Spivak, Ch. 22 "Infinite Sequences", Problem 17b: Understanding the solution manual solution

Question

Below is a problem from Chapter 22 "Infinite Sequences" of Spivak's Calculus. I've previously asked another question regarding an incorrect attempt at solving item $b$ of this problem. Now I am interested in the solution manual solution.

17.(a) Prove that if $\lim\limits_{n\to\infty} [a_{n+1}-a_n]=l$, then $\lim\limits_{n\to\infty} \frac{a_n}{n}=l$.

(b) Suppose that $f$ is continuous and $\lim\limits_{x\to\infty} [f(x+1)-f(x)]=l$. Prove that

$$\lim\limits_{x\to\infty} \frac{f(x)}{x}=l$$

Hint: Let $a_n$ and $b_n$ be the $\inf$ and $\sup$ of $f$ on $[n,n+1]$.

Here is the solution manual solution to part $(b)$

The hypothesis $\lim\limits_{x\to\infty} [f(x+1)-f(x)]=l$ implies that $a_n$ and $b_n$, the $\inf$ and $\sup$ of $f$ on $[n,n+1]$, satisfy $\lim\limits_{n\to\infty} [a_{n+1}-a_n]=l$ and $\lim\limits_{n\to\infty} [b_{n+1}-b_n]=l$.

So by part $(a)$, we have $\lim\limits_{x\to\infty} a_n/n=\lim\limits_{x\to\infty} b_n/n=l$, which implies that $\lim\limits_{x\to\infty} f(x)/x=l$.

How do we conclude

$$\lim\limits_{n\to\infty} [a_{n+1}-a_n]=l$$

and

$$\lim\limits_{n\to\infty} [b_{n+1}-b_n]=l$$

from

$$\lim\limits_{x\to\infty} [f(x+1)-f(x)]=l$$

?

Answer 1

Part(b) follows from part (a).

Here I discussed two solutions: one I obtained by following the link given in Spivak's book (solution 1), and an explanation of the solution in the user's manual described in the OP (solution 2). At the end I also add simple solution to part (a) of the problem in the OP.

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Solution 1: First notice that \begin{align} \frac{f(x)}{x}=\frac{f(x)-f(\lfloor x\rfloor)}{x} +\frac{\lfloor x\rfloor}{x}+ \frac{f(\lfloor x\rfloor)}{\lfloor x\rfloor} \end{align} where $\lfloor x\rfloor$ is the largest less or equal to $x$. Since $\lim_{x\rightarrow\infty}f(x+1)-f(x)=\ell$, $f(\lfloor x\rfloor)/\lfloor x\rfloor \xrightarrow{x\rightarrow\infty}\ell$ by part (a). Thus, it is enough to show that \begin{align} \lim_{x\rightarrow\infty}\frac{f(x)-f(\lfloor x\rfloor)}{x}=0\tag{0}\label{zero} \end{align} Being that that $f$ is bounded in any compact interval, for any $n\in\mathbb{N}$ $\alpha_n=\inf_{x\in[n,n+1]}f(x)$ and $\beta_n=\sup_{x\in[n,n+1]}f(x)$ are finite. Then, $|f(x)-f(\lfloor x\rfloor)|\leq \beta_n-\alpha_n$ for $n=\lfloor x\rfloor$.

Give $\varepsilon>0$, there is $N$ large enough such that $x\geq N$ implies \begin{align} f(x)+\ell-\varepsilon<&f(x+1)<f(x)+\ell+\varepsilon\tag{1}\label{one}\\ f(x+1) -\ell-\varepsilon &< f(x)<f(x+1)-\ell+\varepsilon\tag{2}\label{two}& \end{align} From \eqref{one} and \eqref{two} we obtain \begin{align} \alpha_n+\ell-\varepsilon&\leq \alpha_{n+1}\leq \beta_{n+1}\leq \beta_n+\ell+\varepsilon,\tag{3}\label{three}\\ \alpha_{n+1}-\ell-\varepsilon&\leq \alpha_n\leq \beta_n\leq \beta_{n+1}-\ell+\varepsilon,\tag{4}\label{four} \end{align} for all $n\geq N$ and so, \begin{align} \alpha_n-\beta_n-2\varepsilon\leq \beta_{n+1}-\alpha_{n+1}\leq \beta_n-\alpha_n+2\varepsilon\tag{5}\label{five} \end{align} Iterating this inequality yields

