In calculus of variation: why are minimizing sequences bounded?

Question

Assume the usual variational setting: Let $\mathcal{A} \subset W^{1,q}$ be the set of admissible functions and \begin{equation} I: \mathcal{A} \to \mathbb{R} \end{equation} the functional that needs to be minimized. One then chooses a minimizing sequence $(u_k)_{k \in \mathbb{N}}$, i.e. \begin{equation} I[u_k] \to \inf_{w \in \mathcal{A}} I[w]. \end{equation}

The next step is to show that the minimizing sequence is bounded, so one can show that it converges in the weak topology. My question is:

Assuming you are able to verify that $I[u_k]$ is uniformly bounded for $k \in \mathbb{N}$, how can you show that $(u_k)_{k \in \mathbb{N}}$ is bounded in $W^{1,q}$.

This step seams trivial, as it is frequently omitted in the books I read. Yet, I can't produce a proof for that. Any help is appreciated.

Answer 1

As @Martini has pointed out in the comments, the property you need is coercivity. Indeed, if you assume that your function is coercive , then any minimizing sequence must be bounded, because if not, then $I[u_n]\to\infty$ by coerciviness.

It is important to note that only coerciviness does not guarantee that you functional can be minimized, as this example shows.

Answer 2

Prove that $I[u]\geq F(|u|_{W^{1,q}})$ for some $F:[0,\infty)\rightarrow [0,\infty)$ non-decreasing such that $F(0)=0$ and $\lim_{t\rightarrow\infty} F(t)=\infty$. In your specific case, this is possible if $q=2$ and $W^{1,q}=H^1$.