upper bound of limsup for one specific subsequence implies the same bound for continuous limsup I was reading a paper about large  deviations and they have some probability $p(\varepsilon)$ which depends on the parameter $\varepsilon >0 $ and they want to take limit when $\varepsilon \to 0$. They prove that for a specific sequence $(\varepsilon_n)$ which decreases to $0$ that $$\limsup_{n \to \infty} \varepsilon_n^{1/2} \log(p(\varepsilon_n)) \leq K$$
where $K$ is some constant. And then they say that since $\varepsilon \mapsto p(\varepsilon) $ is non-increasing then the same bound is obtained for the continuous limsup, i.e. 
$$\limsup_{\varepsilon \to 0} \varepsilon^{1/2} \log(p(\varepsilon)) \leq K$$
I don't know why this is true
Any help will be appreciated 
 A: Here are the details; probably easiest done as a kind of proof by contradiction.
Suppose that the 'continuous' limsup is $\gt K$ and hence is also $\gt K+\delta$ for some $\delta>0$. Then, by the definition of the continuous limsup, it follows that for every $x\gt 0$, $$ \sup_{0\lt \varepsilon\lt x} \varepsilon^{1/2} \log (p(\varepsilon))\gt K+\delta$$ So, in particular, for every $n$ we have $$ S_1 (n):=\sup_{0\lt \varepsilon\leq \varepsilon_n} \varepsilon^{1/2} \log (p(\varepsilon))\gt K+\delta$$ How does this quantity compare to $S_2 (n):=\sup_{m\geq n} \varepsilon_m^{1/2}\log (p(\varepsilon_m))$? They are both sup's of the same function $f(z)=z^{1/2}\log(p(z))$, but the second sup is over a subset of the values used in the first sup. So clearly $S_2(n)\leq S_1(n)$. On the other hand, the fact that $p$ is non-increasing implies that $f$ is also non-increasing, so the value of $f(\varepsilon)$ for $\varepsilon_n\lt\varepsilon\lt\varepsilon_{n-1}$ will be 'sandwiched' between $f(\varepsilon_n)$ and $f(\varepsilon_{n-1})$. It follows that $S_2(n)= S_1(n)$. Hence $\limsup_{n \to \infty} \varepsilon_n^{1/2} \log(p(\varepsilon_n)) \geq K+\delta\gt K$, which contradicts $\limsup_{n \to \infty} \varepsilon_n^{1/2} \log(p(\varepsilon_n)) \leq K$
