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Let $f:[0,\infty)\to\mathbb R$ be a function in $C^2$ such that $\lim_{x\to\infty} (f(x)+f'(x)+f''(x)) = a.$ Prove that $\lim_{x\to\infty} f(x)=a$

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marked as duplicate by Paramanand Singh calculus Jul 13 '17 at 4:53

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  • $\begingroup$ Related: math.stackexchange.com/questions/42277/… $\endgroup$ – Li Chun Min Jul 13 '17 at 0:36
  • $\begingroup$ I know Hardy's old problem. but there is some difference.. $\endgroup$ – merow Jul 13 '17 at 0:49
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    $\begingroup$ Yeah. I just realize that. From Bill's answer know that if $f(x)+2f'(x)+f"(x)\to a$ then $f(x)\to a$…but it isn't 2 here $\endgroup$ – Li Chun Min Jul 13 '17 at 0:51
  • $\begingroup$ yes. I think so too.. $\endgroup$ – merow Jul 13 '17 at 0:53
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    $\begingroup$ If you let $h(x) := f''(x) + f'(x) + f(x)$, then I think the solution by variation of parameters is something like $f(x) = -\frac{2}{\sqrt{3}} e^{-x/2} \cos(\frac{\sqrt{3}}{2} x) \int e^{x/2} \sin(\frac{\sqrt{3}}{2} x) h(x) \, dx + \frac{2}{\sqrt{3}} e^{-x/2} \sin(\frac{\sqrt{3}}{2} x) \int e^{x/2} \cos(\frac{\sqrt{3}}{2} x) h(x) \, dx$. $\endgroup$ – Daniel Schepler Jul 13 '17 at 1:28
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Note that with $\alpha = e^{i \pi/3}$ and $\beta = e^{-i \pi/3}$ we have $\alpha \beta = 1$ and $\alpha + \beta = 1$ and , therefore,

$$\tag{1}f(x) + f'(x) + f''(x) = \alpha\beta f(x) + ( \alpha + \beta)f'(x) + f''(x) \\ =\alpha[ \, \beta f(x) + f'(x) + (\beta f(x) + f'(x))' \, ]$$

One can prove the lemma (when the real part of $\gamma$ is positive):

$$\gamma f(x) + f'(x) \to \delta \implies f(x) \to \delta/\gamma$$

To prove the lemma use the Hardy - L'Hospital trick

$$\lim_{x \to \infty}f(x) = \lim_{x \to \infty}\frac{e^{\gamma x}f(x)}{e^{\gamma x}} = \lim_{x \to \infty}\frac{e^{\gamma x}(\gamma f(x) + f'(x))}{\gamma e^{\gamma x}} = \frac{\delta}{\gamma}.$$

Note that at this stage to appy L'Hospital's rule we don't need to assume anything about the existence of the limit of $f(x)$ in the numerator, only that the limit of the denominator is $+\infty.$

Now by (1) and the lemma we have

$$f(x) + f'(x) + f''(x) \to a \implies \beta f(x) + f'(x) \to a/\alpha,$$

and using the lemma again,

$$f(x) \to a/(\alpha \beta) = a$$

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  • $\begingroup$ While this is very smart, I do have reservations regarding the use of L'Hospital's Rule for complex valued functions of a real variable. I believe that the particular usage of the rule in current context can be justified. Also I am not aware of any answer to this question which avoids complex numbers. +1 $\endgroup$ – Paramanand Singh Jul 13 '17 at 4:51
  • $\begingroup$ @ParamanandSingh: Thank you. Well I added that we need real part of $\gamma$ to be positive. $\endgroup$ – RRL Jul 13 '17 at 4:55
  • $\begingroup$ You may have a look at the linked question (I have closed the current question as a duplicate of that). $\endgroup$ – Paramanand Singh Jul 13 '17 at 4:57
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Let $h(x)=f(x)+f'(x)+f''(x)-a$, so $h(x)\to0$ as $x\to\infty$. Let $\omega=e^{2\pi i/3}$ and $$ f_1(x)=\frac1{\omega-\bar\omega}\left(e^{\omega x}\int_0^xe^{-\omega t}h(t)\;dt-e^{\bar\omega x}\int_0^xe^{-\bar\omega t}h(t)\;dt\right). $$ It can be verified that $$ f_1(x)+f_1'(x)+f_1''(x)=h(x). $$ Thus setting $f_2(x)=f(x)-f_1(x)-a$, we have $$ f_2(x)+f_2'(x)+f_2''(x)=0. $$ Then $f_2(x)$ must be a linear combination of $e^{\omega x}$ and $e^{\bar\omega x}$, implying $f_2(x)\to0$ as $x\to\infty$. It remains to show $f_1(x)\to0$ as $x\to\infty$.

