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Let $f$ be a smooth function of one-variable. Suppose that the limit $\lim_{T\rightarrow \infty}\frac{1}{T}\log f(T)$ exists. Now I am trying to see wether or not we will have that if we do a time shift this won't make a difference i.e. if $\lim_{T\rightarrow \infty}\frac{1}{T}\log f(T+c)= \lim_{T\rightarrow \infty}\frac{1}{T}\log f(T)$, where $c$ is a constant.

Intuitively I would think this is true , but I am not being able to prove it.

All that I was able to see is that if $c>0$ then $\lim_{T\rightarrow \infty }\frac{1}{T}\log(f(T+c))=\lim_{T\rightarrow \infty}\frac{1}{T-c}\log(f(t))\geq \lim _{T\rightarrow \infty}\frac{1}{T}\log(f(t))$

Any help is appreciated. Thanks in advance.

Someone
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2 Answers2

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Since $$ \frac{1}{T}\log f(T + c) = \Bigl(1 + \frac{c}{T}\Bigr)\cdot\frac{1}{T + c}\log f(T + c), $$ for all sufficiently large $T$, and $\lim_{T\to\infty}(1 + c/T) = 1,$ it’s enough to show that $$ \lim_{T\to\infty} \frac{1}{T + c}\log f(T + c) = \lim_{T\to\infty} \frac{1}{T}\log f(T). $$

Let $L = \lim_{T\to\infty} \frac{1}{T}\log f(T)$. From the definition of a limit at $\infty$, we have for any $\varepsilon > 0$ a real number $K$ such that $$ T > K \implies \Bigl\lvert \frac{1}{T}\log f(T) - L\Bigr\rvert < \varepsilon. $$

For the same $\varepsilon$, if $T > K - c$, then $T + c > K$. Since $T + c$ satisfies the previous condition, it follows that

$$ \Bigl\lvert\frac{1}{T + c}\log f(T + c) - L\Bigr\rvert < \varepsilon. $$

Hence, the shifted function converges to the same limit.

shoteyes
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Let $$ \phi(T)=\frac{\log f(T)}{T-C} \, . $$ and define $$ \phi(\infty)=\lim_{T \to \infty}\phi(T) \, . $$ Then, $\phi$ is continuous at $\infty$, and so for all $C\in\Bbb{R}$, $$ \lim_{T\to\infty}\phi(T+C)=\phi\left(\lim_{T\to\infty}T+C\right)=\phi(\infty)=\lim_{T\to\infty}\phi(T) \, . $$ See this thread for more details.

Joe
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