# Show that $f''(x)+g(x)f'(x)-f(x)=0$ has at least one root

Let $f$ be continuous on $[a,b]$, with continuous first and second derivatives on $[a,b]$. Suppose $f$ has at least 3 distinct zeroes in $[a,b]$. Show that $f''(x)+g(x)f'(x)-f(x)=0$ has at least one root in $[a,b]$, where $g$ is any continuous function over $[a,b]$.

Say $f(\alpha)=f(\beta)=f(\gamma)=0;\;\alpha,\beta,\gamma\in(a,b)$ and $\alpha<\beta<\gamma$ since they are distinct.
Hence, by Mean Value Theorem, I know $f'(\zeta_1)=0$ where $\zeta_1\in(\alpha,\beta)$; $f'(\zeta_2)=0$ where $\zeta_2\in(\beta,\gamma)$. And also for $f''(\zeta_3)=0$; where $\zeta_3\in(\zeta_1,\zeta_2)$.

I know I have to use intermediate value theorem, but I have no idea on how to use my result to do that. Thank you.

• Please read a LaTeX manual. You should use the dollar signs for the math only, and not for all the text. – TMM Dec 9 '12 at 14:23
• Please only use $\LaTeX$ to typeset the mathematical parts. (If I somehow changed something meaningful, I apologise, but I do not think anything was lost in the move from $\LaTeX$ to text.) – user642796 Dec 9 '12 at 14:30
• That's ok,Thank you^^ – Vulcan Dec 9 '12 at 14:30
• $g(x)=\frac{f(x)-f''(x)}{f'(x)}$, if $f'(x)\neq 0$. If you choose $g(x):=\sup\{\frac{f(x)-f''(x)}{f'(x)}\}+1$, then $g$ is a constant function, continuous, but there is no solution of your equation. Maybe I misinterpret something? – vesszabo Dec 9 '12 at 15:07
• @vesszabo: the supremum will be $\infty$ – Simon Markett Dec 9 '12 at 15:11

The crucial idea is that $g$ can be any continuous function. For example it can be something like $Cf'$ for $C$ really big. Then the function $f''+gf'-f=f''+C(f')^2-f$ will be dominated by the non-negative summand $C(f')^2$. So it is morally clear that we have to have a careful look a the locus where $f'=0$.
So let $x<y$ be two zeros of $f'$ such that there is precisely one zero of $f$ but no other zeros of $f'$ inbetween; without loss of generality we may assume that $f'>0$ on the interval $(x,y)$. Then $f(x)<0$, $f(y)>0$, $f''(x)>0$ and $f''(y)<0$. Therefore $f''(x)+gf'(x)-f(x)>0$ and $f''(y)+gf'(y)-f(y)<0$. By mean value we are done.