Exercise 1.1.15

Question

Let

$$
X:=\left\{\left(a_{n}ight)_{n=0}^{\infty}: \sum_{n=0}^{\infty}\left|a_{n}ight|<\inftyight\}
$$

be the space of absolutely convergent sequences. Define the $l^{1}$ and $l^{\infty}$ metrics on this space by

$$
\begin{aligned}
&d_{l^{1}}\left(\left(a_{n}ight)_{n=0}^{\infty},\left(b_{n}ight)_{n=0}^{\infty}ight):=\sum_{n=0}^{\infty}\left|a_{n}-b_{n}ight| \
&d_{l}\left(\left(a_{n}ight)_{n=0}^{\infty},\left(b_{n}ight)_{n=0}^{\infty}ight):=\sup _{n \in \mathbb{N}}\left|a_{n}-b_{n}ight| .
\end{aligned}
$$

Show that these are both metrics on $X$, but show that there exists sequences $x^{(1)}, x^{(2)}, \ldots$ of elements of $X$ (i.e., sequences of sequences) which are convergent with respect to the $d_{l \infty}$ metric but not with respect to the $d_{l^{1}}$ metric. Conversely, show that any sequence which converges in the $d_{l^{1}}$ metric automatically converges in the $d_{l \infty}$ metric.

