Showing that this Coercivity condition implies uniform boundedness of a minimising sequence.

Question

The following problem is in Dacorogna's book "Introduction to the Calculus of Variations": Let $\Omega\subset\mathbb{R}^n$ be open and bounded with a Lipschitz boundary. Let $f\in C(\mathbb{R}^n\times \mathbb{R}\times\overline{\Omega}), f=f(\xi,u, x)$ satisfy

$(\mathcal{H})\quad$ there exists $p>q\geq 1$ and $\alpha_1>0, \alpha_2, \alpha_3\geq 0$ such that \begin{equation} f(\xi, u, x)\geq \alpha_1|\xi|^p-\alpha_2|u|^q-\alpha_3, \quad \forall\ (\xi, u, x)\in\mathbb{R}^n\times \mathbb{R}\times\overline{\Omega}. \end{equation}

Consider the problem: \begin{equation} (\mathcal{P})\quad\text{inf}\left\{I(u)=\int_{\Omega} f(Du, u, x)\ \mathrm{d}x\ \Big| \ u\in u_0+ W_{0}^{1, p}(\Omega) \right\}=m \end{equation} where $u_0\in W^{1, p}(\Omega)$ with $I(u_0)<\infty$.

I am trying to show that if $\{u_k\}_{k=1}^{\infty}\subset u_0+W^{1, p}_0(\Omega)$ is a minimising sequence, that is \begin{equation} I(u_k)\rightarrow m\quad \text{as } k\rightarrow\infty, \end{equation} then $\{Du_k\}_{k=1}^{\infty}$ is uniformly bounded in $L^p(\Omega)$.

My argument is as follows:

Assume $\alpha_1> q/p$ and that $\alpha_2>0$. For $k$ sufficiently large we have \begin{align} m+1\geq I(u_k)&\geq \alpha_1\|Du_k\|_p^p-\alpha_2\|u_k\|_q^q- \alpha_3\mathcal{L}^n(\Omega)\\ &\geq\alpha_1 \|Du_k\|_p^p-\alpha_2C\|Du_k\|_q^q- \alpha_3\mathcal{L}^n(\Omega)\quad\text{by Poincare's inequality}\\ &\geq\alpha_1 \|Du_k\|_p^p-\gamma_1\|Du_k\|_p^q- \gamma_2\quad \text{since } L^p(\Omega)\hookrightarrow L^q(\Omega)\\ &\geq\alpha_1 \|Du_k\|_p^p-\frac{q}{p}\|Du_k\|_p^p- \gamma_3\quad\text{by Young's inequality}\\ &\geq\left[\alpha_1-\frac{q}{p}\right]\|Du_k\|_p^p-\gamma_3 \end{align} where all the constants above are positive. How can I arrive at a similar conclusion, that is \begin{equation} m+1\geq C\|Du_k\|_p^p-\gamma \end{equation} where $C, \gamma>0$ if $\alpha_1\leq q/p$?

Answer 1

It is not necessary to impose condition on $\alpha_1$, in fact, you can argue like this. By the second line of your inequality, we have that $$\tag{1}m+1\geq \alpha_1\|Du_k\|_p^p-\gamma_1\|Du_k\|_p^q-\gamma_2$$

Consider the polynomious $p(x)=\alpha_1 x^p-\gamma_1 x^q-\gamma_2$ for $x\geq 0$. $(1)$ implies that $$\tag{2}p(\|Du_k\|_p)\leq m+1$$

but $\alpha_1>0$ and $q<p$, so for $(2)$ to be true in necessary that the leading term is bounded, i.e. $\|Du_K\|_p$ must be bounded.

Answer 2

There is also another way:

We will assume $\gamma_1>0$. Young's inequality with $\varepsilon$ is \begin{equation} ab\leq\varepsilon a^P+C(\varepsilon)b^Q\quad(a, b, \varepsilon>0). \end{equation}Letting $a\equiv\|Du_k\|_p^q, \ b\equiv 1, \ P\equiv p/q, \ Q\equiv P/P-1$ and using this in Young's inequality above, we obtain \begin{equation*} \|Du_k\|_p^q\leq\varepsilon\|Du_k\|_p^p+C(\varepsilon) \end{equation*} Using this estimate in \begin{equation} m+1\geq\alpha_1 \|Du_k\|_p^p-\gamma_1\|Du_k\|_p^q-\gamma_2 \end{equation} gives \begin{equation*} m+1\geq(\alpha_1-\gamma_1\varepsilon)\|Du_k\|_p^p-\gamma_2-\gamma_1C(\varepsilon) \end{equation*} and by setting $\varepsilon=\alpha_1/\gamma_12$ we can conclude that $\|Du_k\|_p$ is bounded.