I recently started to study about Direct Methods in Calculus of Variation, so I'd like to apologize in advance if what I 'm about to ask is quite elementary.
It 's not that obvious to me why $\;I(u_0)\lt \infty\;$ since $\;u_0 \in u_0+W^{1,2}_0 (Ω)\;$. I don't understand why $\;I(u_0)\;$ is finite. Is this an assumption we make without loss of generality , in order to obtain that functional $\;I\;$ is bounded from above?
I googled this before I ask here but I didn't find something helpful. It seems it's quite trivial what I'm asking. Any help through this would be valuable!
Thanks!
The function $u_0$ belongs to $W^{1,2}(\Omega)$, which means that its weak gradient is in $L^2(\Omega)$. This is why $I(u_0)$ is finite.