September | 2021 | Tao Analysis Solutions

This post was written by Berke Özgür Arslan, who has generously offered to help Issa complete the blog. It has been edited and converted into the WordPress blog post format by Issa.

Exercise statement

Prove part (c) of Proposition 6.4.12. (Since this is a long exercise, Issa has decided to split it up across different blog posts.)

Proposition 6.4.12. Let $(a_{n})_{n=m}^{\infty}$ be a sequence of real numbers, let $L^+$ be the limit superior of this sequence, and let $L^-$ be the limit inferior of this sequence (thus both $L^+$ and $L^-$ are extended real numbers).

(a) For every $x > L^+$ , there exists an $N \geq m$ such that $a_n < x$ for all $n \geq N$ . (In other words, for every $x > L^+$ , the elements of the sequence $(a_{n})_{n=m}^{\infty}$ are eventually less than $x$ .) Similarly, for every $y < L^-$ there exists an $N \geq m$ such that $a_n > y$ for all $n \geq N$ .

(b) For every $x < L^+$ , and every $N \geq m$ , there exists an $n \geq N$ such that $a_n > x$ . (In other words, for every $x < L^+$ , the elements of the sequence $(a_{n})_{n=m}^{\infty}$ exceed $x$ infinitely often.) Similarly, for every $y > L^-$ and every $N \geq m$ , there exists an $n \geq N$ such that $a_n < y$ .

(d) If $c$ is any limit point of $(a_{n})_{n=m}^{\infty}$ , then we have $L^- \leq c \leq L^+$ .

(e) If $L^+$ is finite, then it is a limit point of $(a_{n})_{n=m}^{\infty}$ . Similarly, if $L^-$ is finite, then it is a limit point of $(a_{n})_{n=m}^{\infty}$ .

(f) Let $c$ be a real number. If $(a_{n})_{n=m}^{\infty}$ converges to $c$ , then we must have $L^+ = L^- = c$ . Conversely, if $L^+ = L^- = c$ , then $(a_{n})_{n=m}^{\infty}$ converges to $c$ .

Hints

This is a tricky exercise and there are a few ways to approach it. One way is to split by cases based on whether $L^+$ and $L^-$ are finite, $+\infty$ , or $-\infty$ (it should also be possible to split by cases depending on whether the original sequence $(a_n)_{n=m}^\infty$ is bounded or not). For this approach, you may want to decide whether the sequences $(a_N^+)_{N=m}^{\infty}$ and $(a_N^-)_{N=m}^{\infty}$ are increasing or decreasing, then to compare their terms, and then to use Proposition 6.3.8 followed by a version of Corollary 5.4.10 for sequences of real numbers. For the cases in which $L^+$ and $L^-$ may not be finite, use part (a) of the proposition.

How to think about the exercise

The concepts of limit superior and limit inferior may be difficult to grasp when one encounters their definitions for the first time. A very helpful picture to keep in mind is the piston analogy that Tao gives before the statement of Proposition 6.4.12. As we keep removing elements $a_1$ , then $a_2$ , and so on, the supremum piston can only slip leftward. This suggests that the sequence $a_1^+, a_2^+, a_3^+, \ldots$ is a decreasing sequence. One of the few results we have so far about decreasing sequences is Proposition 6.3.8 , so it is a good bet that we will want to make use of that result.

So the first thing to try in order to tackle this proof is to treat the sequences $(a_{N}^+)_{N=m}^{\infty}$ and $(a_{N}^-)_{N=m}^{\infty}$ like we would treat any regular sequence. We are dealing here with the infimum and the supremum of these sequences respectively; so it would be nice to use Proposition 6.3.8 and convert the infimum and supremum of these sequences into their limits. But this is only possible when $(a_{N}^+)_{N=m}^{\infty}$ is decreasing and bounded below, and $(a_{N}^-)_{N=m}^{\infty}$ is increasing and bounded above. For the boundedness, luckily, the cases in which these sequences are not bounded correspond to the cases in which $L^+$ and $L^-$ are either $+\infty$ or $-\infty$ , so we can treat these cases separately, and it is seen easily that these cases do not violate the proposition.

