# Uniform boundedness of weak solution

Let $$u_n\in W_{0}^{1,p}(\Omega)$$ be a positive weak solution of the equation: $$-\Delta_p u=\frac{f_n(x)}{(u+\frac{1}{n})^\delta}\text{ in }\Omega.$$

Let $$p=N$$ and $$f\in L^m(\Omega)$$ for some $$m>1$$. Then how $$u_n$$ is uniformly bounded in $$W_{0}^{1,p}(\Omega)$$. This result is proved in Lemma 4.5 in the following article: https://link.springer.com/content/pdf/10.1007%2Fs00030-016-0361-6.pdf

Can you help me with it. It is written as a simple fact.

Here $$\Omega$$ is a bounded smooth domain in $$\mathbb{R}^N$$, $$f$$ is a nonnegative but not identically zero function in $$\Omega$$.

Okay, as you written the result is proved in the article for $$1 < p < N$$. They say that the case $$p=N$$ can be proved similarly. You are now asking about this case.

Remember that $$\Omega \subset \mathbb{R}^N$$ and $$\Delta_N u = \text{div}(|\nabla u|^{N-2} \nabla u).$$ The Sobolev inequality tells us that $$W^{1,N}_0$$ is embedded in $$L^q$$ for all $$q\geq 1$$.

Then

\begin{aligned}\|\nabla u_n\|_{L^N}^N = \int |\nabla u_n|^{N-2} \nabla u \cdot \nabla u= -\int \Delta_N u_n \cdot u_n &=\int \frac{f_n u_n}{(u_n+\frac{1}{n})^\gamma} \\ &\leq \int f_n u_n^{1-\gamma} \\ &\leq \|f\|_{L^m} \|u_n^{1-\gamma}\|_{L^{m'}} \\ &=\|f\|_{L^m} \|u_n\|_{L^{(1-\gamma)m'}}^{1-\gamma} \\ &\leq C \|f\|_{L^m} \|\nabla u_n\|_{L^N}^{1-\gamma} \end{aligned}

Hence

$$\|\nabla u_n\|_{L^N} \leq (C \|f\|_{L^m})^{\frac{1}{N+\gamma-1}}.$$

• Ok. I agree but one confusion is that suppose we have given $m>1$ and $\gamma\in(0,1)$. Then to us ethe above embedding theorem one need that $(1-\gamma)m'\geq 1$. Why this is true? – Mathlover Jan 15 at 19:09
• Please help me with this argument. I have understood all the arguments except this. Thank you very much. – Mathlover Jan 15 at 19:13
• I have got. If this is less than 1 then using Holder inequality one can get the uniform bound. Thanks. – Mathlover Jan 18 at 7:05