If $f(x) + f'(x) + f''(x) \to A$ as $x \to \infty$, then show that $f(x) \to A$ as $x \to \infty$

This problem is an extension to the simpler problem which deals with $f(x) + f'(x) \to A$ as $x \to \infty$ (see problem 2 on my blog).

If $f$ is twice continuously differentiable in some interval $(a, \infty)$ and $f(x) + f'(x) + f''(x) \to A$ as $x \to \infty$ then show that $f(x) \to A$ as $x \to \infty$.

However, the approach based on considering sign of $f'(x)$ for large $x$ (which applies to the simpler problem in the blog) does not seem to apply here. Any hints on this problem?

I believe that a similar generalization concerning expression $\sum\limits_{k = 0}^{n}f^{(k)}(x) \to A$ is also true, but I don't have a clue to prove the general result.

Solution 1:

You may factorize the polynomial $\lambda^2+\lambda+1 = (\lambda -c)(\lambda-\bar{c})$ (where $c= (1+\sqrt{3}i)/2$ has positive real part) and accordingly the differential operator $D^2+D+id= (D+c\ id)\circ (D+\bar{c}\ id)$. It is then enough to show the following claim: If $c\in\mathbb C$ has positive real part, $g:(a,\infty)\to \mathbb C$ is continuous with $g(x)\to A \in\mathbb C$ then every solution $f$ of $f'(x)+c f(x) =g(x)$ satisfies $f(x)\to A/c$. (Applying this twice you get what you want since $c\ \bar c =|c|^2=1$.)

To prove the claim you just use the formula for the general solution of a linear differential equation: For every $x_0$ the unique solution with given $f(x_0)$ is $$ f(x)= e^{-c(x-x_0)}(f(x_0)+\int_{x_0}^x e^{c(t-x_0)}g(t) dt).$$ If $x_0$ is large enough you can replace $g(t)$ by $A$ and you obtain the desired limit $A/c$ by calculating the integral (to make this precise you need that the real part of $c$ is strictly positive).

This method gives also something for polynomials of higher degree as long as the roots have strictly positive real parts. In Daniel Fisher's example you have a root $-1$.

Solution 2:

The idea in Jochen's answer provides the following:

Generalization. Let $p(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$, and assume that all the roots of $p$ have negative real parts (which implies that $a_0\ne 0$). If $f: (b,\infty)\to\mathbb C$ is $n$ times differentiable and $$ \lim_{t\to\infty} f^{(n)}(t)+a_{n-1}f^{(n-1)}(t)+\cdots+a_0 f(t)=A, $$ then $$ \lim_{t\to\infty}f(t)=\frac{A}{a_0}. $$