The question states:
"Let $f$ and $g$ be integrable on $[a, b]$.
a) Prove that $|\int_a^b{fg}|\leq{\sqrt{\int_a^b{f^2}}}\sqrt{\int_a^b{g^2}}$. Hint: Consider separately the cases $0=\int_a^b(f-\lambda{g})^2$ for some $\lambda\in\mathbb{R}$ and $0<\int_a^b(f-\lambda{g})^2$ for all $\lambda\in\mathbb{R}$.
b) If equality holds, must $f=\lambda{g}$ for some $\lambda\in\mathbb{R}$? What if $f$ and $g$ are continuous?
c) Show that Theorem 1-1(2) [the Cauchy-Schwarz inequality] is a special case of (a)."
My question is about part (b). Can't we just let $g=0$ and let $f$ always be positive in both cases so that equality holds but there is no $\lambda$ such that $f=\lambda{g}$? Did Spivak mean to require that $f$ and $g$ be non-zero? If they are both non-zero and not necessarily continuous though, we can just let $g=0$ everywhere except on a set of measure $0$, so equality holds but $f$ need not equal $\lambda{g}$ for some $\lambda$. If $g$ is continuous, it can't be $0$ everywhere except on a set of measure $0$ and still have $\int_a^bg^2=0$. We can still let $g=\frac{1}{\lambda}f$ on only part of the interval though (say $[a, (a+b)/2]$) and $0$ everywhere else, in which case we still have equality, but on $[(a+b)/2, b]$ $\ \ f$ need not equal $\lambda{g}$ for some $\lambda$. Did Spivak maybe mean for $f$ and $g$ to be non-zero everywhere, or everywhere except a set of measure $0$?
Does anyone know what the original intent for part (b) of this exercise and its solution was?