$\limsup_{x \to \infty} f(x) < 0 \implies \exists c,K>0$ s.t. $x > K \Rightarrow f(x) < -c$?

Question

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ and suppose $\limsup_{x \to \infty} f(x) < 0$.

Is is it then true that there exists $c,K>0$ s.t. for $x > K$ we have that $f(x) < -c$?

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My intuition tells me yes since the result holds for if we looking at $\lim$ and ordinary sequences.

I assume we go to the definition of $\limsup$ and write $$ 0>\limsup_{x \to \infty} f(x) = \lim_{y \to \infty} \sup_{x \geq y} f(x) $$ so there must be some step $L>0$ for which $y > L \implies \sup_{x \geq y} f(x) < 0$.

But from this how do we go to $K > 0$ s.t. $f(x) < 0$? I assume we use the Archimedian property of the real numbers to get that it is smaller that $-c$.

Can someone help me?

Answer 1

You are almost there. Remember that $\sup_{x\geq y}f(x)$ is just a number for a given $y$, so with your $y$, we know that $\sup_{x\geq y}f(x)=-c<0$ for some $c>0$, exactly by the Archimedian property. Just pick $K=y$ (of course, you can pick this $y$ positive) and you are done.

Answer 2

$\limsup_{x \to \infty} f(x) < 0$ is given. First suppose $\lim_{y \to \infty} \sup_{x \geq y} f(x) =-\infty.$

Hence, for any $c>0$ you may choose, it is true that there exists $K>0$ such that for $x > K$ we have $f(x) \leq \sup_{x \geq K}f(x) < -c.$

Otherwise let $\lim_{y \to \infty} \sup_{x \geq y} f(x) =-3c \in \mathbb{R}.$

Note $c>0$, and there exists $K>0$ such that for $x > K$ we have $f(x) \leq \sup_{x \geq K}f(x) < -c.$