Let us begin with the proof of Theorem 1.4.
Proof of Theorem 1.4
Set \(f_{n}=\frac{f}{1+\frac {1}{n}}\), obviously, \(f_{n}\rightarrow f\) in \(L^{1}(\Omega)\) as \(n\rightarrow\infty\). Let \(\phi=T_{k}(u_{n})\) as a test function in (2.4), we have
$$\begin{aligned} &\frac{a_{N,s}}{2} \int_{\Omega}\int_{\Omega}\frac {(u_{n}(x)-u_{n}(y))(T_{k}(u_{n}(x))-T_{k}(u_{n}(y)))}{|x-y|^{N+2s}} \\ &\quad {}-\lambda \int_{\Omega}\frac{u_{n}T_{k}(u_{n})}{|x|^{2s}+\frac{1}{n}}+ \int _{\Omega}u^{p}_{n}T_{k}(u_{n})= \int_{\Omega}f_{n} T_{k}(u_{n}). \end{aligned}$$
(3.1)
Since for any \(\sigma\in\mathbb{R}\), \(\sigma=T_{k}(\sigma)+G_{k}(\sigma)\),
$$\begin{aligned} &\bigl(u_{n}(x)-u_{n}(y)\bigr) \bigl(T_{k} \bigl(u_{n}(x)\bigr)-T_{k}\bigl(u_{n}(y)\bigr) \bigr) \\ &\quad =\bigl(T_{k}\bigl(u_{n}(x)\bigr)-T_{k} \bigl(u_{n}(y)\bigr)\bigr)^{2} \\ &\qquad{} +\bigl(T_{k} \bigl(u_{n}(x)\bigr)-T_{k}\bigl(u_{n}(y)\bigr) \bigr) \bigl(G_{k}\bigl(u_{n}(x)\bigr)-G_{k} \bigl(u_{n}(x)\bigr)\bigr). \end{aligned}$$
(3.2)
Moreover, by Lemma 4 in [19], we know that
$$ \bigl(T_{k}\bigl(u_{n}(x) \bigr)-T_{k}\bigl(u_{n}(y)\bigr)\bigr) \bigl(G_{k} \bigl(u_{n}(x)\bigr)-G_{k}\bigl(u_{n}(y)\bigr) \bigr)\geq0. $$
(3.3)
Therefore, (3.1)–(3.2) lead to
$$ \frac{a_{N,s}}{2} \int_{\Omega}\int_{\Omega}\frac {(T_{k}(u_{n}(x))-T_{k}(u_{n}(y)))^{2}}{|x-y|^{N+2s}}+ \int_{\Omega}u^{p}_{n} {T_{k}(u_{n})} \leq k \int_{\Omega}|f|+\lambda \int_{\Omega}\frac{u^{2}_{n}}{|x|^{2s}}. $$
(3.4)
Recall that
$$\begin{aligned} \begin{aligned}[b] \int_{\Omega}\frac{u^{2}_{n}}{|x|^{2s}}&= \int_{\Omega}\frac{ (T_{k}(u_{n})+G_{k}(u_{n}) )^{2}}{|x|^{2s}} \\ &= \int_{\Omega}\frac {T^{2}_{k}(u_{n})}{|x|^{2s}}+ \int_{\Omega}\frac{G^{2}_{k}(u_{n})}{|x|^{2s}}+2 \int _{\Omega}\frac{T_{k}(u_{n})G_{k}(u_{n})}{|x|^{2s}}. \end{aligned} \end{aligned}$$
(3.5)
On the other hand, using \(G_{k}(u_{n})\) as a test function in (2.4), we have
$$\begin{aligned}& \frac{a_{N,s}}{2} \int_{\Omega}\int_{\Omega}\frac {|G_{k}{u_{n}(x)}-G_{k}{u_{n}(x)}|^{2}}{|x-y|^{N+2s}}-\lambda \int_{\Omega}\frac {u_{n}G_{k}(u_{n})}{|x|^{2s}} \\& \quad {}+ \int_{\Omega}u^{p}_{n} G_{k}(u_{n}) \leq \int_{\Omega}f G_{k}(u_{n}). \end{aligned}$$
(3.6)
Moreover, \(u_{n} G_{k}(u_{n})=G^{2}_{k}(u_{n})+T_{k}(u_{n})G_{k}(u_{n})\), thus this fact combined with (3.6), implies that
$$\begin{aligned}& \frac{a_{N,s}}{2} \int_{\Omega}\int_{\Omega}\frac {|G_{k}{u_{n}(x)}-G_{k}{u_{n}(x)}|^{2}}{|x-y|^{N+2s}}-\lambda \int_{\Omega}\frac {G^{2}_{k}(u_{n})}{|x|^{2s}}+ \int_{\Omega}u^{p}_{n} G_{k}(u_{n}) \\& \quad \leq k \int_{\Omega}f_{n}+k\lambda \int_{\Omega}\frac{G_{k}(u_{n})}{|x|^{2s}}. \end{aligned}$$
(3.7)
Applying the Young inequality on the right-hand-side of (3.7), we get
