$f=\underset{+\infty}{\mathcal{O}}\bigr(f''\bigl)$ implies that $f=\underset{+\infty}{\mathcal{O}}\bigr(f'\bigl)$.

Question

Let $f\in\mathcal{C}^2(\Bbb{R},\Bbb{R})$ be a positive function such that $f=\underset{+\infty}{\mathcal{O}}\bigr(f''\bigl)$ does it implies that $f=\underset{+\infty}{\mathcal{O}}\bigr(f'\bigl)$?

First intuitively I would say that because of the first hypothesis $f''$ is positive but I don't know how can I prove this.

More precisely, I want to prove that if there are $N,C$ so that $|f(x)|\leq C|f''(x)|$ for all $x>N$, then there are $M,B$ so that $|f(x)|\leq B|f'(x)|$ for all $x>M$.

Perhaps I can use the taylor expression? $ f(x+h)=f(x)+hf'(x)+\frac{h^2}{2}f''(x)+O(h^3)$ and $f(x-h)=f(x)-hf'(x)+\frac{h^2}{2}f''(x)+O(h^3)$. Then we get that $$f''(x)=\lim_{h\to0} \frac{f(x+h) - 2 f(x) + f(x-h)}{h^{2}}.$$ Any ideas?

Answer 1

Answer: Yes.

Since $\left|\frac{f(x)}{f''(x)}\right|\le k_1^2$ for large $x$, we know that if $f(x)\ne0$, $$ \left|\frac{f''(x)}{f(x)}\right|\ge\frac1{k_1^2}\tag{1} $$ In an interval where $f(x)\ne0$, $\frac{f''(x)}{f(x)}$ is continuous and either always positive or always negative.

If $f(x_0)=0$, then, since $f(x)\ge0$, we must have $f'(x_0)=0$ and $f''(x)\ge0$ in a neighborhood of $x_0$. Thus, $f''(x)$ does not change sign if $f(x)$ should vanish.

Therefore, $\frac{f''(x)}{f(x)}$ is either always positive or always negative where $f(x)\ne0$.

Note that $$ \left(\frac{f'(x)}{f(x)}\right)'=\frac{f''(x)}{f(x)}-\left(\frac{f'(x)}{f(x)}\right)^2\tag{2} $$

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Case $\boldsymbol{1}$: Suppose that $\frac{f''(x)}{f(x)}\le-\frac1{k_1^2}$ for large $x$.

Integrating $(2)$ implies that $$ \frac{f'(x)}{f(x)}\le k_2-\frac{x}{k_1^2}\tag{3} $$ Let $x_0\gt k_1^2k_2$, then $(3)$ implies that $f'(x_0)\lt0$. By assumption, $f''(x)\le0$; therefore, for $x\ge x_0$, we have $f'(x)\le f'(x_0)$. Thus, for $x\gt x_0-\frac{f(x_0)}{f'(x_0)}$, $$ \begin{align} f(x) &\le f(x_0)+f'(x_0)(x-x_0)\\ &\lt f(x_0)-f'(x_0)\frac{f(x_0)}{f'(x_0)}\\ &=0\tag{4} \end{align} $$ which contradicts the positivity of $f(x)$.

Therefore, this case cannot happen.

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Case $\boldsymbol{2}$: Suppose that $\frac{f''(x)}{f(x)}\ge\frac1{k_1^2}$ for large $x$ when $f(x)\ne0$.

Let $u(x)=\frac{f'(x)}{f(x)}$, then $(2)$ implies that $$ u'(x)+u(x)^2\ge\frac1{k_1^2}\tag{5} $$ $(5)$ says that $u(x)$ is increasing if $|u(x)|\lt\frac1{k_1}$. In fact, $$ \int_{-\frac1{2k_1}}^{\frac1{2k_1}}\frac{\mathrm{d}u}{\frac1{k_1^2}-u^2}=k_1\log(3)\tag{6} $$ $(6)$ says that $|u(x)|\le\frac1{2k_1}$ can be true for only a finite span of $x$. Thus, for all sufficiently large $x$, $$ \left|\frac{f(x)}{f'(x)}\right|\lt2k_1\tag{7} $$ which says that $f=O(f')$.

Answer 2

This is a partial answer addressing strict positivity of the functions. The main conclusion is this: either $f$ vanishes identically or we can assume the following for $x>N$ (possibly after increasing $N$):

• $f>0$
• $f''>0$
• $f'$ has constant sign and $\lim_{x\to\infty}f'(x)\in[0,\infty]$ exists

Note that it is possible that $f'<0$, for example with $f(x)=e^{-x}$.

Proposition: The set of zeros of $f$ after $N$ is of $(N,\infty)$ and a closed interval (which may contain zero or one points or may be unbounded).

Proof: Now denote $Z=\{x>N;f(x)=0\}$. Since $f$ is continuous, $Z$ is closed in the subspace topology of $(N,\infty)$. If $Z$ contains at most one point, the claim is true, so let us assume that there are at least two points. It suffices to show that $Z$ is connected.

Suppose $N<a<b$ are such that $f(a)=f(b)=0$. Since $f\geq0$, we know that $f'(a)=f'(b)=0$ and $f''(a)>0$ and $f''(b)>0$. Now $f''$ cannot be positive on $[a,b]$ (otherwise $\int_a^bf''(x)dx>0$ and thus $0=f'(b)>f'(a)=0$) so by continuity of $f''$ there is $c\in(a,b)$ so that $f''(c)=0$. It now follows from the estimate $|f(c)|\leq C|f''(c)|$ that $f(c)=0$.

Now suppose $Z$ was not connected. Then there is a point $w\in (N,\infty)\setminus Z$ with points of $Z$ on both sides. Since $Z$ is closed, we can define $a=\max(N,w)\cap Z$ and $b=\min(w,\infty)\cap Z$. We now have $N<a<b$ with $f(a)=f(b)=0$. By the previous paragraph there is $c\in(a,b)\cap Z$, contradicting $w\notin Z$, maximality of $a$ or minimality of $b$. Therefore $Z$ is indeed connected. $\square$

Corollary: If $f$ has arbitrarily large zeros, then $f$ is eventually identically zero. $\square$

Because of the corollary we can choose a bigger $N$ so that for all $x\geq N$ we have $0<f(x)\leq C|f''(x)|$. (This only excludes the case where $f$ is identically zero, but that is not very interesting.)

It follows from continuity that $f''$ has constant sign. Therefore $f'$ can only change sign once, so it will eventually have constant sign.

Proposition: $f'$ has a positive limit and $f''$ is positive (after big enough $N$).

Proof: Since $f''(x)$ has a constant sign (for $x>N$), we know that $f'$ is monotone and therefore has a limit at infinity (possibly $\pm\infty$). If the limit is negative, then there is $M>N$ and $\epsilon>0$ so that $f'(x)\leq-\epsilon$ for all $x\geq M$. Thus for $x\geq M$ we get $f(x)\leq f(M)-\epsilon(x-M)\to-\infty$ as $x\to\infty$, which contradicts positivity of $f$. Therefore the limit $L=\lim_{x\to\infty}f'(x)$ is positive.

It follows from this that if $f''$ is negative, then $f''(x)\to0$ as $x\to\infty$. By the estimate $f(x)\leq C|f''(x)|$ this means that also $f$ tends to zero at infinity. As we observed above, $f'$ has a constant sign (eventually). Since $f>0$ and $f\to0$, we have $f'<0$. But now $f''<0$ implies that the limit $L$ of $f'$ is negative, a contradiction. Therefore $f''>0$. $\square$