prove that $f’(x)e^{\lambda x}$ is increasing if and only if $f’(x)+\lambda f(x)$ is increasing. Where $f\in C^1(0,\infty)$.

Question

prove that $f’(x)e^{\lambda x}$ is increasing if and only if $f’(x)+\lambda f(x)$ is increasing. Where $f\in C^1(0,\infty)$, and $\lambda$ is a real number.

I have tried to prove it by taking $0<x_1<x_2$ to make difference and to control each other. But it seems not that easy, I also have tried to prove it by contradiction, it failed as I need to find a small interval that $f’$ is monotone. But it isn’t always satisfied.

Can anyone help me to figure it out or just give me some intention? Very appreciate it!

Answer 1

Let $p(x)=f'(x)+\lambda f(x)$

Assume $p(x)$ be non-monotonic function

$p'(x)=f''(x)+\lambda f'(x)<0$ for some $x=x_°\in \mathbb{R}$

Hence $$r(x)=e^{\lambda x}p'(x)=e^{\lambda x}f''(x)+e^{\lambda x}\lambda f'(x)<0$$ for some $x=x_°\in \mathbb{R}$

Now consider $q(x)=f'(x)e^{\lambda x}$

$$q'(x)=f''(x)e^{\lambda x}+\lambda f'(x)e^{\lambda x}$$

Now $q'(x)=r(x)<0$ for some $x=x_°\in \mathbb{R}$

It means $q(x)$ is non-monotone if $p(x)$ is non-monotone,Similarly,It can be shown $q(x)$ is monotone if $p(x)$ is monotone

This proves that $q(x)$ is monotone if and only only if $p(x)$ is monotone

Now consider $q(x)$ to be monotonically increasing,This can be similarly proved to be true if and only if $p(x)$ is monotonically increasing,

Answer 2

I have decided to undelete this partial answer (it was a bit incorrect) but I hope the following observations contribute something. We have two statements:

i) $f'(x)e^{\lambda x}$ is increasing

ii) $f'(x) + \lambda f(x)$ is increasing

Case 1: $\lambda >0 $ then i) implies ii). Fix $x_0 >0$, consider $x\geq x_0$, we have $ f'(x)e^{\lambda x} \geq f'(x_0)e^{\lambda x_0} $, hence

$$ f(x) - f(x_0) = \int^{x}_{x_0}f'(u)du \geq f'(x_0) \int^{x}_{x_0} e^{\lambda(x_0-u)}du = \frac{f'(x_0)}{\lambda} \left(1-e^{\lambda(x_0-x)} \right), $$

which implies

$$ f'(x) - f'(x_0) + \lambda (f(x) - f(x_0)) \geq f'(x) - f'(x_0)e^{\lambda(x_0-x)} \geq 0 $$

Case 2: $\lambda <0$ then ii) implies i). For $x \geq x_0$, we have

$$-f'(x) \leq -f'(x_0) + \int^x_{x_0}\lambda f'(u)du, $$ by Gronwall inequality (requires $\lambda < 0$), it holds that

$$ -f'(x) \leq -f'(x_0)e^{-\int^x_{x_0} \lambda du} = -f'(x_0) e^{\lambda(x_0-x)} \implies f'(x)e^{\lambda x} \geq f'(x_0)e^{\lambda x_0}. $$

The cases left are

1. $\lambda <0 $, i) $\implies$ ii)
2. $\lambda >0 $, ii) $\implies$ i)

Answer 3

(This is highly inspired by LNT's answer.)

Part 1: If $F(x) = f'(x)e^{\lambda x}$ is increasing then $G(x) = f'(x) + \lambda f(x)$ is also increasing.

Proof: Assume that $F$ is increasing and let $0 < a < b$. Then $$ f(b)-f(a) = \int_a^b f'(x) \, dx = \int_a^b F(x) e^{-\lambda x} \, dx = F(c) \int_a^b e^{-\lambda x} \, dx $$ for some $c \in (a, b)$, using the general form of the first mean value theorem for definite integrals. It follows that $$ \lambda( f(b)-f(a)) = F(c) \lambda \int_a^b e^{-\lambda x} \, dx = \underbrace{F(c)}_{\ge F(a)} e^{-\lambda a} - \underbrace{F(c)}_{\le F(b)} e^{-\lambda b} \\ \ge F(a) e^{-\lambda a} - F(b) e^{-\lambda b} = f'(a) - f'(b) $$ and therefore $$ G(b) - G(a) = f'(b)-f'(a) + \lambda( f(b)-f(a)) \ge 0 \, , $$ i.e. $G$ is increasing.

Part 2: If $G(x) = f'(x) + \lambda f(x)$ is increasing then $F(x) = f'(x)e^{\lambda x}$ is also increasing.

Proof: This can be concluded from part 1. Assume that $G$ is increasing and define the function $g\in C^1(0,\infty)$ as $g(x) = f(x)e^{\lambda x}$. Then $$ g'(x)e^{-\lambda x} = f'(x) + \lambda f(x) = G(x) $$ is increasing. We can apply part 1 to $g$ and $-\lambda$ and it follows that $$ g'(x) - \lambda g(x) = f'(x) e^{\lambda x} = F(x) $$ is also increasing.

Remark: “Increasing” here means “weakly increasing“ or “nondecreasing.” But the same proof works for strict monotony, i.e. we also have the equivalence

$f'(x)e^{\lambda x}$ is stricly increasing if and only if $ f'(x) + \lambda f(x)$ is stricly increasing.