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Here's Prob. 5, Chap. 3 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

For any two real sequences $\left\{ a_n \right\}$, $\left\{ b_n \right\}$, prove that $$\lim\sup_{n\to\infty}\left( a_n + b_n \right) \leq \lim\sup_{n\to\infty} a_n + \lim\sup_{n\to\infty} b_n,$$ provided the sum on the right is not of the form $\infty-\infty$.

My effort:

Let us put $$ c_n \colon= a_n + b_n \tag{Definition A} $$ for all $n \in \mathbb{N}$, and let's put $$ \begin{align} a^* & \colon= \lim\sup_{n\to\infty} a_n, \\ b^* & \colon= \lim\sup_{n\to\infty} b_n, \\ c^* & \colon= \lim\sup_{n\to\infty} c_n. \end{align} \tag{Definitions B} $$

We need to show that $$ c^* \leq a^* + b^*. \tag{0} $$

So let's suppose that $$ c^* \not\leq a^* + b^*. $$ Then we have $$ c^* > a^* + b^*, $$ and so $$ c^*- b^* > a^*,$$ and let's take a real number $x$ such that $$ c^*- b^* > x > a^*. \tag{0} $$

Then as $ x > a^*$, so by Theorem 3.17 (b) in Baby Rudin, we can find a natural number $N_1$ such that $$ x \geq a_n \tag{1}$$ for all $n > N_1$.

Now from (0) above, as $$ c^*-x > b^*, $$ so we can find a real number $y$ such that $$c^*-x > y > b^*. \tag{2} $$

Then as $y > b^*$, so again by Theorem 3.17 (b) in Baby Rudin, we can find a natural number $N_2$ such that $$ y \geq b_n \tag{3}$$ for all $n>N_2$.

Now from (2) above we can conclude that $$ c^* > x+y, \tag{4} $$ and hence from (1) and (3) we also have $$ x+y \geq a_n + b_n = c_n, $$ for all $n > \max \left\{\ N_1, N_2 \ \right\}$, which in turn implies that $$ x+y \geq \lim\sup_{n \to \infty} c_n, $$ that is [Please Refer to (Definition A) and (Definitions B) above.], $$ x+y \geq c^*, $$ which, in view of (4) above, gives rise to a contradiction to our choice of $c^*$ as the limit superior of the sequence $\left\{c_n\right\}$.

Is this proof correct? If not, then where is it deficient?

Saaqib Mahmood
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  • http://math.stackexchange.com/questions/408135/prove-limsup-limits-n-to-infty-a-nb-n-le-limsup-limits-n-to-infty – Behrouz Maleki Dec 25 '16 at 08:14
  • http://math.stackexchange.com/questions/515077/prove-that-limsup-n-to-inftya-nb-n-leq-limsup-n-to-inftya-n-lim – Behrouz Maleki Dec 25 '16 at 08:19
  • http://math.stackexchange.com/questions/450697/clarifying-a-proof-of-limsup-a-nb-n-le-limsup-a-n-limsup-b-n – Behrouz Maleki Dec 25 '16 at 08:20
  • You prove that $c^>c_n$ for all $n>\max { N_1, N_2}$. That's not a contradiction. $c^$ is a limit superior, it can be greater than all $c_n$. – positron0802 Dec 25 '16 at 09:01
  • @BehrouzMaleki thank you for sharing these pages; I'd appreciate if you could also go through my proof and comment on it. – Saaqib Mahmood Dec 25 '16 at 10:22
  • @antimateria0802 we have seen that $c_n \leq x+y < c^$ for all $n > \max \left( N_1, N_2 \right)$, and $x+y$ is a fixed real number. So no subsequence of $\left{ c_n \right}$ can have $c^$ as its limit (in the extended real number system). – Saaqib Mahmood Dec 25 '16 at 10:25
  • @SaaqibMahmuud Sorry, you're right. – positron0802 Dec 25 '16 at 10:30

1 Answers1

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An easier solution to this is to directly use the definition of $\limsup$.

Note, $\limsup_n(a_n+b_n)=\inf_n\sup_{k\geq n}(a_k+b_k)\le \inf_n (\sup_{k\geq n}a_k+\sup_{k\geq n}b_k)$

Now let $x_n=\sup_{k\geq n}a_k$ and $y_n=\sup_{k\geq n}b_k$. Then note that both $x_n$ and $y_n$ are decreasing, so $(x_n+y_n)$ is also decreasing. Hence $\inf_n (x_n+y_n)=\lim_n (x_n+y_n)=\lim_n x_n+\lim_n y_n=\inf_n x_n+\inf _n y_n$.

Therefore, $\inf_n (\sup_{k\geq n}a_k+\sup_{k\geq n}b_k)=\inf_n\sup_{k\geq n}a_k+\inf_n\sup_{k\geq n}b_k=\limsup_n a_n+\limsup_n b_n$.

Landon Carter
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  • thank you for your answer. Can you please also go through my proof and see if I've made any errors? Rudin doesn't use the definition of the lim sup that you've used in your proof. – Saaqib Mahmood Dec 25 '16 at 11:49
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    Your proof is correct. You can prove as an exercise the equivalence of the two definitions. – Landon Carter Dec 25 '16 at 12:17