# How to show that $f'(x)<2f(x)$ [closed]

This is question B4 in 1999 Putnam Exam. I would appreciate if somebody could help me with this.

Let $f(x),f'(x),f''(x),f'''(x)>0$ , $f'''(x)$ is a continuous function and $f'''(x)<f(x)$ on $\mathbb{R}$. Then show that $$f'(x)<2f(x),~ \forall x\in \mathbb{R}.$$

• That's a great question, +1. Commented Apr 17, 2013 at 13:04
• It would be even greater if we knew where this question comes from. Commented Apr 17, 2013 at 15:49
• @Martin From now on, we can say it comes from MSE in particular. Commented Apr 17, 2013 at 16:45
• @julien By chance, I ran into this problem today and remembered seeing it on here: Putnam 1999. Commented May 1, 2013 at 3:05
• @L.F. Thanks for the info. I should look at these sometimes. Commented May 1, 2013 at 3:09

From the assumptions we know that $f$, $f'$ and $f''$ are positive and increasing, so $$\lim_{x\to-\infty}f(x)\ge 0, \quad\text{and}\quad\lim_{x\to-\infty}f'(x)=\lim_{x\to-\infty} f''(x)=0.$$ It follows that $$f''(x)=\int_{-\infty}^xf'''(t) \, dt,$$ $$f'(x)=\int_{-\infty}^xf''(t) \, dt=\int_{-\infty}^x \left(\int_{-\infty}^tf'''(s) \, ds \right) \, dt=\int_{-\infty}^x(x-s)f'''(s) \, ds,\tag{1}$$ and $$f(x)\ge\int_{-\infty}^xf'(t) \, dt=\int_{-\infty}^x \left(\int_{-\infty}^t (t-s) f'''(s) \, ds \right) \, dt=\frac{1}{2}\int_{-\infty}^x(x-s)^2f'''(s)\,ds.\tag{2}$$ We will try to compare $f$ and $f'$ by comparing $x-s$ and $(x-s)^2$ in Equation (1) and (2). $(x-s)^2$ exceeds $x-s$ only when the latter exceeds some value. In order to use (2), the integration still needs to run to $x$ instead of somewhere to the left of $x$. We will subtract the "over-integration". We do this by adjusting the critical value $\lambda$ of $x-s$ vs $(x-s)^2$. In fact, it turns out we do not have to choose the optimal critical value for this particular problem, but getting the optimal critical value poses no harm.

For $\lambda>0$ to be determined below, as $x-s<\frac{(x-s)^2}{\lambda}$ for $s\in(-\infty,x-\lambda)$, we have $$\begin{eqnarray*} f'(x) &\le &\lambda^{-1}\int_{-\infty}^{x-\lambda}(x-s)^2f'''(s)ds+\int_{x-\lambda}^x(x-s)f'''(s)ds \tag{by (1)}\\ &\le & 2\lambda^{-1}f(x)+\int_{x-\lambda}^x [(x-s)-\lambda^{-1}(x-s)^2]f'''(s) \, ds\tag{by (2)}\\ &<& 2\lambda^{-1}f(x)+f(x)\int_{x-\lambda}^x[(x-s)-\lambda^{-1}(x-s)^2] \, ds\tag{f'''(s)<f(x)}\\ &=&\left(2\lambda^{-1}+\frac{\lambda^2}{6}\right)f(x). \end{eqnarray*}$$

Since $\min_{\lambda>0}(2\lambda^{-1}+\frac{\lambda^2}{6})=\frac{3}{\sqrt[3]{6}}<2$( for $\lambda=\sqrt[3]{6}$), the conclusion follows.

• Without knowing that $\lim_{x\to-infty}f'''(x)=0$, I am not sure how you can conclude that $f''(x)=\int_{-\infty}^xf'''(t)dt$. Commented Apr 17, 2013 at 16:11
• Never mind, I am being silly. Commented Apr 17, 2013 at 16:23
• Very, very nice, +5. Commented Apr 17, 2013 at 16:49
• @julien: Thank you! Are you interested in the best constant $C>0$ in $f'<Cf$? I have no idea.
– 23rd
Commented Apr 17, 2013 at 17:03
• No idea either. But your estimates seem pretty sharp already. All I know is that it is $\geq 1$ by the trivial examples $f(x)=e^{tx}$, for $0<t<1$. Commented Apr 17, 2013 at 17:08