This question is from Calculus of Variations by Rindler (problem 2.2):
Let $\Omega \subset \mathbb{R}^d$ be a bounded Lipschitz domain. Define
$V := \{ u \in W^{1,2}(\Omega) : \int_\Omega u(x)dx = 0 \}$.
Assume that $f : \Omega \times \mathbb{R}^d \rightarrow \mathbb{R}$ is continuously differentiable with \begin{align} \mu \vert A \vert^2 &\leq f(x,A) \quad \quad \quad \, \, \text{for some }\mu > 0 \text{ and all } (x, A) \in \Omega \times \mathbb{R}^d, \quad \ (1) \\ \vert D_A f(x,A)\vert &\leq M(1 + \vert A \vert^2) \quad \text{ for some }M > 0 \text{ and all }(x, A) \in \Omega \times \mathbb{R}^d, \quad (2) \end{align} and that $A \mapsto f(x,A)$ is convex for all $x \in \Omega$.
Let $g \in L^2(\Omega)$.
Show that the functional
$\mathcal{F}[u] := \int_\Omega f(x,\nabla u(x)) - g(x)u(x) $
is coercive on $V$; that is, show there exists $\mu > 0$ such that for all $u \in V$ we have
$\mathcal{F}[u] \geq \mu \vert\vert u \vert\vert_{W^{1,2}}^2 - \mu^{-1}$.
I can deal with the first part of the integrand: we have \begin{align} \mathcal{F}[u] &\geq \mu\int_\Omega \vert\nabla u(x) \vert^2 - \int_\Omega gu \quad\,\,\,\,\text{ by }(1) \\ &\geq \frac{\mu}{1+C_p} \vert\vert u \vert\vert_{W^{1,2}}^2 - \int_\Omega gu \quad\text{ by the Poincare-Wirtinger inequality} \end{align} using the condition that $\int_\Omega u = 0$. I haven't been able to deal with the $-\int_\Omega gu$ term on the end though. I suspect this is where the $g \in L^2$ condition is meant to come into play; I've tried to use Holder's inequality and Young's inequality, but neither of them have got me anywhere and I can't see how I'm meant to tackle this question.
Any suggestions on how to proceed?
Thanks
So (I think) I've managed to solve this using Holder's inequality-- details below in case anyone's interested. I had tried this approach before but mistakenly "worked out'' that from the fourth line it wasn't possible to obtain $a_2$ and $c$.
\begin{align*} \mathcal{F}[u] &\geq \mu\int_\Omega \vert\nabla u(x) \vert^2 - \int_\Omega gu \, \text{ by } (1)\\ &\geq \frac{\mu}{1+C}\int_\Omega \vert\nabla u(x) \vert^2 + \frac{\mu}{1+C} \int_\Omega u^2 - \int_\Omega gu \,\text{ by the Poincare-Wirtinger inequality} \\ &\geq \frac{\mu}{1+C}\int_\Omega \vert\nabla u(x) \vert^2 + \frac{\mu}{1+C} \int_\Omega u^2 - \left(\int_\Omega g^2\right)^{1/2} \left(\int_\Omega u^2\right)^{1/2} \text{ by Holder's inequality} \\ &= \frac{\mu}{1+C}\int_\Omega \vert\nabla u(x) \vert^2 + a_1x^2 - bx \,\text{ setting }x=\left(\int_\Omega u^2\right)^{1/2} \\ &\geq \frac{\mu}{1+C}\int_\Omega \vert\nabla u(x) \vert^2 + a_2x^2 - c \, \text{ for some }a_2, c > 0\quad (*) \\ &\geq m \vert\vert u \vert\vert_{W^{1,2}}^2 - c \, \text{ setting }m=\min \left( \frac{\mu}{1+C}, a_2 \right) > 0. \end{align*} A quick computation shows that this is greater than or equal to either $m \vert\vert u \vert\vert_{W^{1,2}}^2 - \frac{1}{m}$ or $\frac{1}{c} \vert\vert u \vert\vert_{W^{1,2}}^2 - c$, which gives us what we wanted.
$(*)$ This step is valid since the inequality $a_1 x^2 - bx \geq a_2 x^2 - c$ is equivalent to $(a_1 - a_2)x^2 - bx + c \geq 0$. If $a_1 > a_2$, this holds so long as $b^2 - 4(a_1 - a_2)c < 0$, which we can guarantee for positive $a_2$, $c$ by taking an arbitrarily small $\varepsilon > 0$ then choosing $a_1 - a_2 = \frac{b^2}{4c} + \varepsilon$ and $c$ large enough so that $a_2$ is positive ($a_1$ is positive since $\mu$ and $C$ are; then when $a_2$ is close enough to $a_1$ it is also forced to be positive).