Is $$V_1 = \{v \in H^1(\Omega) \;:\;f(v) = 0 \text{ on } \partial \Omega\}$$ dense in $H^1(\Omega)$ with the same norm as $H^1(\Omega)?$
Here $f$ is some linear functional so that $V_1$ is also Hilbert.
I don't know how to prove this. Maybe someone can sketch how to prove it for $f(v) = v$ (so $V_1 = H^1_0$) cause I don't know how to do that either.
Edit: This answer addresses the case $f(v)=v$.
I'm far from being an expert, so I've looked around and apparently, your $V_1$ makes sense in particular when $\Omega$ is a bounded open set in $\mathbb{R}^n$, with $C^1$ boundary. In this case, one can define the trace operator $$ T:u\longmapsto u_{|\partial \Omega} $$ on $C^1(\overline{\Omega})$ functions by restriction to the boundary (it would not make sense on $H^1$ directly, as these functions are defined up to a measure zero set, and the boundary is such a set).
Wth the hypotheses above, we can prove that $T$ is bounded from $C^1(\overline{\Omega})$ to $L_2(\partial\Omega)$. So it can be extended by density to a continuous operator on $H_1$ and your $V_1$ is the kernel of the trace operator, usually denoted by $H^1_0(\Omega)$.
By construction, $H^1_0(\Omega)$ is closed in $H_1(\Omega)$ as the trace operator is continuous. Clearly, $H^1_0(\Omega)$ is not the whole of $H_1(\Omega)$, as it does not contain, for instance, the nonzero constant functions. So with the assumptions above, $H^1_0(\Omega)$ is not dense in $H_1(\Omega)$.