# Why chose $H^{-1}(\Omega)$ instead of $L^2(\Omega)$?

For given $$f \in L^2(\Omega)$$ Poisson's equation reads $$- \Delta u=f \quad \text{on }\Omega.$$ So the variational problem becomes: For given $$f \in H^{-1}(\Omega)$$ find $$u \in H_0^1(\Omega)$$ such that $$\int_{\Omega} \nabla u \cdot \nabla \varphi \, \mathrm dx=\int_{\Omega}f\varphi \, \mathrm dx.$$ for all $$\varphi \in H_0^1(\Omega).$$

Why don't we keep $$f \in L^2(\Omega)$$?

• The notation $\int_\Omega f \, \varphi \, \mathrm{d}x$ does not make sense for $f \in H^{-1}(\Omega)$. – gerw Jun 12 at 11:25

For the poisson equation the natural space of the weak formulation is $$H^{-1}$$. However it is wrong or at least bad style to write it as an integral. It is advised to use the bracket notation $$(f,\phi)_{H^{-1}}$$.

Notice that for all functions $$f\in L^2$$ there exists an associated element $$f^*\in H^{-1}$$ given by $$(f^*,\phi)_{H^{-1}}=\int f \phi dx$$

To this end it is common to simply use $$f$$ instead of $$f^*$$. But it is not possible the other way round: not every element of $$f\in H^{-1}$$ can be represented via an integral.

• Thanks a lot - it's the Riesz representation theorem right? – Tesla Jun 12 at 12:10
• The standard Riesz theorem applies only for Hilbert spaces so in this case for $L^2$ and it dual. But we get the chain $H^{-1}\subset (L^2)^* \simeq L^2\subset H^1$ – maxmilgram Jun 12 at 12:21