Poincaré inequality using $H^1$ seminorm

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.