Boundedness of sequence in Lebesgue space 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 $L^{r}(\Omega)$.
Thank you in advance
 A: 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 $L^{r}(\Omega)$.
