 Solution

Assume the usual nondegenerate interval $a<b$. The integral is bilinear and symmetric because the functions and weight are real. It is nonnegative on $(f,f)$. If $f$ is nonzero at a point, continuity makes $|f|$ bounded below by a positive number on a nontrivial relative interval nearby, and the positive continuous weight is also bounded below there. Hence ${\int }_a^b{f}^{2}w>0$. Thus the formula is positive definite and defines an inner product.

The functions are orthogonal precisely whenA singleton interval would make this integral zero for every function, so the condition $a<b$ is necessary.