$I(u)= \int_{-1}^1x^2(u'(x))^2dx$ How do I prove that the condition $\exists$ $c_1 >0, c_2 \ge 0 $ ,$f(x,y,z)\ge c_1|z|^2-c_2 $ does not hold?

Question

Consider the functional $I(u)=\int_{-1}^1 f(x,u(x),u'(x))dx= \int_{-1}^1x^2(u'(x))^2dx$ , where

$f(x,y,z)=f(x,u(x),u'(x))=x^2(u'(x))^2$

and the admissible class of functions $\Phi =\{u \in W^{1,2}((-1,1))|u(-1)=-1, u(1)=1\}$.

I am trying to prove that the following coercivity condition is not satisfied:

$\exists$ constants $c_1 >0, c_2 \ge 0 $ such that $f(x,y,z)\ge c_1|z|^2-c_2 \forall (x,y,z)\in [-1,1]\times \mathbb{R}\times \mathbb{R}$

My textbook only says "coercivity does not hold because that $f(x,y,z)\ge c_1|z|^2-c_2$ is only satisfied for $c_1=0$" which is not admissible but that doesn't prove to me is not satisfied for some other $c_1 >0, c_2 \ge 0 $. How do I prove that?

I think I would have to prove that

$\forall$ constants $c_1 >0, c_2 \ge 0 $ : $f(x,y,z)< c_1|z|^2-c_2 \forall (x,y,z)\in [-1,1]\times \mathbb{R}\times \mathbb{R}$

At most I can see that $f(x,y,z)=x^2z^2<z^2$, because $x\le1$, but that doesn't help showing that it is true $\forall$ constants $c_1 >0, c_2 \ge 0 $, but only for $c_1>=1$

Answer 1

So you have a function $f : [-1,1] \times \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ defined as $f(x,y,z) := x^2 z^2$.

As you recalled, the coercivity condition is

$\exists c_1 > 0, \exists c_2 \geq 0, \forall (x,y,z) \in [-1,1] \times \mathbb{R} \times \mathbb{R} , \quad f(x,y,z) \geq c_1|z|^2 - c_2$

Now, beware that, correctly negating the quantifiers (unlike in your question), disproving the coercivity condition amounts to prove that

$\forall c_1 > 0, \forall c_2\geq 0, \exists (x,y,z) \in [-1,1] \times \mathbb{R} \times \mathbb{R} , \quad f(x,y,z) < c_1 |z|^2 - c_2$

So, let us take some (now fixed) $c_1 > 0$ and $c_2 \geq 0$. We need to find an $x \in [-1,1]$, $y \in \mathbb{R}$ and $z \in \mathbb{R}$ such that $x^2 z^2 < c_1|z|^2 - c_2$. The left-hand side is small when $|x|$ is small, so let us choose $x = 0$. We now need to find $z$ such that $0 < c_1|z|^2 - c_2$. Any $z$ such that $|z| \geq \sqrt{c_2/c_1}$ will do.