Let me explain my setting: I want to solve a second order, linear, elliptic PDE on some domain $\Omega\subset\mathbb{R}^2$ open and bounded. I discretize my domain using quadrilateral elements with equidistant mesh-size $h$ in both directions, that are aligned to the axes. My FE-Space is $$V_h = \{u:\Omega\to\mathbb{R}|\,u_{|Q} = p(x)q(y),\,\text{ for some } p,q\in\mathcal{P}_1,\,\forall Q\in\mathcal{Q}_h\}$$
In the method I will be using, I am 'correcting' the basis functions chosen for $V_h$ in order to get piecewise constants. When estimating the error done in this projection, I directly get (no estimates done, so this cannot result in other terms) something of the form $$\|\nabla\partial\varphi_i\|_{L^2(Q)}$$ $\varphi_i$ being one of the basis function for $V_h$ and $Q\in\mathcal{Q}_h$ some element.
The problem is now the following: The standard, tensor product basis for a quadrilateral $Q\in\mathcal{Q}_h$ has for one of the local nodes the form $$\varphi_i(x,y)= \frac{x}{h}\frac{y}{h}$$ and the second derivative of this function scales like $\tfrac{1}{h^2}$. This basically kills my approach where I need a good estimate of the above norm, also if $h\to 0$.
My Idea: Why not scale the basis functions? If I drop the assumption that $\phi_i(v_i) = 1$ where $v_i$ is the local vertex my basis function corresponds to, precisely $$\tilde\varphi_i(x,y)=xy,$$ obviously this solves the current problem, since the second derivative of this function scales as 1.
My question: Since the values of the above basis function $\tilde\varphi_i$ are now not in the range $[0,1]$ anymore, but in $[0,h^2]$, will this cause problems when solving the linear system later?
I think the matrix will eventually get singular due to machine precision if $h$ is small enough.
What do you think? Maybe you have even tried this, or know some reference where this is addressed..
Thank you very much for all your responses or hints!
Denis