Apologies for my use of angular brackets, I don't know how to do it properly.
Suppose $p(t) = a_{0}+a_{1}t^2+a_{2}t^2, q(t) = b_{0}+b_{1}t+b_{2}t^2 \in P_{2}(\mathbb{R}).$
Define $\langle p \;| \;q \rangle:=p(0)q(0)+p(1)q(1)+p(2)q(2).$ Is this an inner product on $P_{2}(\mathbb{R})$?
Now, to show whether or not this is an inner product we resort to our axioms.
$(1)$ $\langle p \;|\; p \rangle.\;$ is non-negative. For all $p\in P_{2}(\mathbb{R}).$
$(2)$ If $\langle p \;|\; p \rangle = 0$ this implies $p = 0.$ For all $p \in P_{2}(\mathbb{R}).$
$(3)$ $\langle ap+q\;|\;r\rangle = a \langle p\;|\;r \rangle+\langle q\;|\;r\rangle$ For all $p,q, r \in P_{2}(\mathbb{R}), a \in \mathbb{R}.$
$(4)$ $\langle p \;|\; q\rangle = \overline{\langle q \;|\; p\rangle}$ For all $p,q \in P_{2}(\mathbb{R}).$
I am able to show true/false for all axioms except for axiom $(2)$. So my attempt at axiom $(2)$, so we show the when the inner product of an element in our space with itself is $0$, then the element must be the zero polynomial? Correct?
If $\langle p \;|\; p\rangle = 0$, that is $[p(0)]^2+[p(1)]^2 +[p(2)]^2 = 0$. Clearly, equality holds when $p(0) =p(1) = p(2) = 0$. How would I argue this is indeed the zero polynomial?