Limit Supremum - limit of a product

Question

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.

Answer 1

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.)

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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.)