Is the following bilinear form positive

Question

Let $B:V\times V\rightarrow \mathbb R$ where $V$ is all the polynomials with degree $\leq2$.

Determine whether the following is positive $\langle f,g\rangle =\int^1_0f(x)g(x)xdx$.

I tried refuting it, but I'm not really sure its not true. I can't see how to prove it though.

P.S. The definition of positive bilinear form is for every $v\in V \neq 0$,

$0\leq \langle v,v\rangle $

Answer 1

For $0 \ne f(x) \in V$, consider $\langle f, f \rangle = \int_0^1 (f(x))^2 x dx$. Since $f(x) \ne 0$, we have $(f(x))^2 x \ge 0$ everywhere on $[0, 1]$ and $(f(x))^2 x > 0$ somewhere in the interior $(0, 1)$ of this interval. If $(f(x_0))^2 x_0 > 0$ for $x_0 \in (0, 1)$, by continuity of $(f(x))^2 x$ there exists an open interval $(a, b) =I \subset [0, 1]$ with $x_0 \in I$ and $(f(x))^2 x > m >0$ on $I$ for some real $m$ with $0< m < (f(x_0))^2 x_0$. Then we must have $ \int_0^1 (f(x))^2 x dx \ge m(b - a) > 0$, showing that $\langle f, f \rangle > 0$; thus $\langle \cdot, \cdot \rangle$ is positive. QED.

Hope this helps. Happy New Year,

and as always,

Fiat Lux!!!

Answer 2

Hint:

You said you want to check if $\langle f,f\rangle\geq 0$ holds.

That would be $\int_0^1f(x)^2x dx\geq 0$, by the definition you gave.

Is the integral of $f(x)^2x$ on $[0,1]$ nonnegative?! It would be if, say $f(x)^2x$ were a nonnegative function...