Explain why the following map is NOT a scalar product

Question

$\langle,\rangle:V \times V \to \mathbb{R}. V = \{f : [0,1] \to > \mathbb {R}\ : f \text { is continuous}\},$ $$\langle f,g\rangle =f(0)g(0).$$

I know that a scalar product on $V$ is a map $\langle,\rangle:V \times V \to K$ which associates to each pair $(v,w)$ a scalar denoted by $\langle v, w \rangle$, satisfying the following:

• Linearity in the first variable

• Hermitian symmetry(or symmetry in the real case)

• Positivity

Answer 1

Try to write down the proof that the mapping satisfies all three properties.

Hint:

Focus on the positivity. The mapping satisfies the first two conditions, but it is not positive for all nonzero functions.

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Just to repeat the property of positivity that needs to hold:

For all $x\in X$, $\langle x, x\rangle \geq 0$, and if $\langle x, x\rangle = 0$, then $x=0$.

Answer 2

The last property actually has two parts or at least often is stated that way. First part is $\langle u, u\rangle\ge 0$ which certainly holds, but the second part is $\langle u, u\rangle = 0\Leftrightarrow u=0$ which isn't quite true.

However the last can be achieved by collapsing vectors by using equivalence classes of the relation $\langle u-v, u-v\rangle=0$ which would mean that they are represented by the value of the function at $0$ which basically makes the space $\mathbb R$.