In order to see whether $(a_{N}^-)_{N=m}^{\infty}$ is increasing or not, we need to refer to its definition. For some number $M$ , we know that $a_M^-$ is the infimum of the terms $a_M, a_{M+1}, a_{M+2}, \ldots$ starting with the index $M$ . Therefore, informally, we can say that the infimum of a sequence starting from a smaller index, say, 5, cannot be greater than of a sequence starting from a greater index, say, 10. This is because the sequence starting from index 5 contains the sequence starting from index 10, and it is “easier” to find an even smaller infimum when we have more elements to choose from. (If that was difficult to follow, you can also consider a piston moving to the right and slipping rightward.) Then we must conclude that the sequence of infima is an increasing sequence. You should figure out an analogous argument for the sequence of suprema. Then we just use Proposition 6.3.8 to conclude that limsup and liminf are just the limits of these sequences. (This also justifies the notation $\limsup$ and $\liminf$ : $\limsup$ is the limit of the suprema, i.e. $\displaystyle \limsup_{n\to\infty} a_n = \lim_{n\to \infty} \sup(a_N)_{N=n}^\infty$ , and similarly for $\liminf$ .)

After this, we need to make another important observation. If we pick yet again any number $M$ , then $a_M^-$ is the infimum of the sequence $(a_n)_{n=M}^{\infty}$ , and $a_M^+$ is its supremum. Clearly, we have $a_M^- \leq a_M^+$ , and by the analogue of Corollary 5.4.10 for real sequences this means that their limits will behave accordingly; and we get the desired result.

While considering the cases where $L^+$ and $L^-$ may not be finite, the following table might help you to keep track of what you need to show. Check marks indicate that their corresponding cases already satisfy the proposition because $+\infty$ is always larger than any extended real number, and $-\infty$ is smaller than any extended real number, so there is nothing to show. In the proof, we show that the cases indicated with a question mark cannot exist. Try to see why the assertions (1) “If $L^+ \in \mathbf{R}$ , then $L^- \neq +\infty$ ”, and (2) “If $L^+ = -\infty$ , then $L^- = -\infty$ as well” exclude these cases.

	$L^-$
$L^+$		$\mathbf R$	$+\infty$	$-\infty$
	$\mathbf R$	shown	?	✓
	$+\infty$	✓	✓	✓
	$-\infty$	?	?	✓

Model solution 1

Consider first the inequality $L^+ \leq \sup(a_{n})_{n=m}^{\infty}$ . Now, $L^+$ is defined to be the infimum of the sequence $(a_{N}^+)_{N=m}^{\infty}$ where $a_{N}^+ := \sup(a_{n})_{n=N}^{\infty}$ , so it should be smaller than any particular element of this sequence. Therefore $L^+ \leq a_{N}^+$ for all $N \geq m$ , and in particular for $N=m$ , we have $L^+ \leq a_{m}^+ = \sup(a_{n})_{n=m}^{\infty}$ . A similar argument gives the inequality $\inf(a_{n})_{n=m}^{\infty} \leq L^-$ . Thus we only need to show that $L^- \leq L^+$ to finish part (c).

We first assume that both $L^+$ and $L^-$ are real numbers, i.e. neither of them are $+ \infty$ or $- \infty$ . In this case, we first want to show that the sequences $(a_N^-)_{N=m}^{\infty}$ and $(a_N^+)_{N=m}^{\infty}$ are sequences of real numbers, i.e. that none of the elements are $+\infty$ or $-\infty$ . Consider first the sequence $(a_N^+)_{N=m}^{\infty}$ . We show each of the possibilities that an element is $-\infty$ or $+\infty$ leads to a contradiction:

If some element $a_M^+ = -\infty$ , then since $L^+$ is the infimum of this sequence we must have $L^+ \leq a_M^+ = -\infty$ , a contradiction of Definition 6.2.3 (a real number cannot be less than or equal to $-\infty$ ).
Suppose some element $a_M^+ = +\infty$ . We first show that the original sequence $(a_n)_{n=m}^\infty$ is not bounded above. If there were some upper bound $K \in \mathbf R$ for the whole sequence, then this $K$ would also be an upper bound for the sequence $(a_n)_{n=M}^\infty$ starting at $n=M$ . Since $a^+_M$ is the least upper bound of this sequence and $K$ is merely an upper bound, we have $+\infty = a^+_M \leq K$ , a contradiction of the fact that $K$ is a real number. Thus the original sequence $(a_n)_{n=m}^\infty$ is not bounded above. But this means that $(a^+_N)_{N=m}^\infty = (+\infty, +\infty, +\infty, \ldots)$ , so the infimum of this sequence is $L^+ = +\infty$ , a contradiction.