Since $h(x)$ is continuous and has a limit as $x\to\infty$, it is bounded. Suppose $|h(x)|<K$ for all $x\geq0$. Consider $\epsilon>0$. Pick $L$ such that $|h(x)|<\epsilon$ for $x>L$. Pick $M$ such that $Ke^{(L-M)/2}<\epsilon$. For $x>M$ we have $$\begin{eqnarray*} |f_1(x)|&\leq&\frac2{|\omega-\bar\omega|}e^{-x/2}\int_0^xe^{t/2}|h(t)|\;dt\\ &=&\frac2{|\omega-\bar\omega|}e^{-x/2}\left( \int_0^Le^{t/2}|h(t)|\;dt+ \int_L^xe^{t/2}|h(t)|\;dt\right)\\ &\leq&\frac2{|\omega-\bar\omega|}e^{-x/2}\left( K\int_0^Le^{t/2}\;dt+ \epsilon\int_L^xe^{t/2}\;dt\right)\\ &\leq&\frac4{|\omega-\bar\omega|}\left( Ke^{(L-M)/2}+\epsilon\right)\\ &\leq&\frac8{|\omega-\bar\omega|}\epsilon. \end{eqnarray*}$$ Since $\epsilon>0$ was arbitrary, $f_1(x)\to0$ as $x\to\infty$ as required.

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Regards @merow . If $ \lim_{x \rightarrow \infty} f(x) $ exists, here is one argument : $ \lim_{x \rightarrow \infty}f(x) = \lim_{x \rightarrow \infty}f(x+h) = \lim_{x \rightarrow \infty}f(x-h) $ for any finite value $h$.

First we have $$ \lim_{x \rightarrow \infty}f(x) + \lim_{x \rightarrow \infty} \left[ f'(x) + f''(x) \right] = a $$ and $$ \lim_{x \rightarrow \infty}f(x) + \lim_{x \rightarrow \infty} \left[ \lim_{h \rightarrow 0} \frac{(f(x+h)-f(x))}{h} + \lim_{h \rightarrow 0} \frac{(f(x+h)-2f(x)+f(x-h))}{h^{2}} \right] = a $$ $$ \lim_{x \rightarrow \infty}f(x) + \lim_{x \rightarrow \infty} \left[ \lim_{h \rightarrow 0} \frac{(hf(x+h)-hf(x))}{h^{2}} + \frac{(f(x+h)-2f(x)+f(x-h))}{h^{2}} \right] = a $$ $$ \lim_{x \rightarrow \infty}f(x) = \lim_{h \rightarrow 0} \left[ a - \frac{(h \lim_{x \rightarrow \infty }f(x+h)-h\lim_{x \rightarrow \infty}f(x) )}{h^{2}} - \frac{(\lim_{x \rightarrow \infty } f(x+h)+\lim_{x \rightarrow \infty }f(x-h) - 2 \lim_{x \rightarrow \infty}f(x) )}{h^{2}} \right] $$ we may also see that $ \lim_{x \rightarrow \infty}f(x) = \lim_{x \rightarrow \infty}f(x+h) = \lim_{x \rightarrow \infty}f(x-h) $, for any finite value $h$. Using this, the numerators on the right side will be $0$, and we get $$ \lim_{x \rightarrow \infty}f(x) = \lim_{h \rightarrow 0} a = a$$ Hope this will be useful, but this can be improved. Thanks.

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    $\begingroup$ You can't assume that $\,\lim_{x \to \infty}f(x)\,$ exists. That's part of what you have to prove. $\endgroup$ – dxiv Jul 13 '17 at 3:04
  • $\begingroup$ I don't think the two limits can always be exchanged like this. Indeed a similar argument would show that $\lim_{x\to\infty}f(x)=0$ implies $\lim_{x\to\infty}f'(x)=0$, but this is false for $f(x)=\frac1x\sin(x^2)$. $\endgroup$ – stewbasic Jul 13 '17 at 3:05

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