\begin{align} \beta_N-\alpha_N - (n-N+1)2\varepsilon\leq \beta_{n+1}-\alpha_{n+1}\leq (\beta_N-\alpha_N)+ (n-N+1)2\varepsilon\tag{6}\label{six} \end{align} It follows that for all $x\geq N+1$, $$\frac{|f(x)-f(\lfloor x\rfloor)|}{x}\leq \frac{\beta_N-\alpha_N}{x}+\frac{\lfloor x\rfloor -N}{x}2\varepsilon$$ This means that $\limsup_{x\rightarrow\infty}\frac{|f(x)-f(\lfloor x\rfloor)|}{x}\leq2\varepsilon$ for any $\varepsilon>0$, and \eqref{zero} follows.

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Solution 2: As for what is in the solution's manual, notice that from \eqref{three} and \eqref{four} \begin{align} \ell-\varepsilon&\leq \alpha_{n+1}-\alpha_n\leq \ell+\varepsilon\\ \ell-\varepsilon&\leq \beta_{n+1}-\beta_n\leq \ell+\varepsilon \end{align} for all $n\geq N$. This implies that $$\lim_n(\beta_{n+1}-\beta_n)=\ell=\lim_n(\alpha_{n+1}-\alpha_n)$$ By part (a) of the problem $$\lim\frac{\alpha_n}{n}=\ell=\lim_n\frac{\beta_n}{n}$$ wich also implies that $\lim_{x\rightarrow\infty}\frac{f(x)}{x}=\ell$.

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Part(a) follows by taking Cesaro sums: since $a_{n+1}-a_n\xrightarrow{n\rightarrow\infty}\ell$, $\ell=\lim_n\frac1n\sum^{n-1}_{k=0}(a_{k+1}-a_k)=\lim_n\frac{a_n-a_0}{n}=\ell$.

Answer 2

Applying the general property $$\inf_{x\in A} (a(x)+b(x))\ge\inf_{x\in A} a(x)+\inf_{x\in A} b(x)$$ to $A=[n,n+1]$, $a(x)=f(x+1)-f(x)$ and $b=f$, we get: $$a_{n+1}\ge\inf_{x\in[n,n+1]}\left(f(x+1)-f(x)\right)+a_n.$$ Similarly, $$a_n\ge\inf_{x\in[n,n+1]}\left(f(x)-f(x+1)\right)+a_{n+1}.$$ These two inequalities are summarized in $$\inf_{x\in[n,n+1]}\left(f(x+1)-f(x)\right)\le a_{n+1}-a_n\le-\inf_{x\in[n,n+1]}\left(f(x)-f(x+1)\right).$$ From this, using the equality $-\inf_{x\in[n,n+1]}\left(f(x)-f(x+1)\right)=\sup_{x\in[n,n+1]}\left(f(x+1)-f(x)\right)$ and the hypothesis $\lim\limits_{x\to\infty}\left(f(x)-f(x+1)\right)=l$, we conclude: $$\lim\limits_{n\to\infty}\left(a_{n+1}-a_n\right)=l.$$ The similar result for $b_n$ may be proved the same way, or deduced from the result for $a_n$ by changing $f$ to $-f.$

Answer 3

Given that $a_n:=\inf f[n,n+1], b_n=\sup f[n,n+1] .$

Well-definedness: $a_n,b_n$ are well defined since $f$ is continuous and $[n,n+1]$ is a compact set (this is same as closed and bounded in $\mathbb R$).

