# Inner product space axiom

So, I just have a small question.

Given that $\langle f,g\rangle=\langle g,f\rangle$, $\forall f, g\in V$. One of the Inner Product Space axioms.

So, I prove by saying $\langle f,g\rangle = \int f(x)g(x)dx$ and $\langle g,f\rangle = \int g(x)f(x)dx$. So these are obviously equivalent, but how do I justify that the order of $g(x)$ and $f(x)$ in the integral doesn't matter?

Thanks in advance :)

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This property is true only if $f, g$ are real valued functions. –  Mhenni Benghorbal Mar 12 '13 at 0:18
@MhenniBenghorbal: And if these are complex valued functions? –  Asaf Karagila Mar 12 '13 at 2:09

## 2 Answers

The commutative law of multiplication.

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Thanks so much! It was just a question a tutor posed to us and although you know naturally that they are the same I didn't know how to actually describe it! But that's perfect :) –  Swayy Mar 12 '13 at 0:21

To extend ncmathsadist's answer, the multiplication inside the integral is pointwise multiplication. That is, we calculate $f(x)$ and $g(x)$ and multiply these numbers as real numbers.

The multiplication in the real numbers is commutative, so the pointwise multiplication of real-valued functions is commutative.

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