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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. – kennytm Jul 2 '12 at 19:02

IN general as $f^2\ge0$, we have

\begin{equation} \int_{-1}^{1} f^2(x)dx =0 \Rightarrow f = 0 \quad a.e. \end{equation} But as $f$ is continuous this show that $f$ is zero, because if $f$ were not identically equal to zero, $f^2$ would be positive on some interval (by continuity) and have a nonzero integral.

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