Let $P_2 = \{p:\mathbb{R}\rightarrow\mathbb{R} \mid p(x) = ax^2+bx+c,$ with $ a,b,c \in \mathbb{R}\}$ be a vector space of polynomials up to second degree.
And the dot product $\langle\cdot,\cdot\rangle:P_2\times P_2 \rightarrow \mathbb{R}$ defined as: $$\langle\cdot,\cdot\rangle = \int_{-1}^1p(x)q(x)\,dx$$
How do I show that the dot product is defined on $P_2$?
I've tried to solve the integral as such: $$ \int_{-1}^1p(x)q(x)\,dx = \int_{-1}^1(ax^2+bx+c)(\tilde{a}x^2+\tilde{b}x+\tilde{c})\,dx = \cdots = 2a\left(\frac{\tilde{a}}{5}+\frac{\tilde{c}}{3}\right)+2c\tilde{c} $$ Since $\tilde{a}, c, \tilde{c} \in \mathbb{R}$ and the last bit $$2a\left(\frac{\tilde{a}}{5}+\frac{\tilde{c}}{3}\right)+2c\tilde{c}$$ can attain all values in $\mathbb{R}$ (which is the same for $P_2$), have I then showed what's needed?
No need to calculate it explicitly.
$$\langle p,p\rangle = \int_{-1}^1 p(x)^2 \,dx \ge 0$$ and $\langle p,p\rangle = 0$ if and only if $p(x) = 0$ for all $[-1,1]$ which is equivalent to $p = 0$ (because a nonzero polynomial cannot have an infinite number of zeros).
Linearity directly follows from the linearity of the intergral:
$$\langle \alpha p + \beta q, r\rangle = \int_{-1}^1 (\alpha p + \beta q)(x) r(x) \,dx = \int_{-1}^1 (\alpha p(x) + \beta q(x)) r(x) \,dx \\= \alpha \int_{-1}^1 p(x)r(x)\,dx + \beta \int_{-1}^1 q(x)r(x)\,dx = \alpha \langle p, r\rangle + \beta \langle q, r\rangle$$
Symmetry is obvious:
$$\langle p, q\rangle = \int_{-1}^1 p(x)q(x)\,dx = \int_{-1}^1 q(x)p(x)\,dx = \langle q,p\rangle$$
And there you have it.