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I know that if $\limsup_{x\to\infty}f(x)<L$ then there exist $x_{0}$ such that $f(x)<L$ for all $x\geq x_{o}$.

Is the following true:

If $\limsup_{x\to\infty}f(x)<L$ then, for any $\epsilon>0$, there exist $x_{0}$ such that $f(x)<L+\epsilon$ for all $x\geq x_{o}$.

(If the above is true, does $x_{o}$ depends on $\epsilon$?) Thanks

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Did you mean this, very similar property? If $\limsup_{x\to\infty} f(x)\le L$ then there is an $x_0$ such that $f(x)<L+\varepsilon$ for $x>x_0$. (The same is, of course, true if $\limsup_{x\to\infty} f(x)=L$; it is part of one of equivalent definitions of limit superior that appear in literature.) This is almost the same as you wrote, but $<$ is replaced by $\le$ or $=$. –  Martin Sleziak Apr 17 '12 at 8:24
    
Yes, this is exactly what I want. DO you have a reference for this theorem! –  Romensky Apr 17 '12 at 15:01
    
In many books this is part of definition of limsup. Could you edit the question and put your definition of limsup there? Or at least the name of the book you're studying or link to the lecture notes you're studying? –  Martin Sleziak Apr 17 '12 at 15:29

3 Answers 3

Yes, it is true and $x_0$ is independent of $\epsilon$ - just pick the same $x_0$ as you have in the first statement you said you know is true. Your second statement is a weaker version of your first.

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Yes, it is true trivially since $L<L+\epsilon$.

A stronger statement is (trivially) true also: there exists an $\epsilon>0$ and an $x_0$ such that $f(x)<L-\epsilon$ for all $x\ge x_0$. To see this, just pick $\epsilon>0$ such that $\limsup\limits_{x\rightarrow\infty} f(x)<L-\epsilon$ and apply your first statement to $L-\epsilon$.

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Based on this comment is seems that OP had a slightly different question in mind. (But he did not edited the question. And he did not answered, what definition of the limit superior he is using.) Basically, without knowing what is his definition of $\limsup$ we should be working with, the question cannot be answered. Anyway, I'll mention here several possibilities how $\limsup$ can be defined. I'll also discuss (very briefly) why these definitions are equivalent. Hopefully, this will be helpful for the OP.


Definition 1. Let $S\in\mathbb R$. We say that $S=\limsup_{n\to\infty} f(x)$ if the following two conditions are fulfilled:

  • For each $\varepsilon>0$ there exists $x_0$ such that $x>x_0 \Rightarrow f(x)<S+\varepsilon$. $$(\forall\varepsilon>0)(\exists x_0)(x>x_0 \Rightarrow f(x)<S+\varepsilon).$$

  • For each $\varepsilon>0$ and $M$ there exists $x>M$ such that $f(x)>S-\varepsilon$. $$(\forall\varepsilon>0)(\forall M)(\exists x>M) f(x)>S-\varepsilon.$$

Of course, limit superior can also be equal to $\pm\infty$. Definition has to be appropriately modified for these cases; I'll omit this for the sake of simplicity. (I have also ignored the question why such $S$ exists.)

Definition 2. $S=\sup\{L; \text{ there exists a sequence }(x_n)\text{ such that }x_n\to\infty, f(x_n)\to L\}$

Definition 3. $S=\lim\limits_{N\to\infty} \sup\{f(x); x\ge N\}$

All these definition can be modified to get $\limsup_{x\to x_0} f(x)$ instead of $\limsup_{x\to\infty} f(x)$.

Let us show that these definitions are equivalent. I.e., suppose that $S_1$, $S_2$, $S_3$ are the values given by the above definitions. We will assume that $f(x)$ is bounded (for the sake of simplicity), which implies that $S_{1,2,3}$ are finite.

It is easy to see that $\boxed{S_2\le S_3}$, since if $x_n\to\infty$ then $x_n>N$ for large $n$'s and $f(x_n) \le \sup\{f(x); x\ge N\}$.

Using definition 1 we can inductively construct sequence $x_n$ such that $x_n\to\infty$ and $|f(x_n)-S_1|<\frac1n$ (by choosing $\varepsilon=\frac1n$); which implies that $x_n\to S_1$. This implies $\boxed{S_1\le S_2}$.

If $N$ is large enough then, from definition 1, we have $\sup\{f(x); x\ge N\}\le S_1+\varepsilon$. This implies that $S_3\le S_1+\varepsilon$. Since this is true for any $\varepsilon>0$, we get $\boxed{S_3\le S_1}$.

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