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RIESZ-FISCHER THEOREM (matrix form and proof)
Posted:
Jul 15, 1996 10:02 AM
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Dear Emma,
You asked me to reveal the gist of the statement and the proof of the Riesz-Fisher Theorem. Here tyey are.
Everything is in the Hilbert space H of the square summable real sequences. Let M be an infinite square matrix whose rows form an orthonormal set of elements of H (the columns of H need not form an orthonormal set, however, the norm of each column of M is necessarily less than or equal to 1). Let c be an element of H. Then the Riesz-Fischer Theorem can be stated in matrix form as follows.
THEOREM. The equation
(1) Mx = c (x and c are written as column vectors)
has a solution x = k such that k is an element of H, the norm of k is equal to the norm of c and the elements of c are the Fourier coefficients (w.r.t the rows of M) of k.
PROOF. Let M' be the transpose of M. Clearly,
(2) MM' = the infinite unit matrix
Thus, from (2) it follows that
(3) (MM') c = c
But then with a little effort (since the product of infinite matrices need not be associative, although for the special case (3) it is) from (3) we obtain
(4) M(M'c) = c
But then it can be readily verified that (4) implies that k = M'c
is the desired solution x = k mentioned above.
P.S. I did not give details since I did not want to obfuscate the issue.
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