Working on counterexamples to the direct method if a condition is missing I have the following problems:
$1. F:W^{1,2}([0,1]) \rightarrow \mathbb{R_\infty}, F(u)= \int_{0}^{1} (x*u(x))^2 dx$ if $u(0)=1, u(1)=0$, $F(u)=\infty$ else. Show: the functional cannot be coercive, because it is bounded from below and weakly lower semi- continuous on a Hilbert space, but doesn't have a minimizer.
I was able to show the coercivity and the weakly lower semi-continuity. Now I have the sequence $f_n(x)= 1-nx$ for $0 \leq x \leq \frac{1}{n}$ and $0$ else, for which $F$ converges to $0$, but $Fu=0 \Rightarrow u'=0$ a.e. and this should be a contradiction. But why can't the minimizer u be $u(0)=1, u(x)=0$ else? This function is weakly differentiable with weakly differential $u'=0$, isn't it?
A similar problem appears within the second part:
$2. F:W^{1,1}([0,1]) \rightarrow \mathbb{R_\infty}, F(u)= \int_{0}^{1} \sqrt{u(x)^2+u'(x)^2} dx$ if $u(0)=0, u(1)=1$, $F(u)=\infty$ else. We should show that $W^{1,1}([0,1])$ is not a Hilbert space, because the functional is coercive, bounded from below and weakly lower semi- continuous and has no minimizer.
Again, I was able to show these facts but with my sequence $f_n(x)= 1+n(x-1)$ for $1-\frac{1}{n} \leq x \leq 1$ and $0$ else, for which $F$ converges to $1$ and $Fu=1 \Rightarrow u=0$ a.e. and this should be a contradiction, but I again don't see why $u(0)=0$ and $u(x)=1$ else is not a solution. But such functions can't be in $W^{1,1}([0,1])$, because then for $u(1)=1$ and $u(x)=0$ else $F(u)$ would be zero, but it holds that $F(u) \geq \int_{0}^{1} u'(x) dx=1$ which is a contradiction.
Can you help me with why such functions are not weakly differentiable? Or where is my mistake?
It is tricky. In the definition of $F$ we have to evaluate $u(0)$ and $u(1)$. This implies that by writing $W^{1,2}([0,1])$ we already select the continuous representation, which exists by the Sobolev embedding theorem. This is an extremely important point. In that way, the function with $u(0)=1$ and $u(x)=0$ else is not in $W^{1,2}([0,1])$, since it is not continuous. Instead, you have to select $u=0$ everywhere as the function in $W^{1,2}([0,1])$ and you get the contradiction. The same for the other example.