# What is the difference between $L^2$ norm and $\ell^2$ norm?

I can find no precise definitions on the internet for the $L^2$ and $\ell^2$ norms. Certain websites keep switching between the two. Can someone please help me?

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The first is for function spaces, the latter for spaces of sequences. –  anon Jun 22 '12 at 6:31
I meant, could you please define them for me? It it is square root of the integral of the function or the square root of the summation of the discrete sums of the function? –  nada Jun 22 '12 at 6:32
The $\ell^2$ norm is a special case of the $L^2$ norm on a measure space (en.wikipedia.org/wiki/Measure_(mathematics)) for the counting measure. –  Qiaochu Yuan Jun 22 '12 at 13:16

Regarding the switching I would like to add that $L^2([0,1])$ and $\ell^2$ are isomorphic as Hilbert spaces (as they are both separable and infinite-dimensional). That means: If $(f_n)_{n\in\mathbb N}$ is an orthonormal basis of $L^2([0,1])$, for example the basis $\{\exp(2\pi i n \,\cdot\,) : n \in \mathbb Z\}$, then \begin{align*} L^2([0,1]) &\to \ell^2\\ f &\mapsto (\left<f, e_n\right>)_n \end{align*} is an isometric isomorphism of Hilbert spaces, so especially $\|f\|_{L^2} = \|(\left<f, e_n\right>)_n\|_{\ell^2}$.
The scalar product on $L^2$ is given by $\langle f,g\rangle=\int_X \bar{f}{g} \ d\mu$, whereas the scalar product on $\ell^2$ is given by $\langle x,y\rangle =\sum_{i \in \mathbb{N}} \bar{x_i} y_i$. In both cases the norm is given as usual in Hilbert spaces by $\lVert f\lVert=\sqrt{\langle f,f\rangle}$.
So $\ell^2$ is a special case of $L^2$ with $X=\mathbb{N}$ and the counting measure $\mu$.
$\ell^p$ is sometimes used to denote the spaces of functions over other discrete spaces, e.g. $\ell^2(\mathbf Z)$. –  tomasz Jun 22 '12 at 13:08