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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.

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    $\begingroup$ It doesn't work for all $n$, $$\sin x + \sin' x + \sin'' x + \sin''' x \equiv 0.$$ $\endgroup$
    – Daniel Fischer
    Mar 25, 2014 at 11:28
  • $\begingroup$ @DanielFischer: Cool Daniel! This counterexample was simple enough to guess but I did not think enough about it. I have updated my post. $\endgroup$
    – Paramanand Singh ♦
    Mar 25, 2014 at 11:35

2 Answers 2

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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 $-1$.

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  • $\begingroup$ This is really amazing. I was not expecting anything based on differential equations. But the argument you have presented is very simple and clear enough. +1 $\endgroup$
    – Paramanand Singh ♦
    Mar 25, 2014 at 13:43
  • $\begingroup$ By the way your solution needs the continuity of $g$ which implies the continuity of $f''$. I will add that in my question. I don't know if this can be avoided but let's assume it to be so. $\endgroup$
    – Paramanand Singh ♦
    Mar 25, 2014 at 13:48
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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 $$ \lim_{t\to\infty}f(t)=\frac{A}{a_0}. $$

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  • $\begingroup$ Is it easy to prove ? $\endgroup$
    – user119228
    Apr 2, 2014 at 20:17
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    $\begingroup$ Factoring $p$, and using Jochen's idea $n$ times. $\endgroup$
    – Yiorgos S. Smyrlis
    Apr 2, 2014 at 20:18
  • $\begingroup$ Thank you, it's a nice result. $\endgroup$
    – user119228
    Apr 2, 2014 at 20:20
  • $\begingroup$ it's not positive real parts ? $\endgroup$
    – user119228
    Apr 2, 2014 at 20:21
  • $\begingroup$ The real parts of the roots are negative. In the original exercises the roots are $-\frac{1}{2}\pm \frac{i\sqrt{3}}{2}$. $\endgroup$
    – Yiorgos S. Smyrlis
    Apr 2, 2014 at 20:23

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