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$$ \limsup \left(f(h)+g(h)\right) \leq \limsup f(h)+ \limsup g(h).$$

How can we prove this? Any help would be appreciated.

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What have you tried? –  Nate Eldredge Oct 3 '11 at 0:22
    
$ h \in $ what? –  Mark Oct 3 '11 at 0:35
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Thomas, it is very rare for a thread on math.stackexchange to have no activity for 2 hours, plenty of us are willing to help you as long as you show us some effort and work you have done! –  Ragib Zaman Oct 3 '11 at 2:46
    
To expand on Mark's comment: What is $h$ in your formula? Is this a sequence, so that $h=1,2,3,\ldots$? –  Arturo Magidin Oct 3 '11 at 2:53
    
In general, to prove that lim sup (whatever) is less than or equal to (something), it suffices to prove that for every $\epsilon > 0$, eventually (whatever) is less than (something + $\epsilon$). Here "eventually" means when $h$ is sufficiently large, if you want $h\to\infty$, or else $|h-a|$ is sufficiently small, if you want $h\to a$. –  Greg Martin Oct 3 '11 at 3:53
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2 Answers 2

The general statement $$ A\subset B\Rightarrow\sup\limits_{x\in A}f(x)\le\sup\limits_{x\in B}f(x)\tag{1} $$ seems intuitively obvious because the proof is very simple:

Suppose that $a=\sup\limits_{x\in A}f(x)$, then for any $\epsilon>0$, there is an $x_\epsilon\in A$ so that $f(x_\epsilon)>a-\epsilon$. Since $A\subset B$, $x_\epsilon\in B$ and we have that $\sup\limits_{x\in B}f(x)>a-\epsilon$. Because $\epsilon$ was arbitrary, we get that $$ \sup_{x\in B}f(x)\ge a=\sup_{x\in A}f(x) $$

Let $A_m=\{n\in\mathbb{Z}:n>m\}$, then $\sup\limits_{n>m}f(n)=\sup\limits_{n\in A_m}f(n)$.

Note that $$ \sup_{n\in A_m}f(n)+\sup_{n\in A_m}g(n)=\sup_{(n_1,n_2)\in A_m\times A_m}f(n_1)+g(n_2)\tag{2} $$ and $$ \sup_{n\in A_m}f(n)+g(n)=\sup_{\substack{(n_1,n_2)\in A_m\times A_m\\n_1=n_2}}f(n_1)+g(n_2)\tag{3} $$ and $$ \{(n_1,n_2)\in A_m\times A_m:n_1=n_2\}\subset A_m\times A_m\tag{4} $$ $(1)$ and $(4)$ say that $(3)\le(2)$, that is $$ \sup_{n>m}f(n)+g(n)\le\sup_{n>m}f(n)+\sup_{n>m}g(n)\tag{5} $$ Now take the $\lim\limits_{m\to\infty}$ of both sides of $(5)$ to get $$ \limsup_{n\to\infty}f(n)+g(n)\le\limsup_{n\to\infty}f(n)+\limsup_{n\to\infty}g(n)\tag{6} $$

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This question would definitely need some more clarification from OP. (Does he mean limit superior of sequences or of functions? What is his definition of $\limsup$ and which properties of $\limsup$ does he already know?)

It is very improbable that he gets back to his question after such a long time. So I've decided to post some answer, so that the question is not left unanswered and it can be useful for other people.


If the question is about limit superior of a sequence, i.e. $$\limsup_{n\to\infty} (f(n)+g(n))\le \limsup_{n\to\infty} f(n) + \limsup_{n\to\infty} g(n),$$ where $f,g \colon \mathbb N \to \mathbb R$, then the question was already answered here.

So I'll assume that the question is about the inequality $$\limsup_{x\to a} (f(x)+g(x))\le \limsup_{x\to a} f(x) + \limsup_{x\to a} g(x),$$ where $f,g \colon \mathbb R\to\mathbb R$ and $a\in \mathbb R\cup\{\pm\infty\}$.

Since we know that for any real function $f$ $$\limsup_{x\to a} f(x) = \sup\{ \limsup x_n; (x_n)\text{ is a sequence such that }\lim_{n\to\infty} x_n=a \text{ and } (\forall n\in\mathbb N) x_n\ne a\}$$ we can get this result easily from the subadditivity of limit superior for sequences.

We know that for every such sequence $$\limsup_{n\to\infty} (f(x_n)+g(x_n))\le \limsup_{n\to\infty} f(x_n) + \limsup_{n\to\infty} g(x_n),$$ and if we apply $\sup$ to both sides of this equation, we get the desired result.

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