existence of a minimizer for functional 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.
 A: 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.
