Minimization problem in Sobolev spaces This is a homework problem and I don't know how to solve it:
Consider the following minimization problem on the real-valued sobolev space $H^{1,2}(\Omega)$ with dimension $n=1$ and $\Omega=(0,1)$:
$$F(u) = \int_0^1 \left(u^2+ \text{min}\left\{(u'-1)^2, (u'+1)^2\right\}\right) \ \mathrm{d}x \ \to\ \min$$
($u'$ denotes the derivative of $u$).
a) How to prove, that $F\colon H^{1,2}(\Omega) \to \mathbb R$ is continuous?
b) How to prove, that $\displaystyle\inf_{u\in H^{1,2}(\Omega)} F(u) = 0$, but there is no $u \in H^{1,2}(\Omega)$ with $F(u)=0$ ?
I think, $F$ is not linear mapping (because of quadratic terms in the integral). So it is maybe not the right way to show that the operator norm is bounded. Do I have to prove the continuity of $F$ by the definition (with $\varepsilon$ and $\delta$) or is there another possibility? And I don't know how to determine the infimum.
 A: Some hints, since it's a homework problem I won't give the full solution. 
For the first question, since for all $a,b\in\mathbb R, 2\min (a,b)=a+b-|a-b|$, we have 
\begin{align*}
F(u)&=\int_{[0,1]}\left(u^2+\frac 12((u'+1)+(u'-1)^2-|(u'-1)^2-(u'+1)^2|)\right)dx\\\
&=\int_{[0,1]}\left(u^2+\frac 12(2u'^2+2-4|u'|)\right)dx\\\
&=\int_{[0,1]}u^2dx+\int_{[0,1]}u'^2dx -2\int_{[0,1]}|u'|dx+1.
\end{align*}
Since in any Banach space $x\mapsto \lVert x\rVert$, and $F(u)=\lVert u\rVert_{H^{1,2}(\Omega)}^2-2\int_{[0,1]}|u'|dx$, you only have to show that $L(u):=\int_{[0,1]}|u'|dx$ is continuous from $H^{1,2}$ to $\mathbb R$. But $L$ is linear and Cauchy-Schwarz inequality gives you the result.
For the second question, if $F(u)=0$ for a $u$ then $u=0$, but we can't have $\min ((u'-1)^2,(u'+1)^2)=0$ since $u'=0$. To show that the infimum is indeed $0$, fix $n\geq 1$. We cut $(0,1)$ into $2n$ of length $\frac 1{2n}$, and use a piece wise affine function with $u_n\left(\frac{2k}{2n}\right)=0$ and $u_n\left(\frac{2k}{2n}\right)=\frac 1n$.
Check that this function is indeed in $H^{1,2}$ and compute $F(u_n)$.
A: This problem reminds me the classical Bolza example (named after Oskar Bolza) in the CoV, because $\inf F =0$ can be proved using sawtooth functions.
Obviously $F[u]\geq 0$, thus $\inf F \geq 0$. Now, take:
$$\phi (x):=\begin{cases} 1 -|x-1| &\text{, if } |x-1|\leq 1 \\ 0 &\text{, otherwise}\end{cases}$$
and, for $N\in \mathbb{N}$, define $u_N:[0,1]\to \mathbb{R}$ via:
$$u_N(x):= \frac{1}{2N}\ \sum_{n=0}^{N-1}\phi (2Nx-2n)\; ;$$
then $u_N$ is a nonnegative continuous function mapping $[0,1]$ onto $[0,1/(2N)]$ which is a.e. differentiable and, in particular:
$$u_N^\prime (x)=\pm 1\quad \text{for a.e. } x\in [0,1]\; ;$$
hence:
$$\min \Big\{(u_N^\prime (x)-1)^2,(u_N^\prime (x)+1)^2\Big\} =0 \quad \text{for a.e. } x\in [0,1]$$
and therefore:
$$F[u_N]=\int_0^1 u^2 +\int_0^1\min \Big\{(u_N^\prime (x)-1)^2,(u_N^\prime (x)+1)^2\Big\} \leq \frac{1}{4N^2}\; .$$
The latter inequality yields $\inf F =0$.
For the continuity, you can argue as follows.
You notice that using the reverse triangular inequality and other elementary tricks you get:
$$\begin{split} |F[u]-F[v]| &\leq \Big| \lVert u\rVert_2^2 -\lVert v\rVert_2^2\Big|\\
&\phantom{\leq} + \left| \int_0^1 \bigg( \min \Big\{ |u^\prime -1|,|u^\prime +1|\Big\}\bigg)^2 -\int_0^1\bigg( \min \Big\{ |v^\prime -1|,|v^\prime +1|\Big\}\bigg)^2\right|\\
&= (\lVert u\rVert_2+\lVert v\rVert_2)\ \Big| \lVert u\rVert_2 -\lVert v\rVert_2 \Big|\\
&\phantom{\leq} +\left| \left\lVert \min \Big\{ |u^\prime -1|,|u^\prime +1|\Big\}\right\rVert_2^2 -\left\lVert \min \Big\{ |v^\prime -1|,|v^\prime +1|\Big\} \right\rVert_2^2\right|\\
&\leq (\lVert u\rVert_2+\lVert v\rVert_2)\ \Big| \lVert u\rVert_2 -\lVert v\rVert_2 \Big|\\
&\phantom{\leq} +\left( \left\lVert \min \Big\{ |u^\prime -1|,|u^\prime +1|\Big\}\right\rVert_2 +\left\lVert \min \Big\{ |v^\prime -1|,|v^\prime +1|\Big\}\right\rVert_2 \right)\times \\
&\phantom{\leq +}\times \left| \left\lVert \min \Big\{ |u^\prime -1|,|u^\prime +1|\Big\}\right\rVert_2 -\left\lVert \min \Big\{ |v^\prime -1|,|v^\prime +1|\Big\}\right\rVert_2  \right|\\
&\leq (\lVert u\rVert_2+\lVert v\rVert_2)\ \lVert u-v\rVert_2\\
&\phantom{\leq} +\left( \left\lVert \min \Big\{ |u^\prime -1|,|u^\prime +1|\Big\}\right\rVert_2 +\left\lVert \min \Big\{ |v^\prime -1|,|v^\prime +1|\Big\}\right\rVert_2 \right)\times \\
&\phantom{\leq +} \times \left\lVert \min \Big\{ |u^\prime -1|,|u^\prime +1|\Big\} -\min \Big\{ |v^\prime -1|,|v^\prime +1|\Big\}\right\rVert_2\end{split}$$
for each $u,v\in W^{1,2}(0,1)$.
Now, it is easy to prove that function:
$$\mathbb{R}\ni s\mapsto \min \Big\{ |s-1|,|s+1| \Big\} \in \mathbb{R}$$
is Lipschitz with constant $L=1$ (draw a picture), hence:
$$\tag{1} \begin{split}
|F[u]-F[v]| &\leq (\lVert u\rVert_2+\lVert v\rVert_2)\ \lVert u-v\rVert_2\\
&\phantom{\leq} +\left( \left\lVert \min \Big\{ |u^\prime -1|,|u^\prime +1|\Big\}\right\rVert_2 +\left\lVert \min \Big\{ |v^\prime -1|,|v^\prime +1|\Big\}\right\rVert_2 \right)\ \lVert u^\prime -v^\prime\rVert_2 \; .
\end{split}$$
Finally, if you fix $v$ and let $u\stackrel{W^{1,2}}{\longrightarrow} v$ in (1) you obtain $|F[u]-F[v]|\to 0$ as you wanted.
