Real function inner product space

[Mr Davis 97]

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?

 

[fresh_42]

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? $\int_a^b \dfrac{1}{x}\dfrac{1}{x}\,dx = -\dfrac{1}{b}+\dfrac{1}{a}\,.$

 

[Mr Davis 97]

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 $1$

 

[fresh_42]

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!

 

[Mr Davis 97]

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?

 

[fresh_42]

So is what Wolfram said incorrect?

Do you have a link?

 

[martinbn]

http://mathworld.wolfram.com/InnerProduct.html

They are being sloppy. They don't mean all functions.