Hilbert Spaces

[Camillo]

Foreword
An interesting example of infinite dimensional Hilbert space is given by the set of sequences $x=(x_1,x_2,...x_k,.....)$, being $x_k$ complex numbers.
This set can be structured as Vector Space putting :
$lambda*x = (lambda*x_1,lambda*x_2,..., lambda*x_k,....) $ ;$ x+y = (x_1+y_1,x_2+y_2,....x_k+y_k,....)$

The subset $l^2 $ of the sequences which verify the condition :

$sum_(k>=1) |x_k|^2 < +oo $
is a Vector Subspace of the previous one ; the scalar product can be defined as follows :

$ = sum_(k>=1) x_ky_k^* $ being $ y_k^* $ the coniugate of $y_k $.

Let's try this exercise :
In the $l^2$ space of real sequences $x =(x_n )_(n=1)^oo $ such that $||x||_2 = sqrt(sum|x_n|^2) < +oo $ let's consider the infinite dimensional vectors :

$x=(2,0,0,.......) ; y = (1/k)_(k=1)^oo $

a) Calculate the angle formed by the vectors $ x $ and $y $ .

b) Orthormalize the sistem $(x,y) $ that is find an orthonormal system that generates the same space as $(x,y) $.

[Camillo]

Quite correct

[_luca.barletta]

a)

The angle formed is:
$theta = arccos(()/(||x||*||y||))$ (1)
where
$\=sum_(k>=1) x_ky_k=x_1y_1=2$
$||x||=2$
$||y||=sqrt(sum_(k>=1)y_k^2)=sqrt(sum_(k>=1)1/k^2)=pi/sqrt(6)$
substituting into (1) one get:
$theta=arccos(sqrt(6)/pi)$

b)

Gram-Schmidt orthonormalization can satisfy the request:
$a1=x/(||x||)={1,0,0,...}$
$a2=(y-\*a1)/(||y-\*a1||)$
where
$\=sum_(k>=1)a_ky_k=a_1y_1=1$
$a1=a1={1,0,0,...}$
$y-\*a1={0,1/k}_(k=2)^(+infty)$
$||y-\*a1||=sqrt(sum_(k>=2)1/k^2)=sqrt(pi^2/6-1)$
finally
$a2=1/sqrt(pi^2/6-1){0,1/k}_(k=2)^(+infty)