# Poincaré inequality using $H^1$ seminorm

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

• Can you please clarify what norms and seminorms are this? Dec 11, 2012 at 13:52
• I edited my post. I think should be clear now. Dec 11, 2012 at 17:41
• What is the definition of $|v|_{H^1}$? What is $v'$ and what is that $p$ in $C_p$? Dec 11, 2012 at 17:47
• Well, certainly I don't know what's the definition of $|v|_{H^1}$ I just know that is a semi norm --- $v'$ is just the first derivative of a function $v$ (any function) --- That $p$ is just a letter to say that $C_p$ = Poincaré Constant. Dec 12, 2012 at 15:13
• Are your functions defined in $\mathbb{R}$? Dec 12, 2012 at 16:18

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 $$f$$ 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 $$f\sim g \iff f\left(x\right) = g\left(x\right)\;\mathrm{a.e. in }\,\Omega$$.

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 $$f$$" but meaning "the class of equivalence whose representant is the function $$f$$". So when I say $$f$$ I'm really talking about every function $$g$$ such that $$f$$ and $$g$$ 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 $$f\left(x\right)=1$$ for all $$x\in\Omega$$. 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 $$f\left(x\right)=1$$ for all $$x\in\Omega$$ as a counter-example)

I hope everything is clearer now.

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.