To prove: $\lim_n (a_{n+1}-a_n)=l=\lim_n (b_{n+1}-b_n),$ which by definition of limit is equivalent to proving that "Given any $\epsilon>0$, there exists $N\in \mathbb N$ such that $n\ge N\implies -\epsilon\lt a_{n+1}-a_n-l\lt \epsilon.$"

To achieve this, first fix an $\epsilon>0$.

Let's break down rest of the solution in steps:-

Step (1): (Choosing large $x$) Since $\lim_{x\to \infty}(f(x+1)-f(x))=l$, there exists $N\in \mathbb N$ such that $x\ge N\implies -\epsilon/2\lt f(x+1)- f(x)-l\lt \epsilon/2.$

Step(2):(Estimating $a_{n+1}-a_n$ from above) Note that for all $n\ge 2N,$

By definition of infimum, there exists $x_n\in [n,n+1]$ such that $a_n+\epsilon/2\gt f(x_n).$ It follows that $$\begin{align} a_{n+1}-a_n-l \le &a_{n+1}-f(x_n)-l+\epsilon/2\\ \le& f(x_n+1)-f(x_n)-l+\epsilon/2\\ \overbrace{\le }^{\text{ By step $(1)$}}& \epsilon/2+\epsilon/2=\epsilon \end{align}$$

Step (3):(Estimating $a_{n+1}-a_n$ from below) Note that for all $n\ge 2N,$

By definition of infimum, there exists $y_n\in [n+1,n+2]$ such that $a_{n+1}+\epsilon/2\gt f(y_n).$ It follows that $$\begin{align} a_{n+1}-a_n-l \ge & f(y_n)-a_n-l-\epsilon/2\\ \ge& f(y_n)-f(y_n-1)-l-\epsilon/2\\ \overbrace{\ge }^{\text{ By step $(1)$ again}}& -\epsilon/2-\epsilon/2=-\epsilon \end{align}$$

It follows by step (2) and (3) that: Given any $\epsilon>0$, there exists $N\in \mathbb N$ such that $n\ge 2N\implies -\epsilon\lt a_{n+1}-a_n-l\lt \epsilon.$ This proves the statement for $(a_{n+1}-a_n)$. The statement is similar for $(b_{n+1}-b_n)$, which I leave up to you.

Answer 4

Perhaps this answer does not fit, because I prefer not to follow Spivak's suggestion.

Let $g(x)=f(x)-lx.$ Then $\displaystyle\lim_{x\to \infty} [g(x+1)-g(x)]=0.$ We are done once we get $$\lim_{x\to \infty}{g(x)\over x}\overset{?}{=}0\quad (*)$$ Fix $\varepsilon >0.$ There exists a number $x_0\ge 1$ such that $$|g(x+1)-g(x)|<\varepsilon,\qquad x+1>x_0$$ For $x>x_0$ let $n_x$ denote the smallest nonnegative integer such that $x-n_x\le x_0.$ Hence $x-n_x+1>x_0.$ Thus $$ n_x<x-x_0+1\le x$$ We have $$\displaylines{g(x)=[g(x)-g(x-1)]+[g(x-1)-g(x-2)]\\ \qquad \qquad \qquad +\ldots +[g(x-n_x+1)-g(x-n_x)]+g(x-n_x)}$$ Therefore $$|g(x)|\le n_x\varepsilon +\max_{0\le t \le x_0}|g(t)|$$ Hence for $\displaystyle c:=\max_{0\le t\le x_0}|g(t)|$ we have $${|g(x)|\over x}\le {n_x\over x}\varepsilon+{1\over x}c\le \varepsilon +{1\over x}c,\quad x>x_0-1$$ Finally $${|g(x)|\over x}<2\varepsilon,\qquad x>x_0-1,\ x>{c\over \varepsilon}$$ This completes the proof of $(*).$