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Let $f\in C^2(a,b)$ such that $f+f'+f''\geq0$. Prove that $f$ has a lower bound.

$My\quad Attempt$

$1.\quad$ Suppose that $f$ has no lower bound at x=b, so there is a sequence$\{x_n\}$ which converges to b($\lim_{n\to\infty}{x_n}=b$), and $f'(x_n)<0,f(x_n)<-n$.

$2.\quad\forall k\in \mathbb N,\exists N\in\mathbb N,n>N,x_n>x_k$, then prove that $$\int_{\{x|x\in(x_k,x_n)\land f(x)>0\}}{}f^2(x)dx\leq C.$$

$3.\quad$ Prove: $$0\leq\int_{x_k}^{x_n}{(f+f'+f'')}dx\leq0+f(x_k)-f(x_n)+C\to-\infty$$ $\quad\quad\quad\quad\quad\quad\quad$ This contradicts the problem

So that's my idea, but I can't do it from step 2. And my idea might be wrong.

Edit in 2019/2/16

I solve the question if $q=0$. There are some Chinese characters in my answer. I hope it doesn't bother you. enter image description here

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2 Answers 2

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Lemma. Given $a<b,$ any $g\in C^2(a,b)$ satisfying $g''+g\geq 0$ is bounded below.

Proof. We can apply (basically) the Picone identity, i.e. the derivative of a Wronskian. Specifically, for $s,t$ satisfying $\max(a,b-\pi)<t<s<b$ define $h_s(t)=g(t)\cos(t-s)-g'(t)\sin(t-s).$ Then

$$h'_s(t)=(g(t)\cos(t-s)-g'(t)\sin(t-s))'=-(g(t)+g''(t))\sin(t-s)\geq 0$$ because $\sin(t-s)<0.$ So $$g(s)=h_s(s)\geq h_s(t)\geq -|g(t)|-|g'(t)|.$$

Picking any $\max(a,b-\pi)<t<b,$ this shows that $g$ is bounded below on $(t,b).$

The function $G\in C^2(a,b)$ defined by $G(x)=g(a+b-x)$ satisfies $$G''(t)+G(t)=g''(a+b-t)+g(a+b-t)\geq 0.$$ So the same argument shows that $G$ is bounded below on $(t,b).$ This means $g$ is bounded below on $(a,a+b-t),$ and we previously showed that $g$ is bounded below on $(t,b).$ By continuity $g$ is bounded below on the whole interval $(a,b).$

Corollary. Given $a<b,$ any $f\in C^2(a,b)$ satisfying $f''+f'+f\geq 0$ is bounded below.

Proof. The transformation $g(t)=e^{t/\sqrt 3}f(\tfrac{2}{\sqrt 3}t)$ gives $$g''(t)+g(t)=e^{t/\sqrt 3}(\tfrac{4}{3}f''(\tfrac{2}{\sqrt 3}t)+\tfrac{4}{3}f'(\tfrac{2}{\sqrt 3}t)+\tfrac{4}{3}f(\tfrac{2}{\sqrt 3}t))\geq 0$$

for $a<\tfrac{2}{\sqrt 3}t<b.$ The function $g$ therefore satisfies $g''+g\geq 0$ on $(\tfrac{\sqrt 3}{2}a,\tfrac{\sqrt 3}{2}b).$ By the lemma, there exists a real number $C$ such that $g(t)\geq C$ for all $t\in (\tfrac{\sqrt 3}{2}a,\tfrac{\sqrt 3}{2}b).$ Therefore

$$f(t)=e^{t/2}g(\tfrac{\sqrt3}2t)\geq e^{a/2}C.$$

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  • $\begingroup$ @Dap Your auxiliary function $h_{s}$ is like a stroke of genius to me. :) Could you give me some references about how to apply this Picone identity? $\endgroup$
    – Eric Yau
    Feb 16, 2019 at 7:48
  • $\begingroup$ @EricYau: this is a variant of the Sturm comparison theorem. The most general statement I know is "New comparison theorems in Riemannian geometry" by Y. Han and coauthors, Theorem 6.1. $\endgroup$
    – Dap
    Feb 16, 2019 at 10:13
  • $\begingroup$ @Dap But for $(a,t)$,does $f$ has a lower bound? $\endgroup$
    – Li Taiji
    Feb 16, 2019 at 12:06
  • $\begingroup$ @Dap the $g(x)$ is bounded and $f(x)$ is bounded are equivalent?But I still don't think your proof can prove $f$ has a lower bound in$ (a,b)? Could you please write more detail? $\endgroup$
    – Li Taiji
    Feb 17, 2019 at 2:31
  • $\begingroup$ @梦里年华似烟花: ok, I've added some detail and made the structure of the argument explicit $\endgroup$
    – Dap
    Feb 18, 2019 at 6:43
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Proof: Let $F(x) = e^{\frac{x}{2}}f(x)$,then

