# $f+f'+f''\geq0$ implies that $f$ has a lower bound

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.

Lemma. Given $$a any $$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 $$\max(a,b-\pi) define $$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 $$\max(a,b-\pi) this shows that $$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 $$a any $$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 $$a<\tfrac{2}{\sqrt 3}t The function $$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

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

• @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? Feb 16, 2019 at 7:48
• @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.
– Dap
Feb 16, 2019 at 10:13
• @Dap But for $(a,t)$,does $f$ has a lower bound? Feb 16, 2019 at 12:06
• @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? Feb 17, 2019 at 2:31
• @梦里年华似烟花: ok, I've added some detail and made the structure of the argument explicit
– Dap
Feb 18, 2019 at 6:43

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 $$(a,b)$$.