$L^{2}[-\pi,\pi]$ is unitarily isomorphic to $l^2(\Bbb C)$ So I have countable orthonormal basis of $L^2[-\pi,\pi]$ as $\{e^{inx}\}_{n \in \Bbb Z}$ and countable orthonormal basis of $l^2(\Bbb C)$ as $\{a_n\}_{n \in \Bbb Z}$ such that $a_i=\{0,0,...,0,1,0,...0,0,...\}$ where "$1$" is in $i$-th place.
My motive is to find a Linear map $T:L^2[-\pi,\pi] \to l^2(\Bbb C)$ which is one-one , onto, isometry and with the property $TT^\star=T^\star T =I$ where $T^\star$ is adjoint of $T$. (i.e. $<Tx,y>=<x,T^\star y>$)
I define the linear map as $T(e^{inx})=a_n$.
As soon as I do this my worries begin.


*

*How would I show that $T$ is a linear map?


So let $f,g \in L^2[-\pi,\pi]$. I need to show that $T(f+g)=T(f)+T(g)$.
I think I should write $f$ and $g$ as linear combinations of elements from the set $\{e^{inx}\}_{n \in \Bbb Z}$.Let $f=\sum b_je^{ijx}$ and $g=\sum c_je^{ijx}$.
Then $T(f+g)=?$
Also for a scalar $\alpha \in \Bbb C$ I am struggling to get anywhere in the search of $T(\alpha f)$ to get it equal to $\alpha T(f)$.
I think I might be able to prove the rest that is one-oneness, onto-ness,isometry and the last property "$TT^\star=T^\star T=I$" once I get to prove the linearity.
P.S: Was the way I defined $T$ fine?
 A: Note that it is not a priori clear that a map satisfying $T(e^{inx}) = a_n$ exists. In fact, there exists infinitely many different linear maps $T \colon L^2([-\pi,\pi]) \rightarrow \ell_2(\mathbb{Z})$ but only one of those maps will be the map you want so it is not enough to say "we define $T$ by $T(e^{inx}) = a_n$".
In linear algebra, given an (algebraic) basis $(v_{\alpha})_{\alpha \in I}$ of some vector space $V$ and some choice of elements $(w_{\alpha})_{\alpha \in I}$ in another vector space $W$, there is a unique linear map $T \colon V \rightarrow W$ satisfying $T(v_{\alpha}) = w_{\alpha}$ for all $\alpha \in I$. If $(v_{\alpha})_{\alpha \in I}$ are merely linearly independent but do not form a basis, there exists many linear maps $T \colon V \rightarrow W$ satisfying $T(v_{\alpha}) = w_{\alpha}$ for all $\alpha \in I$ (each obtained by extending $(v_{\alpha})_{\alpha \in I}$ to a basis of $V$ and defining $T$ arbitrary on the extra vectors). In your case, the sequence $\{e^{inx}\}_{n\in\mathbb{Z}}$ do not form an algebraic basis for $L^2([0,2\pi])$ so there are many linear maps satisfying $T(e^{inx})=a_n$.
However, if you work in a Hilbert space $V$ with a countable orthonormal system $\{v_i\}_{i=1}^{\infty}$ and you choose some sequence of elements $\{w_i\}_{i=1}^{\infty}$ in another Hilbert space $W$, then trying to mimick the linear algebra construction and taking into account the topology, you are lead to defining a map $T \colon V \rightarrow W$ by
$$ T \left( \sum_{i=1}^{\infty} c_i v_i \right) = \sum_{i=1}^{\infty} c_i w_i. $$
Any element $v \in V$ can be uniquely written as $v = \sum_{i=1}^{\infty} c_i v_i$ where $(c_i)_{i=1}^{\infty} \in \ell_2(\mathbb{N})$ but a priori there is no guarantee that the right hand side will converge in $W$. However, if the $\{w_i\}_{i=1}^{\infty}$ are an orthonormal system for $W$, you can check that the right hand indeed converges and so you have a well-defined map between $V$ and $W$ which is obviously linear. You can check that $T$ will be an isometry and the inverse of $T$ will be given by
$$ T^{-1} \left( \sum_{i=1}^{\infty} c_i w_i \right) = \sum_{i=1}^{\infty} c_i v_i $$
which will also be $T^{*}$.
A: The map
$$T:L^2\to l^2:f\mapsto\left(a_n=\frac1{\sqrt{2\pi}}\int f(t)\exp(-int)dt\right)_{n\in\mathbb Z}$$
sends any square integrable function to the sequence of its Fourier coefficients, so we have
$$\eqalignno{f(x)&=\sum_{n=-\infty}^{+\infty}a_n\exp(inx)&(1)}$$
for almost every $x$.
The fact that $T$ is an isometry is known as Parseval's identity:
$$\sum_{n=-\infty}^{+\infty}|a_n|^2=\int|f(t)|^2dt$$
(the LHS is the square of the $l^2$ norm of the sequence $(a_n)_n$ while the RHS is the square of the $L^2$ norm of the function $f$)
Between Hilbert spaces, saying that an invertible linear map is isometric or unitary is one and the same thing. By definition, a linear map $U$ is unitary if it is the inverse of its adjoint ($UU^*=I=U^*U$) but by the definition of the adjoint this is the same as saying that $(Ux,Uy)_{l^2}=(x,y)_{L^2}$ for all $x,y\in L^2,$ i.e., $U$ is unitary if and only if it preserves the scalar product.
Now maps that preserve the norm automatically preserve the scalar product because the norm in an inner product space can be expressed as a function of the scalar product; this is known as polarization and in complex inner product spaces it takes the form
$$\langle x,y\rangle=\frac14\left(\|x+y\|^2-\|x-y\|^2+i\|x+iy\|^2-i\|x-iy\|^2\right).$$
Actually, in our case the explicit form of $T^*$ as a mapping from sequences $(a_n)_n$ to functions $f$ is given by formula (1).