$$\begin{align}\frac{3}{4}F(x)+F''(x) &=\frac{3}{4} e^{\frac{x}{2}}f(x)+e^{\frac{x}{2}}(\frac{1}{4}f(x)+f'(x)+f''(x))\\ &=e^{\frac{x}{2}}(f(x)+f'(x)+f''(x))\\ &\geq0 \end{align}$$

(1) $F(x)<0,\forall x\in(a,b).$

$F''(x)\geq -F(x)\geq 0$, so $F(x)$ is convex function. Picking any $x_0\in(a,b)$,we have

$$F(x)\geq F(x_0)+F'(x_0)(x-x_0),\quad\quad\quad\quad\forall x\in(a,b).$$

(2) $\exists x_0\in(a,b),F(x_0)\geq0$

​ We suppose that there's $n$ zero points in $(a,b)$ at most, and $2\leq n<\infty$.

​ For $0\leq n\leq 1$ , we assume that $F(x)<0 ,x\in(a,x_0),F(x)\geq 0,x\in[x_0,b),$

​ In $(a,x_0)$, we know that $F(x)$ has a lower bound from 1

​ In $[x_0,b)$, $F(x) $ has a lower bound obviously.

So we get $n\geq2$, ​$\exists \alpha,\beta\in(a,b),F(\alpha)=F(\beta)=0$, and

$$F(x)\geq 0, \quad \quad\quad\quad\quad\quad x\in(\alpha,\beta).$$

​ Let $\gamma\in(\alpha,\beta)$ is absolute maximum point of $F(x)$ in $[\alpha,\beta]$, then for $t>\gamma,F''(t)\leq0$.

​ Then $\exists s\in[\gamma,t],F'(s)=0$,

$$F'(x)<0,\quad \quad\quad\quad\quad\quad x\in(s,t)$$

​ So $|F'(x)|^2+\frac{3}{4}|F(x)|^2$ is monotone decreasing in $[s,t]$ ,then

$$\begin{align}F'(x)&\geq-\sqrt{|F'(s)|^2+\frac{3}{4}|F(s)|^2-\frac{3}{4}|F(x)|^2}\\ &=-\frac{3}{4}\sqrt{|F(s)|^2-|F(x)|^2}\\ &\geq-\frac{3}{4}\sqrt{|F(\gamma)|^2-|F(x)|^2},\quad x\in(s,t) \end{align}$$

​ We get,

$$F'(x)\geq-\frac{3}{4}\sqrt{|F(\gamma)|^2-|F(x)|^2},\quad x\in[\gamma,\beta]$$

​ that is, $arcsin\frac{F'(x)}{F'(\gamma)}+\frac{3}{4}x $ is monotone increasing in $[\gamma,\beta]$, then $$ \frac{3\beta}{4}=arcsin\frac{F'(\beta)}{F'(\gamma)}+\frac{3}{4}x\geq arcsin\frac{F'(\gamma)}{F'(\gamma)}+\frac{3}{4}x=\frac{\pi}{2}+\frac{3\gamma}{4}​ $$ ​ which implies $\beta-\alpha\geq\beta-\gamma\geq\frac{2\pi}{3}$, we get $2\leq n<\infty$

​ If $F(x)$ doesn't have a lower bound in $[x_0,b)$, $\exists x_1>x_0,F'(x_1)=0$, and $F(x)<0,x\in[x_1,b)$,

$$F(x)\geq F'(x_1)(x-x_1),\quad \quad\forall x\in[x_1,b).$$

​ We get contradictions from the above formula.So $F(x)$ has a lower bound in$[x_0,b)$.

​ We can prove that $F(x)$ has a lower bound in$(a,x_0]$ by the same way.

​ Obviously, $f(x)$ has a lower pound in $(a,b)$.

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