What does the energy norm $\sqrt{err^T*A*err}$ tell?
Where $A$ is the matrix that would be used in solving the problems linear system of equations $Au=b$.
$err=u_{approx}-u_{exact}$
Particularly, I saw an example, where energy norm gets smaller (with increasing grid size), but e.g. L2-error grows. Thus begging the question about what the energy norm is measuring.
Such as here:
https://www.particleincell.com/2012/finite-element-examples/
The Finite Element Method is designed to minimize the Energy Norm
which may not result in the lowest displacement norm L2.
In the particular example where the $L^2$ error grows and the energy norm error decreases, this is most likely due to the fact that the mesh size hasn't reached the asymptotic regime.
As an example, think of the Poisson problem with homogeneous Dirichlet boundary conditions. In this case the energy norm is the $H^1$-seminorm, and by the Poincare inequality on $H^1_0$ one has that the $L^2$ norm is bounded by the $L^2$ norm, and therefore when we reach the asymptotic regime the $L^2$ error must also decrease. Moreover, for symmetric problems the bound from Poincare and Cea's lemma:
$$\|u-u_h\|_{L^2}\le C|u-u_h|_{H^1}\le Ch^p$$
can be improved upon by duality arguments to
$$\|u-u_h\|_{L^2}\le Ch^{p+1}.$$
In general, the energy norm is just a natural norm associated to problem, e.g. for the Poisson problem it is the $H^1_0$-semi norm, and for more general divergence form problems such as $-\nabla\cdot(A\nabla u) = f$, the energy norm is $\| u\|^2=\int(A\nabla u)\cdot\nabla u$.
Assuming the matrix $A$ is positive definite, the energy norm will be a norm on $H^1_0$. If $A$ is indefinite, this is where problems could occur, as the energy norm may no longer be a norm.