I was reading and doing problems from Spivak's Calculus on Manifolds. Q1-6 (a) stumped me a little.
Let $f$, $g$ be integrable on $[a,b]$.
Prove that $$\left| \int_a^b f\cdot g \; \right | \leq \left(\int_a^b f^2\right)^{1/2} \cdot \left (\int_a^b g^2\right )^{1/2}$$
Hint: Consider separately the cases
$0=\int_a^b \left(f-\lambda g\right)^2$ for some $\lambda \in \Bbb R$, and $0<\int_a^b \left(f-\lambda g\right)^2$ for all $\lambda \in \Bbb R$.
The second case is fine. Square out the brackets, use linearity of the integral and you get a quadratic in $\lambda$ with no real roots so the discriminant is negative, that's that.
But for the first case the subtlety is that $f$ and $g$ are integrable not necessarily continuous. So we can have $f\neq \lambda g$.
What I tried to do was the same thing is use the discriminant again but that gives the inequality the other way around, which suggest that if it holds, the polynomial in $\lambda$ can only have repeated roots. But I can't prove the roots are not distinct.
My other approach would be to say that the integrand is positive and that we can split $[a,b]$ into intervals whereby $f-\lambda g=0$ but I'm not sure how to apply integrability to the points $x$ where $f(x)\neq\lambda g(x)$ which I would assume are isolated and possibly finite.
Best Answer
I don't really understand your problem. We simply note that, since the integrand is nonnegative: $$ 0\leq \int (f-\lambda g)^2 dx = \int f^2 dx -2\lambda \int fg dx +\lambda ^2 \int g^2dx = \alpha \cdot \lambda^2 - 2\beta \lambda +\gamma $$ with the obvious choices of $\alpha,\beta,\gamma$. Observe that $\alpha \geq 0$, so that there are two cases:
$\alpha=0$. This implies $\beta=0$ (otherwise, let $\lambda \to \pm \infty$ depending on the sign of $\beta$ to get a contradiction. But for $\beta =0$, the claim is trivial.
$\alpha >0$. Here, we get $$ 0\leq (\sqrt{\alpha}\lambda - \beta/\sqrt{\alpha})^2 +\gamma - \beta^2/\alpha . $$ A suitable choice of $\lambda$ makes the bracket vanish, which easily implies the claim.
EDIT: These considerations show that we do not need to distinguish the cases given in the hint, as long as we are not interested in a criterion for equality, but your are only asking for a proof of the inequality, not for a characterization of the case in which equality holds.