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Please help me solve this problem. He tried to read Kolmogorov, but did not find ways to prove this problem.

Closed due to the fact that off-topic participants aleksandr barakin , Kromster , Stranger in the Q , nick_n_a , AK ♦ Jan 16 at 18:29 .

It seems that this question does not correspond to the subject of the site. Those who voted to close it indicated the following reason:

  • " Learning tasks are allowed as questions only on the condition that you tried to solve them yourself before asking a question . Please edit the question and indicate what caused you difficulties in solving the problem. For example, give the code you wrote, trying to solve the problem "- aleksandr barakin, stranger in the Q, nick_n_a
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  • four
    I found a truly amazing proof of this statement, but the lack of mathematical formulas on RuSO does not allow bringing it here ... :) - Harry
  • one
    This is not homework, this is math, i.e. perfect offtopic programming site. - AK ♦

1 answer 1

To prove orthogonality

  • Prove first that

    <x - P H 0 (x), e i > = 0 , for any i

    This is proved from the definition of P H 0 (x) , the properties of the scalar product

    <a + b, c> = <a, c> + <b, c>
    <Ca, b> = C <a, b>

    and orthonormality of the system (e i ) , i.e. the fact that

    <e i , e j > = 0 | 1 , depending on the equality i = j

  • From this we prove the required assertion for an arbitrary vector y from H 0 , since vector y is a linear combination of vectors from (e i )

Self-adjointness is generally proved to the forehead by writing the expression for <P H 0 (x), y> and rearranging the terms / factors in it by the same properties of the scalar product. That is, it is possible in two or three steps to show that

<P H 0 (x), y> = ∑ <e i , x> <e i , y>

The expression on the right is perfectly symmetrical with respect to x and y .