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Consider a vector space defined as all continuous functions $f:[-1,1] \to \mathbb{R}$ equipped with the following inner product $$\langle f,g \rangle = \int_{-1}^{1} f(x)g(x) \ dx$$

Now $\langle f,f \rangle = 0$ implies that $$\langle f,f \rangle = \int_{-1}^{1} f(x)^2 \ dx = 0$$

But $f$ doesn't have to be $0$ for the above to hold. Any symmetric function on $[-1,1]$ will have inner product $0$. How do we rectify this problem?

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    What? $\langle f,f\rangle$ is an integral of $f^2$ which is always nonnegative.2012-07-02

1 Answers 1