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Does this inequality holds for Poincaré Inequality?

$\|v\|_{L^2} \leqslant C_p |v|_{H^1}$

and $ |v|_{H^1} = \|v'\|_{L^2} $ where $| \cdot |$ denotes the semi norm and $\|\cdot\|$ the norm.

I'm really confused with norms and semi norms in $H^1$ and $L^2$.

Cheers

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    Yes they are defined in $\Re$2012-12-12

2 Answers 2

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The answer is no, which you can verify by calculating $\|v\|_{L^2}$ and $\|v'\|_{L^2}$ for the function $v_n(x)=\min(1,\max(0,n-|x|))$. As $n\to\infty$, one of the norms grows indefinitely while the other remains constant.

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I have good news and bad news. The good news is that you were correct. The bad news is that it was when you said:

I'm really confused with norms and semi norms in $H^1$ and $L^2$.

I'll try to clarify it a bit. First of all...

The basic difference between a norm $\Vert\cdot\Vert$ and a seminorm $\vert \cdot\vert$ is that a norm can only be zero if applied to the zero vector: $\Vert v\Vert = 0 \iff v = 0$ which is not true for a seminorm.

Given a bounded domain $\Omega\subset\mathbb{R}^n$, the space $\mathcal{L}^2\left(\Omega\right)$ is the space of square integrable functions. If you try to define a norm for this space like: $ \vert v \vert_{\mathcal{L}^2\left(\Omega\right)} = \left( \int_\Omega v(x)^2\,\mathrm{d}x \right)^\frac{1}{2}$

you end up with a seminorm. Why? because for every function function $f$ which is zero everywhere except in a set of zero meassure would be $ \vert f \vert_{\mathcal{L}^2\left(\Omega\right)} = \left( \int_\Omega f(x)^2\,\mathrm{d}x \right)^\frac{1}{2}=0$ but the function $f$ is not the null function $0$.

In order to solve this the space $L^2\left(\Omega\right) = \mathcal{L}^2\left(\Omega\right) / \sim $ is defined. The $L^2\left(\Omega\right)$ space is the quotient space of the $\mathcal{L}^2\left(\Omega\right)$ by the equivalence relation $f(x)\sim g(x) \iff f(x) = g(x)\,\mathrm{a.e. in }\,\Omega$.

This is, $L^2\left(\Omega\right)$ is the space of classes of equivalence of functions, so that two functions belong to the same class if they differ at most in a set of zero meassure in $\Omega$.

From now on, I will be talking about "function $f(x)$" but meaning "the class of equivalence whose representant is the function $f(x)$". So when I say $f(x)$ I'm really talking about every function $g(x)$ such that $f(x)$ and $g(x)$ only differ in at most a set of zero meassure.

So in this new space $L^2\left(\Omega\right)$ the aplication:

$ \Vert v \Vert_{L^2\left(\Omega\right)} = \left( \int_\Omega v(x)^2\,\mathrm{d}x \right)^\frac{1}{2}$ is indeed a norm, as $ \Vert v \Vert_{L^2\left(\Omega\right)} = 0 \iff f\sim 0$ in other words, there is only one class of equivalence of functions that has zero norm, and is the class of equivalence of the null function.

Now, $H^1\left(\Omega\right)$... Not being very formal, $H^1\left(\Omega\right)$ is the subspace of functions of $L^2\left(\Omega\right)$ such that their derivative is also in $L^2\left(\Omega\right)$: $H^1\left(\Omega\right)=\left\{v\in L^2\left(\Omega\right);\,\vert\mathbf{grad}\left(v\right)\vert\in L^2\left(\Omega\right)\right\} $

( if you are in $\mathbb{R}$ just replace $\vert\mathbf{grad}\left(v\right)\vert$ by $v'$ )

You can define a norm for this space also, and is defined as $ \Vert v \Vert_{H^1\left(\Omega\right)} = \left( \Vert v \Vert_{L^2\left(\Omega\right)}^2 + \Vert \vert\mathbf{grad}\left(v\right)\vert \Vert_{L^2\left(\Omega\right)}^2 \right)^\frac{1}{2}$

And you can define the seminorm:

$ \vert v \vert_{H^1\left(\Omega \right)} = \left(\int_\Omega \vert \mathbf{grad}\left(v\right)\vert^2\,\mathrm{d}x\right)^\frac{1}{2}$

You can also write the norm like $\Vert v \Vert_{H^1\left(\Omega\right)}^2 =\Vert v \Vert_{L^2\left(\Omega\right)}^2 +\vert v \vert_{H^1\left(\Omega \right)}^2$.

Why is the seminorm a seminorm? I'll just show you and example. Take the function $f(x)=1$. This function is not the zero function (neither it belongs to the class of equivalence of the zero function) but you have

$ \vert f \vert_{H^1\left(\Omega \right)} = 0$

Finally!. The Poincaré inequality is true for a special set of functions. The subspace $H^1_0\left( \Omega \right)$. Again, far from being formal the space is the subpace of $H^1\left( \Omega \right )$ such the functions vanish at the boundary of $\Omega$: $ H^1_0\left(\Omega\right)=\left\{v\in H^1\left(\Omega\right);\,v_{\vert\partial\Omega} =0\right\}$ (not exactly, I can be more precise if you need it)

Now, in this space the seminorm from $H^1\left(\Omega\right)$ is indeed a norm as we have that $\vert v \vert_{H^1\left(\Omega\right)}=0\Rightarrow v\,\text{ is constant in}\,\Omega$ but there is only one function which is constant and is zero at the boundaries, wich is the zero function. So in $H^1_0\left(\Omega\right)$ we have that $\vert v \vert_{H^1\left(\Omega\right)}=0\iff v=0$.

Explained this, the Poincaré inequality is:

$\Vert v \Vert_{L^2\left(\Omega\right)} \le C_p\vert v\vert_{H^1\left(\Omega\right)}\quad\forall v\in H^1_0\left(\Omega\right)$

where $C_p>0$ is a positive constant depending only on $\Omega$.

You can easy find that in general for $v\in H^1\left(\Omega\right)$ the inequality doesn't hold. (try with $f(x)=1$ for example)

I hope everything is clearer now.