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My problem is the following: Show that the mapping $u \rightarrow ||\nabla u||^2 + (fu,u)$ has a minimum $u$ in $M:=\{ w \in H^1(\Omega): ||w||=1\}$ . The function $f$ is in $L^\infty$.

I dont see how to start here. What is needed for a proof? Thanks for every hint!

James T.

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Welcome to Math.SE! Does $(fu,u)$ mean the $L^2$ inner product, $\int_\Omega fu\bar u$? Also, is the norm on $H^1$ given by $\sqrt{\|\nabla u\|^2+\|u\|^2}$? If so, then for $w$ on the unit sphere your functional is equal to $1-\|u\|^2+(fu,u)$. – user53153 Jan 13 '13 at 14:45
Is $\Omega$ an open domain of some euclidean space? – Mercy King Jan 15 '13 at 22:58

Let's assume that $\Omega \subset \mathbb{R}^n$ is bounded and open. Then this problem is amenable to attack by the "direct methods in the calculus of variations." Here's a sketch of what to do.

First let $$ E(u) = \int_\Omega |\nabla u|^2 + f |u|^2$$ and $$ \lVert u\rVert_{H^1}^2 = \int_\Omega |\nabla u|^2 + |u|^2.$$ Since $ f\in L^\infty$ it's easy to see that $E$ is well-defined on $H^1$ (and hence on $M$).

Next show the coercivity of $E$ on $M$: $$ u \in M \Rightarrow E(u) \ge \int_\Omega |\nabla u|^2 - \lVert f\rVert_{\infty} \int_\Omega |u|^2 = \lVert u\rVert_{H^1}^2 -1 -\lVert f\rVert_{\infty}. $$ In particular this means $-\infty < \inf_M E < \infty$.

Next let $\{u_n\}_{n=1}^\infty \subset M$ be a minimizing sequence, i.e. choose them so that $$ u_n \in M \text{ and } E(u_n) \to \inf_{M} E \text{ as }n\to \infty.$$ We might as well assume that the $u_n$ are chosen so that $|E(u_n)| \le \inf_{M} E +1$.

From the above, we then know that $$\lVert u_n \rVert_{H^1}^2 \le E(u_n) +1 + \lVert f\rVert_{\infty} \le \inf_{M} E +2 + \lVert f\rVert_{\infty} \text{ for all }n.$$ This means that $\{u_n\}_{n=1}^\infty$ are uniformly bounded in $H^1$.

Now we use weak-compactness in $H^1$ (bounded sequences have weakly convergent subsequences -- ultimately this follows from Banach-Alaoglu) and the Rellich-Kantorovich compactness theorem ($H^1$ embeds compactly in $L^2$) to extract a subsequence $u_{n_k}$ so that $$ u_{n_k} \rightharpoonup u \text{ weakly in }H^1 \text{ and } u_{n_k} \to u \text{ strongly in }L^2. $$

Next we use weak lower semicontinuity of the norm to see that $$ \int_\Omega |\nabla u|^2 \le\liminf_{k\to \infty} \int_\Omega |\nabla u_{n_k}|^2. $$ Also, by strong $L^2$ convergence, $$ \int_\Omega f|u_{n_k}|^2 \to \int_\Omega f|u|^2 \text{ as } k \to \infty$$ and $$ 1 = \int_\Omega |u_{n_k}|^2 \to \int_\Omega |u|^2 \Rightarrow u \in M.$$ Combining, we see that $u \in M$ and $$ E(u) \le \liminf_{k\to \infty} E(u_{n_k}) = \inf_M E,$$ and hence that $$ E(u) = \inf_M E.$$ So, $u$ is the desired minimizer.

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