Real function inner product space

  • #1
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Main Question or Discussion Point

Wolfram says that an example of an inner product space is the vector space of real functions whose domain is an closed interval [a,b] with inner product ##\langle f, g\rangle = \int_a^b f(x) g(x) dx##. But ##1/x## is a real function, and ##\langle 1/x, 1/x\rangle## does not converge... So how is this an inner product space?
 

Answers and Replies

  • #2
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Wolfram says that an example of an inner product space is the vector space of real functions whose domain is an closed interval [a,b] with inner product ##\langle f, g\rangle = \int_a^b f(x) g(x) dx##. But ##1/x## is a real function, and ##\langle 1/x, 1/x\rangle## does not converge... So how is this an inner product space?
Why shouldn't it converge? ##\int_a^b \dfrac{1}{x}\dfrac{1}{x}\,dx = -\dfrac{1}{b}+\dfrac{1}{a}\,.##
 
  • #3
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Why shouldn't it converge? ##\int_a^b \dfrac{1}{x}\dfrac{1}{x}\,dx = -\dfrac{1}{b}+\dfrac{1}{a}\,.##
Sorry, I meant to replace ##a## with ##-1## and ##b## with ##1##
 
  • #4
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Sorry, I meant to replace ##a## with ##-1## and ##b## with ##1##
But the function ##x \mapsto \dfrac{1}{x}## you mentioned isn't defined on ##[-1,1]##. You also need functions which are at least integrable, usually Lebesgue integrable, or continuous. Real valued alone is too weak, because at least the inner product must be defined!
 
  • #5
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But the function ##x \mapsto \dfrac{1}{x}## you mentioned isn't defined on ##[-1,1]##. You also need functions which are at least integrable, usually Lebesgue integrable, or continuous. Real valued alone is too weak, because at least the inner product must be defined!
So is what Wolfram said incorrect?
 
  • #6
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  • #7
martinbn
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