Delta distribution doesn't belong to any $L^p$ but there is a $H^1_0$ representation of it, why? Let $\Omega := (-1,1)$. I heard that the delta distribution $$\delta\colon H^1_0 \to \mathbb{C},\, \phi \mapsto \phi(0), \,\, \delta \in H^{-1} := (H^1_0(\Omega))'$$ has a Riesz representation in $H^1_0(\Omega)$ and on the other hand doesn't belong to any $L^P$ space. But how is this possible, since $H^1_0(\Omega) \subseteq L^2(\Omega)$? 
I'm relatively new to the topic and I feel quite a bit lost. I did some research on this board and found the two threads
Delta Dirac Function and
Representing the dirac distribution in $H^1(\mathbb R)$ through the scalar product , in wich the two assumptions above are proven.
So what I'm getting wrong, why is this no contradiction? 
 A: The key point is: different functions/distributions may induce the same linear functional on a space, acting in different ways. For example, on $H^1(\mathbb{R})$ we have a continuous linear functional $E_0v = v(0)$ (evaluation at $0$). This functional is induced by the function $u(t)=\frac12 e^{-|x|}$ under the action 
$$
v\mapsto \int_{\mathbb{R}}(uv+u'v')
$$
On the other hand, the same functional $E_0$ is induced by the distribution $\delta_0$ (Dirac delta) under the action
$$
v\mapsto \int_{\mathbb{R}} \delta_0 v
$$
where the   integral isn't really an integral but the action of a distribution. 
So: two different distributions, acting in different ways, yield the same result. The Dirac delta is the distribution that yields $v(0)$ when applied to test function $v$. The function $\frac12 e^{-|x|}$ does not do that: when applied to $v$, it yields $\int_{\mathbb{R}} \frac12e^{-|x|}v(x)\,dx$, not $v(0)$. 
If we use a different way for distributions to induce linear functionals on $H^1$: namely, integrate also against $v'$, then $\frac12e^{-|x|}$ induces $E_0$ functional, and $\delta_0$ induces some unbounded functional.
Functions like $\frac12 e^{-|x|}$ are called a reproducing kernel for a  Hilbert space. 
Here's one more way to look at it: the map $u\mapsto u-u''$ is an isomorphism of $H^1_0(\mathbb{R})$ onto $H^{-1}(\mathbb{R})$ which sends $\frac12 e^{-|x|}$ to $\delta_0$. But "isomorphic" does not mean "the same space".
A: An important issue is this: functions in $H_0^1$ have definite point values, while an element $f$ in $L^p$ is not changed by redefining $f(0)$. So it does not make sense to talk about $\delta$ on $L^p$. A element $f$ of $H_0^1(-1,1)$ is equal a.e. to a unique continuous function $\tilde{f}$ on $(-1,1)$, and $\tilde{f}$ is absolutely continuous with $\tilde{f}'\in L^2$. So $\tilde{f}(0)$ has definite meaning, but $\tilde{f}'(0)$ does not.
