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Define $f \in C^{2}[a,b]$ satisfying $f''+pf'+qf=0$ with $f(a)=f(b)=0$, where $p\in C^{0}[a,b]$ and $q\in C^{0}[a,b]$ are two functions.

If $q\leq0$, can we prove $f\equiv 0$ $\forall x\in [a,b]$ ?

My try:
If $f\not\equiv 0$, without loss of generality, we assume that the maximum of $f$ on $[a,b]$ is greater than zero, while notating $f(x_0)=\displaystyle\max_{[a,b]} f$.

Then we have $f(x_0) > 0$, $f'(x_0) = 0$ and $f''(x_0) \leq 0$.

I figured out that if we alter the condition $q\leq0$ into $q(x)<0$ there evidently exists contradiction.

But how to analyze further with the condition $q\leq0$? Can we still find contradiction if $q(x_0)=0$ and $f''(x_0)=0$ ?

Any ideas would be highy appreciated!

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The following is based on the classical proof of the maximum principle.

Let $L(f)=f''+p\,f'+q\,f$. If $L(f)>0$, then your argument shows that $f$ must be identically $0$.

But we have $L(f)=0$, not $>0$. What can we do? Take $M>0$ such that $M^2+M\,p(x)+q(x)>0$ for all $x\in[a,b]$ and let $\epsilon>0$. Then $$ L(f+\epsilon\,e^{Mx})=\epsilon\,e^{Mx}(M^2+M\,p(x)+q(x))>0\quad\forall x\in[a,b]. $$ Then $$ \max_{a\le x\le b}(f+\epsilon\,e^{Mx})=\max\bigl(f(a)+\epsilon\,e^{Ma},f(b)+\epsilon\,e^{Mb}\bigr)=\epsilon\,e^{Mb}. $$ Letting $\epsilon\to0$ gives the desired result.

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  • $\begingroup$ Your answer does help! I got lost in finding contradiction by analyzing $f(x_0 +s)$ ($s\to 0$), which is not as brilliant as your answer is. $\endgroup$ – Zero Nov 28 '18 at 16:06
  • $\begingroup$ All I did was an adaptation of the proof of the maximum principle. $\endgroup$ – Julián Aguirre Nov 28 '18 at 17:33

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