discrete $L^p$ norms for non-uniform grid

Question

I am reading a book on numerical methods and the square of the discrete $L^2$ norm is defined as $$||x||^2_2=h\sum_1^Nx^2_i$$ Every point gets a "weight", which is $h$, thus this is like an average over the squares of the values at all points. This in fact comes from the approximation of a continuous integral. On the other hand, can I define similar norm where the grid is nonuniform with spacing $h_i$ as $$||x||^2_2=\sum_1^Nh_ix^2_i$$ that seems to me natural as I could approximate a continuous integral this way too but as I don't see that in the books made me suspicious I am missing something! So, if I have non-uniform grid and I want to make some estimates in this norm, how one should define it?

Answer 1

You are exactly right: The norm is defined in such a way that the discrete (vector) norm equals (or at least approximates) the continuous norm of a corresponding function.

When you have non-uniform meshes, the form you give (with the $h_i$ inside the sum) is correct and frequently used in the analysis of non-uniform meshes.

Of course, in 2d, the correct formula would contain a factor of $h^2$ and in 3d of $h^3$.

Answer 2

Given a function $f(x)$, $x \in (a,b)$, let we define the $L^2$ norm as \begin{equation} \lVert f \rVert_2^2 = \int_a^b \lvert f(x)\rvert^2 \, dx. \end{equation} Given a vector $\mathbf f \equiv \{ f_i = f(x_i), \; i=0\dots N\}$, with $x_i = a + i \frac{b-a}{N}$ we define the discrete $L^2$ norm as

\begin{align} \lVert \mathbf f \rVert_{2,\text{d}}^2 &= h \sum_{i=0}^N \lvert f_i \rvert^2, &h = \frac{b-a}{N} \end{align} In your question you assume that this is done because we want that \begin{equation} \lim_{h \rightarrow 0} \: \lVert \mathbf f \rVert_{2,\text{d}} = \lVert f \rVert_2 \end{equation} and you are interpreting the discrete norm as a sort of quadrature rule, and wonder why for non uniform grids a better formula is not used.

This interpretation is not wrong, but not the only one possible. As an engineer working with physical quantities, instead of pure numbers, I prefer to think of the discrete norm as of an $\ell^2$-euclidean norm scaled in such a way to make it dimensionally homogeneous to the continuum $L^2$ norm. So if we can prove that $\lVert f - f^h \rVert_2 \rightarrow 0$, we may expect that $\lVert \mathbf f -\mathbf f^h\rVert_{2,\text{d}} \rightarrow 0$. Without the scaling factor this would not be true.

EDIT:

I deleted my conclusions here. See the answer by Wolfgang.

Note that the scaled euclidean norm is easy to compute, while your proposal is a little bit imprecise (may raise some concerns as a quadrature formula) and expensive to compute.

Bottom line: the discrete $L^2$-norm is not (needs not to be) and approximation to the continuous one, but can be simply interpreted as a scaled $\ell^2$-euclidean norm, dimensionally consistent to continuous $L^2$ norm.