), we know from (2.8) that $a(t)$ is positive and bounded away from zero for all large $t>1$.
Thus, (2.24), (2.25) and (2.17) yield

${\int}_{t}^{T}{\overline{\psi}}^{\prime 2}\mathit{d}s=O(1).$(2.26)

In view of (2.17), it follows from (2.20) and (2.22) that ${\overline{\psi}}^{\prime \prime}(t)=O(1)$, and hence ${\overline{\psi}}^{\prime 2}(t)$ is uniformly Lipschitz for large $t>1$.
Hence, we deduce that

$\underset{t\to \mathrm{\infty}}{lim}{\overline{\psi}}^{\prime}(t)=0.$(2.27)

Now we multiply (2.17) by ${\overline{\psi}}^{\prime \prime}$ and integrate it over $[t,T]$, which leads us to

${\int}_{t}^{T}{({\overline{\psi}}^{\prime \prime})}^{2}\mathit{d}s+\frac{1}{2}{\int}_{t}^{T}a{({\overline{\psi}}^{\prime 2})}^{\prime}\mathit{d}s-{\int}_{t}^{T}b\overline{\psi}{\overline{\psi}}^{\prime \prime}\mathit{d}s+{\int}_{t}^{T}{\overline{\zeta}}^{\beta}{\overline{\psi}}^{\alpha}{\overline{\psi}}^{\prime \prime}\mathit{d}s+O({e}^{-t})=0,$(2.28)

due to (2.18)–(2.20) and (2.22), as before. Note that from (2.8) we have $a(t)=O(1)$ and ${a}^{\prime}(t)=O({t}^{-2})$ as $t\to \mathrm{\infty}$.
Hence, from (2.20) and (2.26), we derive

${\int}_{t}^{T}a{({\overline{\psi}}^{\prime 2})}^{\prime}\mathit{d}s={\left[a{\overline{\psi}}^{\prime 2}\right]}_{t}^{T}-{\int}_{t}^{T}{a}^{\prime}{\overline{\psi}}^{\prime 2}\mathit{d}s=O(1).$(2.29)

On the other hand, since $\overline{\psi}{\overline{\psi}}^{\prime \prime}=\frac{1}{2}{({\overline{\psi}}^{2})}^{\prime \prime}-{\overline{\psi}}^{\prime 2}$, a further integration by parts produces

${\int}_{t}^{T}b\overline{\psi}{\overline{\psi}}^{\prime \prime}\mathit{d}s=\frac{1}{2}{\left[b{({\overline{\psi}}^{2})}^{\prime}\right]}_{t}^{T}-{\int}_{t}^{T}\left(\frac{1}{2}{b}^{\prime}{({\overline{\psi}}^{2})}^{\prime}+b{\overline{\psi}}^{\prime 2}\right)\mathit{d}s=O(1),$(2.30)

where the second equality can be deduced analogously to the derivation of (2.24).
Similarly, we also observe that

${\int}_{t}^{T}{\overline{\zeta}}^{\beta}{\overline{\psi}}^{\alpha}{\overline{\psi}}^{\prime \prime}\mathit{d}s=O(1).$(2.31)

Due to (2.29), (2.30) and (2.31), (2.28) leads us to

${\int}_{t}^{T}{({\overline{\psi}}^{\prime \prime})}^{2}\mathit{d}s=O(1).$(2.32)

Differentiating (2.17) with respect to *t*, we deduce, from (2.20), (2.22) and ${\overline{\psi}}^{\prime \prime}(t)=O(1)$, that ${\overline{\psi}}^{\prime \prime \prime}(t)=O(1)$.
Therefore, ${\overline{\psi}}^{\prime \prime 2}(t)$ is uniformly Lipschitz for large $t>1$, from which combined with (2.32), we obtain

$\underset{t\to \mathrm{\infty}}{lim}{\overline{\psi}}^{\prime \prime}(t)=0.$(2.33)

To this end, we shall pass to the limit in (2.17) with $t\to \mathrm{\infty}$. Note that from (2.9) we have

$\underset{t\to \mathrm{\infty}}{lim}b(t)=\frac{2}{\alpha -1}\left(n-2-\frac{2}{\alpha -1}\right),$

while, from (2.16) and (2.20), it follows that

$\underset{t\to \mathrm{\infty}}{lim}\zeta (t)=\frac{2}{\alpha -1}.$

Although we do not know yet if $\overline{\psi}(t)$ converges as $t\to \mathrm{\infty}$, we still know from (2.20) that it converges along a subsequence. Denoting by ${\overline{\psi}}_{0}$ a limit value of $\overline{\psi}(t)$ along a subsequence, say $t={t}_{j}\to \mathrm{\infty}$, after passing to the limit in (2.17), with $t={t}_{j}$, we obtain, from (2.22), (2.27) and (2.33), that

$\frac{2}{\alpha -1}\left(n-2-\frac{2}{\alpha -1}\right){\overline{\psi}}_{0}-{\left(\frac{2}{\alpha -1}\right)}^{\beta}{\overline{\psi}}_{0}^{\alpha}=0.$