A similar argument shows that $(a_N^-)_{N=m}^{\infty}$ is a sequence of real numbers.

Now let us be reminded that if $A$ and $B$ are sets of real numbers such that $A \subseteq B$ , then $\sup (A) \leq \sup (B)$ , and $\inf (B) \leq \inf (A)$ . Now, for some $N \geq m$ , by definition, we have $a_N^- = \inf (a_n)_{n=N}^{\infty} := \inf \{ a_n : n \geq N \}$ (and a similar definition for the supremum). Thus if $M \geq N$ , then $\{ a_n : n \geq M \} \subseteq \{ a_n : n \geq N \}$ , and we must have $a_M^- \geq a_N^-$ . Thus the sequence $(a_N^-)_{N=m}^{\infty}$ is an increasing sequence. By a similar argument, we see that $(a_N^+)_{N=m}^{\infty}$ is a decreasing sequence.

Since $L^- = \sup (a_N^-)_{N=m}^{\infty}$ is a real number, the increasing sequence $(a_N^-)_{N=m}^{\infty}$ is bounded above, and since every increasing sequence which is bounded above is convergent by Proposition 6.3.8, we have $\lim \limits_{N \to \infty} a_N^- = \sup (a_N^-)_{N=m}^{\infty} = L^-$ . With a similar argument for decreasing sequences which are bounded below we also have $\lim \limits_{N \to \infty} a_N^+ = L^+$ . (Note: it is the application of Proposition 6.3.8 that required the sequences $(a_N^-)_{N=m}^{\infty}$ and $(a_N^+)_{N=m}^{\infty}$ to consist solely of real numbers.)

Now let $N \geq m$ be any number. The infimum of $a_N, a_{N+1}, a_{N+2}, \ldots$ is smaller than $a_N$ , and the supremum of the same sequence is larger than $a_N$ , so we have:

$a_N^- = \inf (a_{n})_{n=N}^{\infty} \leq a_N \leq \sup (a_{n})_{n=N}^{\infty} = a_N^+$

In other words, we have the term-wise comparison $a_N^- \leq a_N^+$ . Since this is true for every $N \geq m$ , we have by the analogue of Corollary 5.4.10 for real sequences that $\lim_{N\to\infty} a_N^- \leq \lim_{N\to\infty} a_N^+$ . To summarize, we have:

$L^- = \liminf \limits_{n \to \infty} a_n = \lim \limits_{N \to \infty} a_N^- \leq \lim \limits_{N \to \infty} a_N^+ = \limsup \limits_{n \to \infty} a_n = L^+$

as desired.

Now, we consider the cases in which $L^+$ and $L^-$ are possibly $+\infty$ or $-\infty$ . If $L^+ = +\infty$ or if $L^- = -\infty$ , there is nothing to show. Thus we only need to show two things: (1) if $L^+ \in \mathbf{R}$ , then $L^- \neq +\infty$ , and (2) if $L^+ = -\infty$ , then $L^- = -\infty$ as well. For (1), suppose for the sake of contradiction that $L^+ \in \mathbf{R}$ , and $L^- = +\infty$ . Since $L^- = +\infty$ , then by part (a) of the proposition, for every number $y \in \mathbf{R}$ , we have that $y < +\infty = L^-$ , so there exists an $M \geq m$ such that $a_n > y$ for all $n \geq M$ . This means that the sequence $(a_n)_{n=m}^{\infty}$ doesn’t have an upper bound. Then for each $N \geq m$ , we must have $a_N^+ = \sup (a_n)_{n=N}^{\infty} = +\infty$ . But then we have $\inf(a_N^+)_{N=m}^{\infty} = L^+ = +\infty$ , a contradiction. Similarly, for (2), if $L^+ = -\infty$ , then again by (a), for every number $x \in \mathbf{R}$ , since $x > -\infty = L^+$ , there exists an $M \geq m$ such that $a_n < x$ for all $n \geq M$ , meaning that the sequence the sequence $(a_n)_{n=m}^{\infty}$ doesn’t have an lower bound, i.e. for each $N \geq m$ , we have $a_N^- = \inf (a_n)_{n=N}^{\infty} = -\infty$ . And then $\sup(a_N^-)_{N=m}^{\infty} = L^- = -\infty$ , as desired.