$$\int_{\Omega}\frac{G_{k}(u_{n})}{|x|^{2s}} \leq\frac{1}{2} \int_{\Omega}\frac {G^{2}_{k}(u_{n})}{|x|^{2s}}+\frac{1}{2} \int_{\Omega}\frac{1}{|x|^{2s}}. $$
Taking into account that \(\lambda<\Lambda_{N,s}\), by the Hardy inequality we obtain
$$\int_{\Omega}\int_{\Omega}\frac {|G_{k}(u_{n}(x))-G_{k}(u_{n}(y))|^{2}}{|x-y|^{N+2s}}\leq C(f,k,\lambda, \Lambda_{N,s}). $$
Therefore \(\{G_{k}(u_{n})\}_{n\in\mathbb{R}}\) is uniformly bounded in \(H_{0}^{s}(\Omega)\), it implies
$$\int_{\Omega}\frac{G^{2}_{k}(u_{n})}{|x|^{2s}}\leq C(f,k,\lambda, \Lambda_{N,s}). $$
Then we get
$$\begin{aligned} \int_{\Omega}\frac{u^{2}_{n}}{|x|^{2s}}&= \int_{\Omega}\frac{ (T_{k}(u_{n})+G_{k}(u_{n}) )^{2}}{|x|^{2s}} \\ &= \int_{\Omega}\frac{T^{2}_{k}(u_{n})}{|x|^{2s}}+ \int_{\Omega}\frac {G^{2}_{k}(u_{n})}{|x|^{2s}}+2 \int_{\Omega}\frac {T_{k}(u_{n})G_{k}(u_{n})}{|x|^{2s}} \\ &\leq C(f,k,\lambda, \Lambda_{N,s}). \end{aligned}$$
(3.8)
Putting together (3.4)–(3.5) and (3.8), it follows that
$$\frac{a_{N,s}}{2} \int_{\Omega}\int_{\Omega}\frac {(T_{k}(u_{n}(x))-T_{k}(u_{n}(y)))^{2}}{|x-y|^{N+2s}}+ \int_{\Omega}{T^{p+1}_{k}(u_{n})} \leq C(f,k,\lambda,\Lambda_{N,s}). $$
We deduce that \(T_{k}(u_{n})\) is uniformly bounded in \(H_{0}^{s}(\Omega)\cap L^{p+1}(\Omega)\). Then we pass to the limit in the approximation problem (2.4); up to a subsequence, there exists a function \(u\in H_{0}^{s}(\Omega)\cap L^{p+1}(\Omega)\).
Now we want to prove that \(u_{n}^{p} \rightarrow u\) in \(L^{1}(\Omega)\). Let \(\psi_{i}(\sigma)\) be defined by
$$ \psi_{i}(\sigma)= \textstyle\begin{cases} 1, & \sigma\geq t, \\ 0, & |\sigma|< t, \\ {-}1, & \sigma\leq-t. \end{cases} $$
(3.9)
Choosing \(\phi=\psi_{i}(u_{n})\) as a test function in (2.4), we get
$$\int_{\Omega}u_{n}^{p}\psi_{i}(u_{n}) \leq \int_{\Omega}f\psi_{i}(u_{n}), $$
which implies that
$$\int_{\{u_{n}>t\}\cap\Omega}u^{p}_{n}\leq \int_{\{u_{n}>t\}\cap\Omega} f. $$
Let E is any measurable subset of Ω. For any \(t>0\) we have
$$\int_{E} u_{n}^{p} \leq t^{p}|E|+ \int_{E\cap\{u_{n}>t\}}u_{n}^{p} \leq t^{p}|E|+ \int_{\{u_{n}>t\}}|f|. $$
The above fact and \(f\in L^{1}(\Omega)\) allow us to say that, for any given \(\varepsilon>0\), there exists \(t_{\varepsilon}\) such that
$$\int_{\{u_{n}>t_{\varepsilon}\}}|f|\leq\varepsilon. $$
Hence
$$\int_{E} u^{p}_{n}\leq t_{\varepsilon}^{p}|E|+\varepsilon. $$
Therefore
$$\lim_{|E|\rightarrow0} \int_{E} u^{p}_{n}\leq\varepsilon. $$
Thus we prove that \(\lim_{|E|\rightarrow0} \int_{E} u^{p}_{n}=0\). Vitali’s theorem implies that \(u_{n}^{p}\rightarrow u^{p}\) in \(L^{1}(\Omega)\) i.e.
$$\lim_{n\rightarrow\infty} \int_{\Omega}u^{p}_{n} = \int_{\Omega}u^{p}. $$
□
Let us show Theorem 1.5, that is, the existence of solution to problem (1.1) in the case where \(f\in L^{m}(\Omega)\) with \(m>\frac{p+1}{p}\).