Thus, in view of (1.4), we have

${\overline{\psi}}_{0}=0\mathit{\hspace{1em}}\text{or}\mathit{\hspace{1em}}{\overline{\psi}}_{0}=A.$(2.34)

Now the continuity of $\overline{\psi}$ implies that $\overline{\psi}(t)$ converges as $t\to \mathrm{\infty}$ (without extracting any subsequence) either to 0 or *A*.
If there are two distinct sequences ${t}_{j}\to \mathrm{\infty}$ and ${t}_{j}^{\prime}\to \mathrm{\infty}$ such that $\overline{\psi}({t}_{j})\to 0$ and $\overline{\psi}({t}_{j}^{\prime})\to A$, then by the intermediate value theorem, there must exist some other ${t}_{j}^{\prime \prime}\to \mathrm{\infty}$ such that $\overline{\psi}({t}_{j}^{\prime \prime})\to \frac{A}{2}$, which violates (2.34).
Thus, the proof is completed.
∎

We are now in a position to prove Theorem 1.1.

#### Proof of Theorem 1.1.

If (2.23) is true, then, in view of (2.15), we observe that

$\overline{u}(r)=A(1+o(1)){r}^{-\frac{2}{\alpha -1}}{\left(\mathrm{log}\frac{1}{r}\right)}^{-\frac{\beta}{\alpha -1}}\mathit{\hspace{1em}}\text{as}r\to 0.$

Hence, from (2.1), we derive (1.3) and (1.4), which establishes the proof for Theorem 1.1 (ii).

Henceforth, let us suppose that

$\underset{t\to \mathrm{\infty}}{lim}\overline{\psi}(t)=0.$

The rest of the argument follows closely that of the proof in [1, Theorem 1.3].

In view of (2.8) and (2.9), we may rephrase (2.17) as

${\overline{\psi}}^{\prime \prime}+({a}_{0}+o(1)){\overline{\psi}}^{\prime}-({b}_{0}+o(1))\overline{\psi}+{\overline{\zeta}}^{\beta}{\overline{\psi}}^{\alpha}+{\u2a0d}_{{\mathbb{S}}^{n-1}}({\zeta}^{\beta}{\psi}^{\alpha}-{\overline{\zeta}}^{\beta}{\overline{\psi}}^{\alpha})\mathit{d}\theta =0,$

with

${a}_{0}={\displaystyle \frac{4}{\alpha -1}}-n+2\hspace{1em}\text{and}\hspace{1em}{b}_{0}={\displaystyle \frac{2}{\alpha -1}}\left(n-2-{\displaystyle \frac{2}{\alpha -1}}\right).$

Thus, the decay of $\overline{\psi}(t)$ is determined by the negative root of

${\lambda}^{2}+{a}_{0}\lambda -{b}_{0}=0.$

Since ${a}_{0}^{2}+4{b}_{0}={(n-2)}^{2}$, the root λ is

$\lambda =-\frac{1}{2}\left({a}_{0}+\sqrt{{a}_{0}^{2}+4{b}_{0}}\right)=-\frac{2}{\alpha -1}.$

Therefore, we have

$\overline{\psi}(t)=O\left({e}^{-\frac{2t}{\alpha -1}}\right)\hspace{1em}\text{as}t\to \mathrm{\infty}.$

In view of (2.15), we obtain

$\overline{u}(r)={\left(\mathrm{log}{\displaystyle \frac{1}{r}}\right)}^{-\frac{\beta}{\alpha -1}}O(1)\hspace{1em}\text{as}r\to 0.$(2.35)

Now if $\beta >0$, then we deduce from (2.35) that $\overline{u}(r)\to 0$ as $r\to 0$, from which, combined with (2.1), it follows that $u(x)\to 0$ as $x\to 0$.
Hence, the origin is a removable singularity.
Similarly, if $\beta =0$, then (2.35) implies that $\overline{u}(r)=O(1)$, and thus the origin is again a removable singularity.

Hence, we are only left with the case $\beta <0$. Since $u(x)=(1+O(r))\overline{u}(r)$, (2.35) implies that

${\int}_{{B}_{1}}{u}^{q}\mathit{d}x\le C{\int}_{{B}_{1}}{\left(\mathrm{log}\frac{1}{|x|}\right)}^{-\frac{\beta q}{\alpha -1}}\mathit{d}x\le C{\int}_{0}^{1}{r}^{n-1}{\left(\mathrm{log}\frac{1}{r}\right)}^{-\frac{\beta q}{\alpha -1}}\mathit{d}r\le C$

for each $q\ge 1$, for some constant $C>0$ depending on *n*, α, β and *q*.
Therefore, ${u}^{\alpha}|\mathrm{log}u{|}^{\beta}\in {L}^{p}({B}_{1})$ for any $p\ge 1$, and in particular for $p>n$.
This implies that $\mathrm{\Delta}u\in {L}^{p}({B}_{1})$ for $p>n$, so $u\in {C}^{1,\alpha}({B}_{1/2})$ for $\alpha =1-\frac{n}{p}$, proving again that the origin is a removable singularity.
Thus, the proof of Theorem 1.1 (i) is completed.
∎

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