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

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?

#### Solutions Collecting From Web of "$f=\underset{+\infty}{\mathcal{O}}\bigr(f''\bigl)$ implies that $f=\underset{+\infty}{\mathcal{O}}\bigr(f'\bigl)$."

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

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.

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 $f=O(f’)$.

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 $f”>0$.
$\square$