= Solution
Linearity of the <integral> in each argument gives a <bilinear form>, and $fg=gf$ gives <symmetry>. Also $(f,f)=\int_{-1}^1 f(x)^2dx\geq0$. If a <polynomial> $f$ is nonzero somewhere on this interval, <continuity> makes $f^2$ strictly positive on a smaller interval, so the <integral> is positive. A <polynomial> vanishing on the whole interval is the zero <polynomial>. Thus the form is <positive-definite> and is an <inner product>.
Apply the <Gram-Schmidt process> to $1,x,x^2$. Odd <functions> have zero <integral>, so $x$ is already <orthogonal> to $1$, and the third <orthogonal polynomial> is $x^2-1/3$. Their squared <norms> are
$$
\|1\|^2=2,\qquad \|x\|^2=\frac23,\qquad
\|x^2-1/3\|^2=\frac25-\frac49+\frac29=\frac8{45}.
$$
Therefore an \b[orthonormal basis] is
$$
\boxed{\left\{\frac1{\sqrt2},\ \sqrt{\frac32}\,x,\ \sqrt{\frac58}\,(3x^2-1)\right\}.}
$$
These are the first three normalized <Legendre polynomials>. Their distinct degrees also prove that they span the three-dimensional <vector space>.
Back to article page
