$f(x)=f^{\prime}(x)+f^{\prime\prime }(x)$ such that $f(a)=f(b)=0$ then prove that $f=0$

Question

$f:[a,b] \to \Bbb R$ twice differentiable function is defined in this way that $f(x)=f^{\prime}(x)+f^{\prime\prime }(x)$ such that $f(a)=f(b)=0$ then prove that $f=0$

I have attempted this problem in this way that $f(a)=f(b)=0$ then by Rolle's theorem $\exists c\in (a,b)$ s.t $f'(c)=0$. Now there are three cases

1. $f"(c)>0$ i.e we have a local minimum then $f(c)=f"(c)>0$ then $\exists \delta>0$ s.t f is decreasing in $(c-\delta,c)$ so there exists a local positive maxima at $d\in (a,c)$. Now at that point $f(d)>0$ but $f"(d)<0$ a contradiction.
2. $f"(c)<0$ similar like case 1.
3. $f"(c)=0$. Here we will get $f(c)=0$ so we will repeat those three cases in $[a,c]$.

Now my questions are:

• Is my proof correct?
• If it is correct, can I make the proof much more rigorous e.g giving a proper proof of this line "so there exist a local positive maxima at $d\in (a,c)$. Now at that point $f(d)>0$ but $f"(d)<0$ a contradiction"? I actually made that statement seeing the picture.
• Can you give me any other proofs?

Answer 1

This is a refined version of your attempt. Try to see how it differs from yours.

You can suppose that $f$ is not constant (otherwise there is nothing to prove).

Let us suppose $f$ takes positive value somewhere in $(a, b) $. Then $f$ takes maximum value at some $c\in(a, b) $ where $f(c) >0,f'(c)=0$ so that $f''(c) >0$ whigh makes $c$ a local minimum. The contradiction completes the proof (as a similar contradiction can be achieved when $f$ takes negative values).

Answer 2

Your proof is fine. To make it more rigorous, for case 3 you can argue that for any $\epsilon \gt 0$, for all $x$, after finite amount of iteration, $\lvert f(x) \rvert \lt \epsilon$ as $f$ is continuous. The amount of iteration is finite is by Heine-Borel theorem.

Alternatively, you may just solve the ODE. The identity polynomial is $$x^2+x-1=0$$ Hence the general solution is $$f(x)=C_1e^{\frac {\sqrt 5 - 1} 2 x}+C_2e^{\frac {-\sqrt 5 - 1} 2 x}$$ By boundary conditions, $$f(x)=0$$.