Existence of minimisers for variational problems: Relaxing the coercivity condition (following the L. Evens book) I'd like to show the existence of minimisers for a variational problem for a particular $L_0$:
$$
I[u]=\int_U L_0(\nabla u,u,x) dx
$$
where $u$ belongs to the so called set of admissible functions
$$
\mathcal{A}_0:=\big\{u \in W^{1,p}(U) :  u=g \text{ on } \partial U \quad \textbf{and} \quad u, \nabla u\text{ are bounded on } L^p(U) \big\}
$$
In the Evens book pp. 443-448 I can find sufficient conditions for such a minimiser to exist when $u$ belongs to the set of admissible functions
$$
\mathcal{A}:=\big\{u \in W^{1,p}(U) :  u=g \text{ on } \partial U\big\}
$$
In particular Evens says we need the condition of convexity of $L_0$ in the $\nabla u$ argument and also a coercivity condition $(1)$ on $L_0$, which is that there exists constants $\alpha>0$, $\beta \geq 0$ such that
\begin{equation}
\tag{1}
L_0(q,z,x)\geq \alpha |q|^p-\beta \quad \forall q \in \mathbb{R}^n, z \in \mathbb{R},x \in U 
\end{equation}
Then we are guaranteed the existence of a minimiser for $I[u]$.
If I understand correctly, Evens requires the coercivity condition so $\mathcal{A}$ is a bounded subset of $W^{1,p}(U)$ and weak compactness of minimising sequences can be established.
My questions are:

*

*$L_0$ being convex should be enough to show existence of a minimiser because my set of admissable functions $\mathcal{A}_0$, is bounded by definition.


*If the answer to 1. Is yes,  then if $\mathcal{A}_0$ is bounded do we need to use weak compactness to conclude a minimiser exists? Is there a more elementary method?  My initial thoughts are that this is necessary, since in general $\mathcal{A}_0$ will still be noncompact under the standard norm topology of  $W^{1,p}(U)$.
Sorry if this is trivial, I am still not fantastically confident with functional analysis.
 A: The space $\mathcal{A}$ of admissible functions is not bounded in $W^{1,p}(U);$ if you simply take $g = 0$ for instance, then let $\phi \in C^{\infty}_c(U)$ be any non-zero function. Then the family $\{ \lambda \phi \}_{\lambda > 0} \subset \mathcal{A}$ is unbounded in $W^{1,p},$ as the sequence divergences along any $\lambda_j \to +\infty.$ For general $g$ you can just consider $g + \lambda \phi$ instead. Also as far as I can tell $\mathcal{A} = \mathcal{A}_0$ under your definition, depending on what you mean exactly by being bounded in $L^p.$
For a concrete example, consider the functional
$$ I[u] = \int_0^{2\pi} u(x) \,\mathrm{d}x $$
over $W^{1,p}_0(0,2\pi).$ The associated integrand $L(x,u,z) = u$ is certainly convex in the $z$ argument, but it has no minimum by considering for instance $u_j(x) = -j \sin(x/\pi).$

The coercivity condition is used to show that if $\{u_j\} \subset \mathcal{A}$ is a sequence of admissible functions such that
$$ \sup_j I[u_j] < \infty, $$
then $\{u_j\}$ is bounded in $W^{1,p}(U)$ in the sense that
$$ \sup_j \lVert u_j \rVert_{W^{1,p}(U)} < \infty. $$
From here one can invoke weak compactness to obtain a limit. The direct method proceeds by taking a minimising sequence and extracting a weak limit using this proceedure, which one then (using convexity) shows it a minimizer in the admissible class $\mathcal{A}.$ This answers (1) negatively.
For (2) the answer is again no and it is essential that one takes a weak limit; bounded sequences in $W^{1,p}(U)$ are rarely strongly compact! For a concrete example consider $u_n(x) = \frac1n \sin(nx)$ in $W^{1,2}_0(0,2\pi);$ while $u_n \to 0$ in $L^2,$ we have $u_n'(x) = \cos(nx)$ satisfies $\lVert u_n' \rVert_{L^2(0,2\pi)} = \pi > 0$ for all $n,$ so it evidently does not converge in $W^{1,2}(0,2\pi).$
This is why convexity is essential also; the above example shows that the concave integrand
$$ I[u] = \int_0^{2\pi} -u'(x)^2 \,\mathrm{d}x, $$
is not weakly lower semi-continuous on $W^{1,2}_0(0,2\pi)$ by considering the above sequence $u_n$ which converges weakly to $0.$
