# How to prove $f\equiv 0$ $\forall x\in [a,b]$?$\quad$($f''＋pf'＋qf＝0$ with $f(a)＝f(b)＝0$)

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!

• Maybe it is meant that $f \in C^2[a,b]$ ? – Rebellos Nov 28 '18 at 13:20
• I'm not sure if this approach is useful but I was thinking of taking an inner product with $f$ and see if you run into a contradiction of any kind that way. – Cameron Williams Nov 28 '18 at 13:20
• @Rebellos Thanks for reminding me. I've modified it – Zero Nov 28 '18 at 13:22
• This is a duplicate of math.stackexchange.com/q/3016693. – Paul Frost Nov 28 '18 at 14:14
• – Paul Frost Nov 30 '18 at 11:30

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.
• 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. – Zero Nov 28 '18 at 16:06