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Prove. The square of the length of a vector in Hilbert space is equal to the sum of the squares of its projections onto a complete system of mutually orthogonal directions. In other words, if


             ole.gif


are a complete orthonormal system of functions in Hilbert space and if a function f is given by


             ole1.gif


then


             ole2.gif


Proof. Denote by rn(x) the difference between f (x) and the sum of the first n terms of its series representation i.e.


             ole3.gif


Now the function rn(x) is orthogonal to each of the functions ole4.gif . Let us show that it is orthogonal to, for example, the function ole5.gif , i.e., show that


             ole6.gif


We have


             ole7.gif


or


ole8.gif


where we employ the fact that, because the functions are orthogonal to each other,


             ole9.gif


Now in 1)


             ole10.gif


and thus 1) becomes


             ole11.gif


Hence, in the equation


ole12.gif


the individual terms on the right side are all orthogonal to each other. Now, by the Pythagorean theorem, the square of the length of f (x) is equal to the sum of the square of the summands on the right side of 2), i.e.


             ole13.gif


Since the system of functions ole14.gif is normalized, we have


ole15.gif


Now we are dealing with functions of a Hilbert space and a Hilbert space consists of those functions f for which the Legesgue integral of | f 2 | exists. This means that as n approaches ole16.gif , the integral


             ole17.gif


converges and the quantity


             ole18.gif


on the right side of 3) approaches zero. Thus formula 3) becomes


             ole19.gif




Source: Mathematics, Its Content, Methods and Meaning. Vol. 3, pp.243-244


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