Difference between l2 norm and L2 norm

Question

What is the difference between the $l^2$ norm and the $L^2$ norm. I can not find a definitive reference. Wikipedia uses them interchangeably.

Answer 1

Both norms are similar in that they are induced by the scalar product of the respective Hilbert space, but they differ because the different spaces are endowed with different inner products:

• For $\mathbb{R}^N$, the Euclidean norm of $v = (v_1,\dots,v_N)^T\in \mathbb{R}^N$ is defined by $$\|v\|_{2}^2 = (v,v)_{2} = \sum_{i=1}^N v_i^2.$$

• For $\ell^2$ (the space of real sequences for which the following norm is finite), the norm of $v = \{v_i\}_{i\in\mathbb{N}} \in \ell^2$ is defined by $$\|v\|_{\ell^2}^2 = (v,v)_{\ell^2} = \sum_{i=1}^\infty v_i^2.$$

• For $L^2(\Omega)$ (the space of Lebesgue measurable functions on a bounded domain $\Omega\subset\mathbb{R}^d$ for which the following norm is finite), the norm of $u\in L^2(\Omega)$ is defined by $$\|u\|_{L^2}^2 = (u,u)_{L^2} = \int_\Omega u(x)^2\,dx.$$

All this is standard, can be found in any introductory textbook on functional analysis, and is probably already known to you. Since the question is tagged error-estimation, you are likely interested in the practical difference in using one or the other, for example for finite element discretization. Let's say you have a finite-dimensional subspace $V_h\subset L^2(\Omega)$ which is the span of a finite number of basis functions $\{\varphi_1,\dots,\varphi_N\}$. Then any $u_h\in V_h$ can be written as $$u_h = \sum_{i=1}^N u_i \varphi_i. \tag{1}$$ Since $V_h\subset L^2(\Omega)$, you can of course measure $u_h$ by the $L^2$ norm. Alternatively, you can identify $u_h$ with the vector $\vec u:=(u_1,\dots,u_N)^T\in\mathbb{R}^N$ (sometimes called coordinate isomorphism) and measure $u_h$ by the Euclidean norm of $\vec u$.

How do the two ways of measuring $u_h$ compare? Plugging in the definition $(1)$ yields $$ \|u_h\|_{L^2}^2 = (u_h,u_h)_{L^2} = \sum_{i=1}^N\sum_{j=1}^N u_iu_j \int_\Omega \varphi_i(x)\varphi_j(x)\,dx = \vec u^T M_h \vec u,$$ where $M_h\in \mathbb{R}^{N\times N}$ is the mass matrix with entries $M_{ij} = \int_\Omega \varphi_i(x)\varphi_j(x)\,dx$. By comparison, we have $$ \|u_h\|_{\ell^2}^2 := \|\vec u\|_2^2 = \vec u^T \vec u.$$

Both norms are therefore equivalent, i.e., there exist constants $c_1,c_2>0$ such that $$ c_1\|u_h\|_{\ell^2}\leq \|u_h\|_{L^2} \leq c_2 \|u\|_{\ell^2}\qquad\text{for all }u_h\in V_h.$$ So in principle, you could use both norms interchangeably -- if the error goes to zero in one norm, it also goes to zero in the other norm, and with the same rate. However, note that while the constants $c_1$ and $c_2$ are independent of $u_h$, they do depend on $V_h$, and in particular on $N$. This is important if you want to compare discretization errors for different spaces $V_h$ with (say) $N_1<N_2$, in which case you should use a norm that does not itself depend on $N_1$ or $N_2$, i.e., the $L^2$ norm. (You can see this by taking $u_h$ as the constant function $u_h=1$ and compare $\|u_h\|_{\ell^2}$ for different $N$ with $\|u_h\|_{L^2}$ -- the former scales as $\sqrt{N}$, while the latter has the same value for every $N$, since the mass matrix compensates for the scaling.)

There's also a third -- intermediate -- alternative, where the mass matrix is approximated by a diagonal matrix $D_h$ (e.g., by taking as diagonal elements of $D_h$ the sum of the corresponding row of $M_h$), and the norm is taken as $\|u_h\|_{D}^2:=\vec u^T D_h\vec u = \sum_{i=1}^N (D_h)_{ii} u_i^2$; this is usually referred to as mass lumping. This norm is also equivalent with both the $\ell^2$ and the $L^2$ norm -- and in this case, the constants $c_1$ and $c_2$ when comparing $L^2$ and mass lumping norm do not depend on $N$.

Answer 2

The 2-norm for sequences is denoted by $\ell^2$. For functions on the real line $L^2$ is the standard notation of the 2-norm.