# Limit Supremum - limit of a product

Few days ago I asked about the meaning of $$\limsup _{h\rightarrow \infty}\;\sup_{x\in \mathbb R}\; A(x,h)$$ being finite, and I got the answer. Now I'm facing another situation. Given that $A(x,h)$ and $B(x,h)$ are two positive functions in $x$ and $h$, is it true that

$$\limsup _{h\rightarrow \infty}\;\sup_{x\in \mathbb R}\; \bigg (A(x,h)B(x,h)\bigg )\leq \bigg (\limsup _{h\rightarrow \infty}\;\sup_{x\in \mathbb R}\;A(x,h)\bigg ) \; \bigg (\limsup _{h\rightarrow \infty}\;\sup_{x\in \mathbb R}\;B(x,h)\bigg )$$

I feel it is correct! The same inequality is true for two bounded nonegative real sequences, as I know.

Note: It is known that both of the limits in the RHS exist and nonnegative.

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Why the new account to ask this question? –  Did Aug 31 '11 at 6:03
The question Joan refers to is math.stackexchange.com/questions/60229/… –  Martin Sleziak Aug 31 '11 at 6:05
@Joan: it would help if you consider registering your account. That way the system can better keep track of your questions and answers when you use different computers or when your IP address change. Thanks. –  Willie Wong Aug 31 '11 at 9:53

First, since $\{A(x,h).B(x,h); x\in\mathbb R\} \subseteq \{A(x,h).B(x',h); x,x'\in\mathbb R\}$, you get that $\sup_x A(x,h).B(x,h)\le \sup_x A(x,h).\sup_x B(x,h)$. (Supremum of a subset is less or equal to the supremum of the whole set.)

This yields $$\limsup_{h\to\infty}\sup_x A(x,h).B(x,h)\le \limsup_{h\to\infty}\left(\sup_x A(x,h).\sup_x B(x,h)\right).$$

So it remains to show that $\limsup_{h\to\infty}\left(\sup_x A(x,h).\sup_h B(x,h)\right) \le \limsup_{h\to\infty} \sup_x A(x,h).\limsup_{h\to\infty} \sup_x B(x,h)$ which is a special case of $$\limsup_{h\to\infty} f(h).g(h) \le \limsup_{h\to\infty} f(h). \limsup_{h\to\infty} g(h),$$ which is true for any positive functions $f$ and $h$ with finite limit superior.

Proof: Just notice that for any positive $\varepsilon>0$ there is an $h_0$ such that $$h\ge h_0 \Rightarrow g(h)\le \limsup_{h\to\infty} g(h)+\varepsilon.$$ Thus \begin{align*}\limsup_{h\to\infty} f(h).g(h) &\le \limsup_{h\to\infty} f(h).(\limsup_{h\to\infty} g(h)+\varepsilon) \\ &= \limsup_{h\to\infty} f(h).\limsup_{h\to\infty} g(h)+ \limsup_{h\to\infty} f(h).\varepsilon.\end{align*} Since $\limsup_{h\to\infty} f(h)<+\infty$ and the last inequality holds for any $\varepsilon$, we get that $$\limsup_{h\to\infty} f(h).g(h) \le \limsup_{h\to\infty} f(h). \limsup_{h\to\infty} g(h).$$

Right now, I am not sure about the case when one of the limits superiors is infinite. I hope I did not make a mistake in the above proof.

Note that the proof is very similar to the proof of subadditivity of limit superior: $$\limsup f(h)+g(h)\le \limsup f(h)+\limsup g(h).$$ (It can even be probably deduced from this inequality, if you prefer this approach.)

EDIT: I found in the book Wieslawa J. Kaczor, Maria T. Nowak: Problems in mathematical analysis, Volume 1 as Problem 2.4.17 the following:

Let $(a_n)$, $(b_n)$ be sequences of nonnegative numbers. Prove that (excluding the indeterminate forms of the type $0.(+\infty)$ and $(+\infty).0$) the following inequalities hold: \begin{align*} \liminf a_n \cdot \liminf b_n &\le \liminf (a_n\cdot b_n) \le \\ \liminf a_n \cdot \limsup b_n &\le \limsup (a_n\cdot b_n) \le \limsup a_n \cdot \limsup b_n \end{align*}

So if you have access to that book, you can look up the proof there. (Although I have only seen some parts from this 3 volume set, I can say that I like the selection of the material.)

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