I am computing the rate of convergence of my finite difference scheme in norm $L_2$. Which is the correct way to compute it? This:
\begin{align} L_2 &= \frac{1}{N}\sqrt{\sum_{j=1}^N(u^{numerical}_j-u^{exact}_j)^2} \;\;\;\;\;\;\;\;\;\;\;\; (1)\\ \end{align}
or this:
\begin{align} L_2 &= \sqrt{\frac{1}{N}\sum_{j=1}^N(u^{numerical}_j-u^{exact}_j)^2} \;\;\;\;\;\;\;\;\;\;\;\; (2)\\ \end{align}
where $u^{numerical}$ is the computed along-x velocity at velocity point $j$, with $j=1..N$, and $u^{exact}$ is the analytical solution. In some publications I saw the second option, but that clashes against the fact that norm $L_2$ should be always smaller than norm $L_1$ (see https://math.stackexchange.com/questions/245052/showing-that-l2-norm-is-smaller-than-l1 ). However here norms are "scaled" so I guess that inequality should hold only for the unscaled version. Note that if one uses (1) the inequality still holds. In Leveque's book a pag 140 the error is defined as:
\begin{align} L_2 &= \sqrt{\Delta{x} \sum_{j=1}^N(u^{numerical}_j-u^{exact}_j)^2} \;\;\;\;\;\;\;\;\;\;\;\; (3)\\ \end{align}
Although this would be strictly true for a finite volume method, if we extend it to a finite difference method and to two dimension, by division by the total active area of the domain one gets equation (2). This question is related to question Why a finite difference scheme would give second order of accuracy in norm L2 but 1.5 with L1 (while 1 with Linf)? , since I see that the order of accuracy is larger than the thoretical value of 2 if I use equation (1), while it is smaller if I use equation (2). Which one is the best formulation?