# Showing that a bilinear form is inner product

Let $V$ be a vector space of all polynomials $\mathbb{R}\to \mathbb{R}$ of degree $2$ or less.Define $B(f,g)= \int_1^{-1}f(x)g(x)dx$ And I need to show that $B(f,f)>0$ for all $f\neq 0$ Here was my approach: Suppose $f=a_2x^2+a_1x+a_0$ then I get $B(f,f)= 2/5 a_2^2+2/3 a_1^2+4/3a_0a_2$ and it is supposed to be positive. But I can't see why. Please help.

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I think you are over complicating things a bit. We have $$B(f,\ f) = \int_{-1}^1f(x)f(x)\ dx=\int_{-1}^1\left[f(x)\right]^2\ dx$$ Now the function $\left[f(x)\right]^2$, being the square of some function, is necessarily non-negative.

This is now a calculus problem, where you have to show that if a continuous, non-negative function is non-zero somewhere within the domain of integration, then it's integral is non-zero.

Finally, once you prove that $B(f,\ f)=0$ implies $f=0$ in $(-1,\ 1)$, use the fact that two polynomials equal on an open set are equal everywhere.

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