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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.

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2 Answers

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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 $F(u_n)$.

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Thank you very much! –  David75 Dec 18 '11 at 9:51
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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.

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Thank you very much. –  David75 Dec 18 '11 at 9:51
    
I've been thinking about your demonstration and now I have a question: Why are the functions $u_N$ differentiable in $[0,1]$? They are sawtooth functions with $N$ pikes (cusp) in the intervall $[0,1]$. –  David75 Dec 20 '11 at 19:44
    
Functions $u_N$ are only almost evrywhere differentiable in $[0,1]$: in fact, each $u_N$ is differentiable in $]0,1[\setminus \{ k/(2N) \}_{k=1,\ldots ,2N-1}$, which is a set of full measure in $[0,1]$. Moreover it is easy to see that: $$u_N^\prime (x)=\sum_{n=0}^{N-1} \phi^\prime (2Nx-2n) = \begin{cases} 1 &\text{, if } n/N <x<(2n+1)/(2N),\ n\in \{0,\ldots, N-1\} \\ -1 &\text{, if } (2n+1)/(2N)<x<(n+1)/N,\ n\in \{0,\ldots, N-1\} ,\end{cases}$$ hence $\min \Big\{(u_N^\prime (x)-1)^2,(u_N^\prime (x)+1)^2\Big\} =0$ for a.e. $x\in [0,1]$. –  Pacciu Dec 22 '11 at 2:42
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