Proof of Theorem 1.5
Define \(\beta=p(m-1)\), that satisfies \(p+\beta=\beta m'\). Using \(\phi =u_{n}^{\beta}\) as a test function in (2.4), we have
$$ \frac{a_{N,s}}{2} \int_{\Omega}\int_{\Omega}\frac{(u_{n}(x)-u_{n}(y))(u_{n}^{\beta}(x)-u_{n}^{\beta}(y))}{|x-y|^{N+2s}} + \int_{\Omega}u_{n}^{p+\beta}= \int_{\Omega}f_{n}u_{n}^{\beta}+ \lambda \int_{\Omega}\frac{u_{n}^{\beta+1}}{|x|^{2s}}. $$
Now, by Lemma 2.2, we get
$$ \bigl(u_{n}(x)-u_{n}(y)\bigr) \bigl(u_{n}^{\beta}(x)-u_{n}^{\beta}(y)\bigr) \geq\frac{4\beta}{(\beta +1)^{2}} \bigl(u_{n}^{\frac{\beta+1}{2}}(x)-u_{n}^{\frac{\beta+1}{2}}(y) \bigr)^{2}. $$
(3.10)
Using Hardy’s inequality, we have
$$ \int_{\Omega}\frac{u_{n}^{\beta+1}}{|x|^{2s}}= \int_{\Omega}\frac{(u_{n}^{\frac {\beta+1}{2}})^{2}}{|x|^{2s}}\leq\frac{a_{N,s}}{2\Lambda_{N,s}} \int _{\Omega}\frac{(u^{\frac{\beta+1}{2}}_{n}(x)-u^{\frac{\beta +1}{2}}_{n}(y))^{2}}{|x-y|^{N+2s}}. $$
(3.11)
We conclude that
$$\begin{aligned}& \frac{a_{N,s}}{2} \biggl(\frac{4\beta}{(\beta+1)^{2}}-\frac{\lambda}{\Lambda _{N,s}} \biggr) \int_{\Omega}\int_{\Omega}\frac{(u_{n}^{\frac{\beta+1}{2}}(x)- u_{n}^{\frac{\beta+1}{2}}(y))^{2}}{|x-y|^{N+2s}}+ \int_{\Omega}u^{p+\beta}_{n} \\& \quad \leq\|f\|_{L^{m}(\Omega)} \biggl( \int_{\Omega}u_{n}^{\beta m'} \biggr)^{\frac{1}{m'}}. \end{aligned}$$
With this choice of β, by Lemma 2.3 we obtain
$$\int_{\Omega}\int_{\Omega}\frac{(u_{n}^{\frac{\beta+1}{2}}(x)- u_{n}^{\frac{\beta+1}{2}}(y))^{2}}{|x-y|^{N+2s}} \leq C\|f\|_{L^{m}(\Omega)}. $$
By Lemma 2.1, we arrive at
$$\biggl( \int_{\Omega}u^{\frac{(\beta+1)2^{*}_{s}}{2}}_{n} \biggr)^{\frac {1}{2^{*}_{s}}} \leq C\|f\|_{L^{m}(\Omega)}. $$
Furthermore,
$$ \frac{(\beta+1)2^{*}_{s}}{2}=\frac{N(p(m-1)+1)}{N-2s}=m_{s,p}. $$
As a consequence there exists a function \(u\in L^{\frac{(\beta +1)2_{s}^{*}}{2}}(\Omega)\).
Finally, we want to prove that \(u^{p}_{n}\rightarrow u^{p} \) in \(L^{1}(\Omega )\). Using (3.10) as a test function in (2.4), we have
$$ \int_{\Omega}u^{p}_{n}\psi_{i}(u_{n}) \leq \int_{\Omega}f\psi_{i}(u_{n}). $$
For any \(t>0\) and \(E\subset\Omega\) is measurable. we get
$$ \int_{E} u_{n}^{p} \leq t^{p}|E|+ \int_{E\cap\{u_{n}>t\}}u_{n}^{p} \leq t^{p}|E|+ \int_{\{u_{n}>t\}}|f|. $$
There exists \(t_{\varepsilon}\) such that
$$ \int_{E} u^{p}_{n}\leq t^{p}_{\varepsilon}|E|+\varepsilon. $$
We see that \(|E|\rightarrow0\) implies
$$ \int_{E} u^{p}_{n}\leq\varepsilon, $$
i.e., the sequence \(u_{n}^{p}\) is equiintegrable. Consequently
$$ \lim_{n\rightarrow\infty} \int_{\Omega}u^{p}_{n}= \int_{\Omega}u^{p}. $$
Thus we have proved the existence result. □