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I am trying to solve exercise 5, Chapter 3 of Rudin. Basically showing that the limit superior of a sum of sequences is less than or equal to the sum of the limit superiors of the individual sequences, as n tends to infinity.

For any two real sequences $\{A_n\}$, $\{B_n\}$, prove that: $\limsup_{n \rightarrow \infty} (A_n+B_n) \le \limsup_{n\rightarrow\infty}(A_n) + \limsup_{n\rightarrow \infty} (B_n)$

My solution is the following: For all n, An <= s*(a) where s*(a) is the limit superior of An as n-->inf For all n, Bn <= s*(b) where s*(b) is the limit superior of Bn as n-->inf So that An + Bn <= s*(a) + s*(b)

Now suppose that s*(a+b), (the limit superior of An + Bn, as n goes to inf) is greater than s*(a) + s*(b). That means that there is an x, such that s*(a) + s*(b) < x < s*(a+b), where x is a subsequential limit of a subsequence of An+Bn. But that means that for some n, An+Bn > s*(a) + s*(b), which is a contradiction. So that s*(a+b)<= s*(a) + s*(b).

Does that proof make sense? I looked at proofs on the net and they seem a lot more complicated.

Apologies for the formating issues.

JRB
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G Ch
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    I want to help but I don't understand what exercise is saying. Can you state it more precisely? – D_S Jul 02 '17 at 01:22
  • Exercise is the following:

    For any two real sequences {An}, {Bn}, prove that: lim sup (An+Bn) <= lim sup (An) + lim sup (Bn) as n tends to inf. provided that the sum on the right is not of the form +inf -inf

    Thank you.

     – G Ch Jul 02 '17 at 01:23
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    Your proof isn't quite correct. If $A_n$ is a sequence of real numbers, it is not true that $A_m \leq \limsup\limits_n A_n$ for all $m$. For example, $$(2.1, 1, 2.01, 1, 2.001, 1, ...)$$ has $2$ as its limit superior. – D_S Jul 02 '17 at 01:27

2 Answers2

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Define $$C_k=\sup\{A_n:n≥k\}, \text { and } D_k=\sup\{B_n:n≥k\} \text{ (both of them are non-increasing)}$$

Then given any $ k$, $A_n + B_n \le C_k + D_k$, for all $n \ge k$

If we take sup for above inequality, we get ($E_k$ is also non-increasing):

$$E_k=\sup\{A_n+B_n : n \ge k\} \le C_k + D_k \text{, for all }k$$

Thus

$$\lim_{k\rightarrow \infty}E_k\le \lim_{k\rightarrow \infty}C_k + \lim_{k\rightarrow \infty}D_k$$

We are done, since by definition:

$$\limsup_{n\rightarrow \infty}A_n=\lim_{k\rightarrow \infty}C_k,$$ $$\limsup_{n\rightarrow \infty}B_n=\lim_{k\rightarrow \infty}D_k,$$ $$\limsup_{n\rightarrow \infty}(A_n+B_n)=\lim_{k\rightarrow \infty}E_k$$

Jay Zha
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  • Thank you. I was not aware of this definition of limsup. This is easier to conceive than the Rudin definition. – G Ch Jul 02 '17 at 13:37
  • @GCh You are welcome. They are equivalent definitions, because $C_k$, $D_k$, $E_k$ are non-increasing sequences. – Jay Zha Jul 02 '17 at 14:25
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Hint: By definition, $\limsup (a_n)$ is the infimum of the numbers $s_k = \sup \{a_k, a_{k+1}, ... \}$ for $k \geq 1$. Verify and make use of the following facts:

1 . If $(s_n), (t_n) \subseteq \mathbb{R}$ are sequences of real numbers, then $\sup (s_n + t_n) \leq \sup (s_n) + \sup (t_n)$, where $(s_n + t_n)$ is the sequence $(s_1 + t_1, s_2 + t_2, ...)$.

2 . If $(s_n), (t_n) \subseteq \mathbb{R} \cup \{\infty\}$ are sequences, then $\inf(s_n + t_n) \geq \inf s_n + \inf t_n$. If both sequences are nonincreasing ($s_1 \geq s_2 \geq \cdots$ and the same for $t_n$), then this is actually an equality.

D_S
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