Model solution 2

(This section was written by Issa.)

There is another, shorter proof of part (c) that does not require breaking up the proof into cases. This proof is very neat, and I encourage you to check it out, but I also believe it is more difficult for beginners to discover. I may write it up here in detail later, but for now please check out this writeup by Sangchul Lee; it is Exercise 8(iii) on page 5.

In this post, I will state an analogue of Corollary 5.4.10 for sequences of real numbers. This result easily follows in the book using Lemma 6.4.13 together with Proposition 6.4.12(f), but having this result is handy for proving parts of Proposition 6.4.12, so in order to avoid circularity/anachronisms, we will prove it here using our bare hands.

Exercise statement

Lemma. Let $(a_n)_{n=m}^\infty$ and $(b_n)_{n=m}^\infty$ be convergent sequences of real numbers such that $a_n \geq b_n$ for all $n \geq m$ . Then $\lim_{n\to\infty} a_n \geq \lim_{n\to\infty} b_n$ .

Model solution 1

Let us write $x:=\lim_{n\to\infty} a_n$ and $y:=\lim_{n\to\infty} b_n$ . We are trying to show that $x \geq y$ . Suppose for the sake of contradiction that $x < y$ . Then $y-x > 0$ so for $\varepsilon := (y-x)/3 > 0$ we can find some $N_1 \geq m$ such that $|a_n - x| \leq \varepsilon$ for all $n \geq N_1$ , and some $N_2 \geq m$ such that $|b_n - y| \leq \varepsilon$ for all $n \geq N_2$ . Thus if we pick $N := \max(N_1, N_2)$ then since $N \geq N_1$ and $N \geq N_2$ we have both $|a_N - x| \leq \varepsilon$ and $|b_N - y| \leq \varepsilon$ . These two inequalities say in particular that $-\varepsilon \leq x-a_N$ and $-\varepsilon \leq b_N - y$ . Thus we have

$\begin{aligned} b_N - a_N &= (b_N - y) + (y - x) + (x - a_N) \\ &\geq -\varepsilon + 3\varepsilon - \varepsilon \\ & = \varepsilon \\ &> 0\end{aligned}$

This means that $b_N > a_N$ , which contradicts the fact that $a_n \geq b_n$ for all $n \geq m$ . Thus we conclude that $x \geq y$ after all.

Model solution 2

The following solution is due to Berke Özgür Arslan.

In this proof, we mimic the proof of Corollary 5.4.10 by using the limit laws, in particular Theorem 6.1.19(d). We want to show that $\lim_{n\to\infty} a_n \geq \lim_{n\to\infty} b_n$ , but this is the same thing as showing $\lim_{n\to\infty} a_n - \lim_{n\to\infty} b_n \geq 0$ . By the limit laws, $\lim_{n\to\infty} a_n - \lim_{n\to\infty} b_n = \lim_{n\to\infty} (a_n - b_n)$ , so it suffices to show that $\lim_{n\to\infty} (a_n - b_n) \geq 0$ .

Thus our goal now is to consider the sequence $(a_n - b_n)_{n=m}^\infty$ , and show that its limit is non-negative. To simplify the exposition, let us define $c_n := a_n - b_n$ . Then since $a_n \geq b_n$ for each $n \geq m$ , we have that $c_n = a_n - b_n \geq 0$ for each $n \geq m$ .

Using our new notation of $c_n$ , our goal is the following: we want to show that if $(c_n)_{n=m}^\infty$ is a sequence such that $c_n \geq 0$ for all $n\geq m$ , then $\lim_{n\to\infty} c_n \geq 0$ .

To show this we prove by contradiction. We suppose $L := \lim_{n\to\infty} c_n < 0$ . Then by the definition of sequence convergence, for any $\varepsilon > 0$ , there exists $N \geq m$ such that $L - \varepsilon \leq c_n \leq L + \varepsilon$ for each $n \geq N$ . But if we let $\varepsilon := - L / 2 > 0$ , then we could find some $n$ such that $3L / 2 \leq c_n \leq L / 2 < 0$ . But since $c_n \geq 0$ for all $n \geq m$ , this is a contradiction. Thus we have $\lim_{n\to\infty} c_n \geq 0$ .