Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
2  
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
1  
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

2 Answers 2

up vote 1 down vote accepted

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 $\|f\|_{L^2} = \|(\left<f, e_n\right>)_n\|_{\ell^2}$.

share|improve this answer

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 $\mu$.

share|improve this answer
    
$\ell^p$ 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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.