I need your help to prove: Let $r>1,\; 0<q<1$ and $0<\lambda<q$. If $u_{\varepsilon}\geq 0$ and satisfies \begin{equation} \left[\int_{\Omega}u_{\varepsilon}^{r}\right]^{\frac{q}{r}} \leq C\left[\int_{\Omega}(u_{\varepsilon}+1)^{r}\right]^{\frac{\lambda}{r}}, \end{equation} then the sequence $u_{\varepsilon}$ is uniformly bounded in $L^{r}(\Omega)$.
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Let $r>1,\; 0<q<1$ and $0<\lambda<q$. We assume that $u_{\varepsilon}\geq 0$ satisfies \begin{equation} \left[\int_{\Omega}u_{\varepsilon}^{r}\right]^{\frac{q}{r}} \leq C\left[\int_{\Omega}(u_{\varepsilon}+1)^{r}\right]^{\frac{\lambda}{r}},\qquad (1) \end{equation}
Since $r>1$ and $u_{\varepsilon}\geq 0$, then Minkowski’s inequality implies that \begin{equation} \begin{split} \left[\int_{\Omega}(u_{\varepsilon}+1)^{r}\right]^{\frac{1}{r}}&= \Vert u_{\varepsilon}+1\Vert_{r} \\ &\leq \Vert u_{\varepsilon}\Vert_{r}+meas(\Omega)^{\frac{1}{r}}.\qquad(2) \end{split} \end{equation} Using the fact $0<\lambda<1$, we have the real inquality: \begin{equation} (t+1)^{\lambda}\leq t^{\lambda}+1,\quad\forall t\geq 0\qquad(3) \end{equation} To prove the inequality $(3)$, it suffice to consider the real function \begin{equation} \varphi(t)=(t+1)^{\lambda}-t^{\lambda}-1,\qquad t\geq 0. \end{equation} Since \begin{equation} \varphi'(t)=\lambda\left[\frac{1}{(1+t)^{1-\lambda}}-\frac{1}{t^{1-\lambda}}\right]\leq 0,\qquad\forall t\geq 0, \end{equation} then, the function $\varphi$ is decreasing. Therefore $$\varphi(t)<\varphi(0)=0;,\qquad\forall t>0$$ Thus, we prove the inequality (3). Since $ u_{\varepsilon}\geq 0$, we can use the inequality (3) with $t=\frac{\Vert u_{\varepsilon}\Vert_{r}}{meas(\Omega)^{\frac{1}{r}}}$ to obtain \begin{equation} \begin{split} \left[\Vert u_{\varepsilon}\Vert_{r}+meas(\Omega)^{\frac{1}{r}}\right]^{\lambda}&= meas(\Omega)^{\frac{\lambda}{r}}\left[\frac{\Vert u_{\varepsilon}\Vert_{r}}{meas(\Omega)^{\frac{1}{r}}}+1\right]^{\lambda} \\ &\leq meas(\Omega)^{\frac{\lambda}{r}}\left[\frac{\Vert u_{\varepsilon}\Vert_{r}^{\lambda}}{meas(\Omega)^{\frac{\lambda}{r}}}+1\right] \\ &\leq \Vert u_{\varepsilon}\Vert_{r}^{\lambda}+meas(\Omega)^{\frac{\lambda}{r}}.\qquad (4) \end{split} \end{equation} Combining (1), (2) and (4), we deduce that \begin{equation} \Vert u_{\varepsilon}\Vert_{r}^{q} \leq C \Vert u_{\varepsilon}\Vert_{r}^{\lambda}+C meas(\Omega)^{\frac{\lambda}{r}},\qquad (5) \end{equation} Since $\lambda<q$, then, using Young's inequality with $\alpha=\frac{q}{\lambda},\;\beta=\frac{q}{q-\lambda}$ on the first term of the right-hand side of (5), we deduce that \begin{equation} \Vert u_{\varepsilon}\Vert_{r}^{q} \leq \frac{\lambda}{q}\Vert u_{\varepsilon}\Vert_{r}^{q}+\frac{q-\lambda}{q}C +Cmeas(\Omega)^{\frac{\lambda}{r}}. \end{equation} Therefore \begin{equation} \frac{q-\lambda}{q}\Vert u_{\varepsilon}\Vert_{r}^{q} \leq \frac{q-\lambda}{q}C +Cmeas(\Omega)^{\frac{\lambda}{r}}.\qquad (6) \end{equation} From (6), we obtain \begin{equation} \Vert u_{\varepsilon}\Vert_{r}^{q} \leq C +\frac{Cq}{q-\lambda}meas(\Omega)^{\frac{\lambda}{r}}. \end{equation} So, $u_{\varepsilon}$ is uniformly bounded in $L^{r}(\Omega)$.