Let $a,b \in \mathbb{R}$ such that $ab > 0$ and consider $f : [a,b] \to [a,b]$ a continuous function. Prove there exists $c \in (a,b)$ such that $cf(c) = ab$ (from Berkely Math 104 Final).
As stated in the exam, I don't believe this is correct, with $a = 1, b = 2, f(x) = 3 - x$ a counterexample. Am I correct?
However, if we revise the question to ask for a $c \in [a,b]$, it is correct. My proof is below.
Let $g(x) = ab/x$ and let $h(x) = f(x) - g(x)$ on $[a,b]$. We seek to prove that $h(x)$ has a zero on $[a,b]$. $f, g,$ and $h$ are continuous. $g(a) = b$, so $f(a) \leq g(a)$, and similarly $f(b) \geq g(b)$. Since $h(a) \leq 0$ and $h(b) \geq 0$, by Bolzano's theorem, $h$ has a zero on $[a,b]$. QED.
- Is my proof correct?
- Is there a better approach?
- Given my approach, can the writing be improved?
Your counterexample is correct. You can generalize it to arbitrary intervals with $$ f(x) = a+b-x \, , $$ then the only solutions to $cf(c) = ab$ are $c=a$ and $c=b$.
Your proof is also correct. You can shorten it slightly by defining $$ h(x) = f(x) - \frac{ab}{x} $$ directly, without the intermediate auxiliary function $g$. Then $h(a) = f(a) - b \le 0$ and $h(b) = f(b) - a \ge 0$, which implies that $h$ has a zero in $[a, b]$.