Let $f, g : \mathbb R \to \mathbb R$, with $f$ twice differentiable, and $f(a) = f(b) = 0, a \leq b$. Show that if $f''(x) + f'(x)g(x) - f(x) = 0$, then $f = 0$ on $[a,b]$. (Source: Spivak Calculus.)
My proof is below. I request verification, critique, and suggestions. Note: Proofs are available; this question is to verify my proof.
Since no conditions are specified on $g$, the equation $f''(x) + f'(x)g(x) - f(x) = 0$ simply means that for all $x$, $$[f'(x) = 0] \implies f''(x) = f(x) \tag{*}.$$
We claim that $f([a,b]) \leq 0$. By the Extreme Value Theorem, $f$ has a maximum at $c \in [a,b]$. If $c = a \lor c = b$ we are done. Otherwise, $c$ is a local maximum, so $f'(c) = 0$ and $f''(c) < 0$, and by $(*)$, $f(x) < 0$.
A similar argument shows that $f([a,b]) \geq 0$, completing the proof.
Additional question: Is this result of any significance or application elsewhere?
Thank you for the feedback, which I'm using to rewrite the proof for clarity. Also see my comment that this type of critique and revision, even if it doesn't change the functional equivalence of the proof, is still very important and is not a duplicate.
We claim that $f([a,b]) \leq 0$. By the Extreme Value Theorem, $f$ has a maximum at $c \in [a,b]$. If $c=a \lor c=b$ we are done. Otherwise, $c$ is a local maximum, so $f′(c)=0$ and $f′′(c)<0$.
Since no conditions are specified on $g$, the equation $f′′(x)+f′(x)g(x)−f(x)=0$ simply means that for any $y$ where $f′(y)=0$, then $f′′(y)=f(y)$. Thus, since $f''(c) < 0$, then $f(c) < 0$.
A similar argument shows that $f([a,b]) \geq 0$, completing the proof.