How similar are $l_2$ and $L_2[a,b]$? The former space is the space of infinite sequence Having a norm similar to Euclidean space. The latter is square integrable functions. Is there isometry or isomorphism between them?
 A: They are isometrically isomorphic. And the proof is very simple: all you need is that both have a countable orthonormal basis. So take orthonormal bases $\{e_n\}$ and $\{f_n\}$ of $\ell^2$ and $L^2$ respectively. 
Now define, on the span of $\{e_n\}$, 
$$
W\sum_{n=1}^m a_n e_n=\sum_{n=1}^m a_n f_n. 
$$
We have 
\begin{align}
\left\|W\sum_{n=1}^m a_n e_n\right\|^2
&=\left\|\sum_{n=1}^m a_n f_n\right\|^2=\sum_{n=1}^m|a_n|^2=\left\|\sum_{n=1}^m a_n e_n\right\|^2.
\end{align}
So $W$ is linear an isometric, and thus it extends to a linear operator $W:\ell^2\to L^2$, still isometric. 
It only remains to see that $W$ is onto: 
$$
\sum_n a_n f_n=W\sum_n a_n e_n. 
$$
Note that the above has nothing to do with the particulars of $\ell^2$ and $L^2$: any two Hilbert spaces of the same dimension are isometrically isomorphic. 
A: I'll explain David Ullrich's comment for the sake of completeness.
For convenience, I'll work with the case $a = 0$ and $b = 1$.  Otherwise, one should modify the basis functions $\{e_{n}\}_{n \in \mathbb{Z}}$ (introduced below) accordingly.
Next, if $f \in L^{2}([0,1])$ and $n \in \mathbb{N}$, define the $n$th Fourier coefficient $\hat{f}(n)$ by
$$\hat{f}(n) = \int_{0}^{1} f(x) e^{- i 2 \pi n x} \, dx.$$
I will prove that the function $\mathcal{F} : f \mapsto \hat{f}$ is an isometric isomorphism from $L^{2}([0,1])$ to $\ell^{2}(\mathbb{Z})$.  
Let $e_{n} : [0,1] \to \mathbb{C}$ be given by 
$$e_{n}(x) = e^{i 2 \pi n x}.$$
First, I will show $\{e_{n}\}_{n \in \mathbb{Z}}$ is a complete orthonormal system in $L^{2}([0,1])$ (i.e. an orthonormal set with dense span).  Orthonormality is a calculus exercise:
$$\int_{0}^{1} e_{n}(x) \overline{e_{m}(x)} \, dx = \int_{0}^{1} e^{i 2 \pi (n - m) x} \, dx = \delta_{nm}.$$
Density is a bit more work.
Let $\mathcal{A} = \text{span} \{e_{n} \, \mid \, n \in \mathbb{Z}\}$.  Then $\mathcal{A} \subseteq C([0,1])$ is a self-adjoint algebra that separates points and vanishes nowhere, i.e. if $x \neq y$, then there is a $f \in \mathcal{A}$ such that $f(x) \neq f(y)$, and if $x \in [0,1]$, then there is a $f \in \mathcal{A}$ such that $f(x) \neq 0$.  I leave it to the reader to verify these claims.  (Hint: Use trigonometric functions.)  By the Stone-Weierstrass Theorem, $\mathcal{A}$ is dense in $C([0,1])$ (with respect to the supremum norm).  Note that density of $\mathcal{A}$ can also be proved using the Fejer kernel (or other "approximations of the identity"), which is certainly worth looking into.
Since the inclusion $i : C([0,1]) \to L^{2}([0,1])$ is continuous with dense image, it follows that $\mathcal{A}$ is dense in $L^{2}([0,1])$.  This proves that $\{e_{n}\}_{n \in \mathbb{Z}}$ is a complete orthonormal system in $L^{2}([0,1])$.  
We conclude that if $f, g \in L^{2}([0,1])$, then
\begin{align*}
\int_{0}^{1} f(x) \overline{g(x)} \, dx &= \sum_{n \in \mathbb{Z}} \langle f, e_{n} \rangle_{L^{2}([0,1])} \langle e_{n}, g \rangle_{L^{2}([0,1])} \\
&= \sum_{n \in \mathbb{Z}} \hat{f}(n) \overline{\hat{g}(n)}
\end{align*}
where the last equality follows from the definition of the Fourier coefficients.  This proves that $\mathcal{F} : f \mapsto \hat{f}$ is an isometric isomorphism.   
The $L^{2}$-limit $f(x) = \sum_{n \in \mathbb{Z}} \hat{f}(n) e^{i 2 \pi n x}$ is one way to introduce the study of Fourier series and harmonic anlaysis.  The RHS is called the Fourier series of $f$.  
A: They are isomorphic as inner-product spaces.
First let us be clear about what $\ell^2$ is. It is the space of sequences $(c_n)$ for which $\displaystyle\sum_n |c_n| < \infty.$
If $c$ is the sequence of Fourier coefficients of a square-integrable function $f$ on $[a,b],$ then the correspondence $c \longleftrightarrow f$ is an isomorphism – a linear isometry. See Riesz–Fischer theorem.
A: You already know that $L^2[a,b]$ has a countable orthonormal basis $(f_n)_{n=1}^\infty$.
Define a linear map $A : L^{2}[a,b] \to \ell^2$ as
$$Af = \Big(\langle f, f_n\rangle\Big)_{n=1}^\infty = \Big(\langle f, f_1\rangle, \langle f, f_2\rangle, \langle f, f_3\rangle, \ldots\Big)$$
for all $f \in L^2[a,b]$.
$A$ is an isometry because of Parseval's identity:
$$\|f\|^2_2 = \sum_{n=1}^\infty \left|\langle f, f_n\rangle\right|^2 = \left\|\Big(\langle f, f_n\rangle\Big)_{n=1}^\infty\right\|^2_2 = \|Af\|^2_2$$
$\operatorname{Im} A$ is clearly dense in $\ell^2$ because it contains the canonical basis $(e_n)_{n=1}^\infty$ for $\ell^2$. Namely $Af_n = e_n$ for all $n \in \mathbb{N}$.
Also, $\operatorname{Im} A$ is complete because $L^2[a,b]$ is complete and linear isometries preserve completeness. In particular, $\operatorname{Im} A$ is a closed subspace of $\ell^2$ so:
$$\ell^2 = \overline{\operatorname{Im} A} = \operatorname{Im} A \subseteq \ell^2 \implies \operatorname{Im} A = \ell^2$$
Therefore, $A$ is an isometric isomorphism between $L^2[a,b]$ and $\ell^2$. 
In general, any separable inner product space is isometrically isomorphic to a dense subspace of $\ell^2$, via the same isomorphism. Furthermore, separable Hilbert spaces are isometrically isomorphic to $\ell^2$.
A: Hint $L^2([a,b])$ is separable
Use this to construct a bijection between ONB.
