Proving that a given mapping is a inner product

Question

How to prove that a mapping $\langle\cdot,\cdot\rangle:\mathbb{R_{3}[x]}\times\mathbb{R_{3}[x]}\rightarrow\mathbb{R}$ is a inner product, where $\langle f,g\rangle=\int\limits_{-1}^{1}f(x)g(x)dx?$

First, for $a,b\in\mathbb{R}$ and $f,g,h\in\mathbb{R_{3}[x]}$, we have

$\langle af+bg,h\rangle=\int\limits_{-1}^{1}(af+bg)(x)h(x)dx=\int\limits_{-1}^{1}(af(x)+bg(x))h(x)dx$

$\qquad\qquad\quad\ =a\int\limits_{-1}^{1}f(x)h(x)dx+b\int\limits_{-1}^{1}g(x)h(x)dx=a\langle f,h\rangle+b\langle g,h\rangle.$

Second, for $f,g\in\mathbb{R_{3}[x]}$, we have

$\langle f,g\rangle=\int\limits_{-1}^{1}f(x)g(x)dx=\int\limits_{-1}^{1}g(x)f(x)dx=\langle g,f\rangle$.

And, third, for $f\in\mathbb{R}_3[x],$ $f\neq0,$ we have to prove $\langle f,f\rangle >0.$

I wrote $f=ax^3+bx^2+cx+d$ and I get $\langle f,f\rangle=\dfrac{2}{7}a^2+\dfrac{4}{5}ac+\dfrac{2}{3}c^2+\dfrac{2}{5}b^2+\dfrac{4}{3}bd+2d^2.$

I don't know how to prove that this is greater than 0. I tried to make a square of a binomial, but it isn't working.

Answer 1

Since, for each $x\in[-1,1]$, $f^2(x)\geqslant0$,$$\langle f,f\rangle=\int_{-1}^1f^2(x)\,\mathrm dx\geqslant0.$$

Answer 2

Note that \begin{eqnarray}\left\langle f,f\right\rangle &=& \int_{-1}^1f(x)f(x)\mathrm{x}\\ &=& \int_{-1}^{1}f^2(x)\mathrm{d}x.\end{eqnarray} As $f^2(x)\geq 0$ for all $x$, the integral $\int_{-1}^{1}f^2(x)\mathrm{d}x$ must be greater than or equal to zero as well. (We're basically taking the limit of a sum of positive values).

Answer 3

The answers given already are the most elegant, but for completeness let me give another way (continuing the approach in the post). Let $f = ax^3 +b x^2+cx +d.$ Then \begin{align*} \langle f, f \rangle &= \frac{2}{7} a^2 + \frac{4}{5} ac + \frac{2}{3} c^2 + \frac{2}{5} b^2 + \frac{4}{3} bd + 2d^2 \\ &= \frac{2}{3} \left( c + \frac{3a}{5}\right)^2 + \frac{8}{175} a^2 + 2 \left(d+\frac{b}{3}\right)^2 + \frac{8}{45} b^2, \end{align*} which is greater than zero (and equal to zero iff $f=0$).