MHB Can we prove that $f'(x)<2f(x)$ for all $x$ using the given conditions?

[Ackbach]

Here is this week's POTW:

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Let $f$ be a real function with a continuous third derivative such that $f(x),f'(x), f''(x), f'''(x)$ are positive for all $x$. Suppose that $f'''(x)\leq f(x)$ for all $x$. Show that $f'(x)<2f(x)$ for all $x$.

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[Ackbach]

Re: Problem Of The Week # 257 - Mar 25, 2017

This was Problem B-4 in the 1999 William Lowell Putnam Mathematical Competition.

No one answered this week's POTW. The solution, attributed to Kiran Kedlaya and his associates, follows:

Spoiler

(based on work by Daniel Stronger) We make repeated use of the following fact: if $f$ is a differentiable function on all of $\mathbb{R}$, $\lim_{x \to -\infty} f(x) \geq 0$, and $f'(x) > 0$ for all $x \in \mathbb{R}$, then $f(x) > 0$ for all $x \in \mathbb{R}$. (Proof: if $f(y) < 0$ for some $x$, then $f(x)< f(y)$ for all $x<y$ since $f'>0$, but then $\lim_{x \to -\infty} f(x) \leq f(y) < 0$.)

From the inequality $f'''(x) \leq f(x)$ we obtain
\[
f'' f'''(x) \leq f''(x) f(x) < f''(x) f(x) + f'(x)^2
\]
since $f'(x)$ is positive. Applying the fact to the difference between the right and left sides, we get
\begin{equation}
(1) \qquad \qquad \frac{1}{2} (f''(x))^2 < f(x) f'(x).
\end{equation}

On the other hand, since $f(x)$ and $f'''(x)$ are both positive for all $x$, we have
\[
2f'(x) f''(x) < 2f'(x)f''(x) + 2f(x) f'''(x).
\]
Applying the fact to the difference between the sides yields
\begin{equation}
(2) \qquad \qquad f'(x)^2 \leq 2f(x) f''(x).
\end{equation}
Combining (1) and (2), we obtain
\begin{align*}
\frac{1}{2} \left( \frac{f'(x)^2}{2f(x)} \right)^2
&< \frac{1}{2} (f''(x))^2 \\
&< f(x) f'(x),
\end{align*}
or $(f'(x))^3 < 8 f(x)^3$. We conclude $f'(x) < 2f(x)$, as desired.

Note: one can actually prove the result with a smaller constant in place of 2, as follows. Adding $\frac{1}{2} f'(x) f'''(x)$ to both sides of (1) and again invoking the original bound $f'''(x) \leq f(x)$, we get
\begin{align*}
\frac{1}{2} [f'(x) f'''(x) + (f''(x))^2] &< f(x) f'(x) + \frac{1}{2} f'(x) f'''(x) \\
&\leq \frac{3}{2} f(x) f'(x).
\end{align*}
Applying the fact again, we get
\[
\frac{1}{2} f'(x) f''(x) < \frac{3}{4} f(x)^2.
\]
Multiplying both sides by $f'(x)$ and applying the fact once more, we get
\[
\frac{1}{6} (f'(x))^3 < \frac{1}{4} f(x)^3.
\]
From this we deduce $f'(x) < (3/2)^{1/3} f(x) < 2f(x)$, as desired.

I don't know what the best constant is, except that it is not less than 1 (because $f(x) = e^x$ satisfies the given conditions).