Since in dimension $1$, $H^1$ is continuously embedded in $C_0$, we know that the Dirac distribution $\delta_0 \in H^1(\mathbb R)'$. Then by Riesz representation theorem we know that there exists a unique $u\in H^1(\mathbb R)$ s.t.
$$v(0) = \int_\mathbb R uv + \int_\mathbb R u'v'$$ for all $v\in H^1(\mathbb R)$.
which seems rather strange to me. Are we able to "see" this function u ? We can get a lot of information by specifying $v$ but I am not able to see what it could look like, apart from the fact that it is an even function (replace $v$ with $v(-\cdot)$).
- Am I wrong somewhere in all this ?
- If not, could it be that there is no simple characterization of $u$ ?
Thank you :)
The Dirac distribution is in $H^1(\mathbb{R})$ represented by
$$u(t) = \frac12 e^{-\lvert t\rvert}.$$
You can pretty easily verify that. How to find it: $H^1$ is under the Fourier transform isometrically isomorphic to the space $L^2(\mathbb{R},1+\lvert \xi\rvert^2)$ of (measurable, of course) functions such that
$$\int_\mathbb{R} \lvert f(\xi)\rvert^2(1+\lvert\xi\rvert^2)\,d\xi < \infty.$$
The representation
$$v(0) = \langle v,u\rangle = C\cdot\int_\mathbb{R} \hat{v}(\xi)\hat{u}(\xi)(1+\lvert\xi\rvert^2)\,d\xi$$
suggests trying $\hat{u}(\xi) = K\cdot\frac{1}{1+\lvert\xi\rvert^2}$, where $C$ and $K$ are normalising constants depending on the used definition of the Fourier transform. So, computing the Fourier transform of $\frac{1}{1+\lvert\xi\rvert^2}$, we know it must be a constant multiple of $e^{-\lvert t\rvert}$. Then one just needs to plug that in to find the constant.