Prakash Anand · Accepted Answer

\begin{proof} Now let us prove that $d_{l^{1}}$ is a metric on $X$.\par First we have to show that $d_{l^{1}}\left(\left(a_{n} ight)_{n=0}^{\infty},\left(b_{n} ight)_{n=0}^{\infty} ight)<\infty$ (We have to show this since our function $d$ is supposed to map into $[0, \infty)$ ).\par First note that $\left(a_{n} ight)_{n=0}^{\infty},\left(b_{n} ight)_{n=0}^{\infty} \in X$ hence they are bounded (otherwise their series would not converge). So we have $\left|a_{n} ight|,\left|b_{n} ight|0$. Thus, $$d_{l^{1}}\left(\left(a_{n} ight)_{n=0}^{\infty},\left(b_{n} ight)_{n=0}^{\infty} ight)=\sum_{n=0}^{\infty}\left|a_{n}-b_{n} ight|>0 .$$ (c) Since $\left|a_{i}-b_{i} ight|=\left|b_{i}-a_{i} ight|$ for all $i$ we clearly have that $$d_{l^{1}}\left(\left(a_{n} ight)_{n=0}^{\infty},\left(b_{n} ight)_{n=0}^{\infty} ight)=d_{l^{1}}\left(\left(b_{n} ight)_{n=0}^{\infty},\left(a_{n} ight)_{n=0}^{\infty} ight).$$ (d) Let $\left(c_{n} ight)_{n=0}^{\infty} \in X$. By the standard triangle inequality in $\mathbb{R}$ we get the following: $$S_{N}=\sum_{n=1}^{N}\left|a_{n}-c_{n} ight| \leq \sum_{n=1}^{N}\left|a_{n}-b_{n} ight|+\sum_{n=1}^{N}\left|b_{n}-c_{n} ight| \leq \sum_{n=1}^{\infty}\left|a_{n}-b_{n} ight|+\sum_{n=1}^{\infty}\left|b_{n}-c_{n} ight|.$$ Since $S_{N}$ is bounded and increasing we can let $N ightarrow \infty$ and get that: $$\sum_{n=1}^{\infty}\left|a_{n}-c_{n} ight| \leq \sum_{n=1}^{\infty}\left|a_{n}-b_{n} ight|+\sum_{n=1}^{\infty}\left|b_{n}-c_{n} ight|.$$ Thus we have the triangle inequality for $d_{l^{1}}$.\par Now we will show that $d_{l} \infty$ is a metric on $X$. First we will show that $$d_{l} \infty\left(\left(a_{n} ight)_{n=0}^{\infty},\left(b_{n} ight)_{n=0}^{\infty} ight)<\infty.$$ Let $x_{n}=\left|a_{n}-b_{n} ight|$ a sequence that is clearly bounded from below, and since $\left|a_{n}-b_{n} ight| \leq\left|a_{n} ight|+\left|b_{n} ight| \forall n$ also bounded above. Hence the sup $x_{n}$ is also bounded, and thus we have $$d_{l}\left(\left(a_{n} ight)_{n=0}^{\infty},\left(b_{n} ight)_{n=0}^{\infty} ight)<\infty.$$ (a) We have $$d_{l \infty}\left(\left(a_{n} ight)_{n=0}^{\infty},\left(a_{n} ight)_{n=0}^{\infty} ight)=\sup _{n \in \mathbb{N}}\left|a_{n}-a_{n} ight|=\sup _{n \in \mathbb{N}} 0=0$$ (b) If the sequences are different, then there is some $i$ such that $a_{i} eq b_{i}$ and so $$\left|a_{i}-b_{i} ight|>0$. Thus, $d_{l} \infty\left(\left(a_{n} ight)_{n=0}^{\infty},\left(b_{n} ight)_{n=0}^{\infty} ight)=\sup _{n \in \mathbb{N}}\left|a_{n}-b_{n} ight|>0$$ (c) Since $\left|a_{i}-b_{i} ight|=\left|b_{i}-a_{i} ight|$ for all $i$ we clearly have that $$d_{l} \infty\left(\left(a_{n} ight)_{n=0}^{\infty},\left(b_{n} ight)_{n=0}^{\infty} ight)=d_{l \infty}\left(\left(b_{n} ight)_{n=0}^{\infty},\left(a_{n} ight)_{n=0}^{\infty} ight)$$ (d) We know that we have the triangle inequality in $\mathbb{R} \forall n$. That is, $$\left|a_{n}-c_{n} ight| \leq\left|a_{n}-b_{n} ight|+\left|b_{n}-c_{n} ight|$$ Now if we take the sup on the right hand side over $n \in \mathbb{N}$ we get: $$\left|a_{m}-c_{m} ight| \leq \sup \left\{\left|a_{n}-b_{n} ight|+\left|b_{n}-c_{n} ight|: n \in \mathbb{N} ight\} \leq \sup _{n \in \mathbb{N}}\left|a_{n}-b_{n} ight|+\sup _{n \in \mathbb{N}}\left|b_{n}-c_{n} ight|$$ Now take the sup over $m \in \mathbb{N}$ on the left hand side, and we get: $$\sup _{m \in \mathbb{N}}\left|a_{m}-c_{m} ight| \leq \sup _{n \in \mathbb{N}}\left|a_{n}-b_{n} ight|+\sup _{n \in \mathbb{N}}\left|b_{n}-c_{n} ight|$$ Giving us the triangle inequality in $d_{l \infty}$.\par To prove that convergence in $l^{1}$ implies convergence in $l^{\infty}$ note the following. Given $\epsilon>0$ there is $N>0$ such that $$\sum_{n=0}^{\infty}\left|a_{n}-b_{n} ight|<\epsilon \quad \forall n>N.$$ Hence $$\left|a_{n}-b_{n} ight|<\epsilon \quad \forall n.$$ So we get that $\sup \left|a_{n}-b_{n} ight|<\epsilon$. And that is what we wanted. Now we will give an example of a sequence that converges with respect to $d_{l^{\infty}}$, but does not converge with respect to $d_{l^{1}}$. Consider: $$ \begin{aligned} &x^{(1)}=(1,0,0,0,0, \ldots) \ &x^{(2)}=\left(\frac{1}{2}, \frac{1}{3}, 0,0,0, \ldots ight) \ &x^{(3)}=\left(\frac{1}{3}, \frac{1}{4}, \frac{1}{5}, 0,0, \ldots ight) \ &\vdots \ &x^{(k)}=\left(\frac{1}{k}, \frac{1}{k+1}, \ldots, \frac{1}{2 k-1}, 0,0, \ldots ight) \end{aligned} $$ So we have: $$\sup \left|x_{n}^{(k)}-0 ight|=\sup \left|\frac{1}{k} ight|=\frac{1}{k} ightarrow 0$$ as $k ightarrow \infty.$ But, $$\sum_{n=0}^{\infty}\left|x_{n}^{(k)} ight|=\sum_{n=0}^{k-1}\left|\frac{1}{n+k} ight| \geq \frac{k}{2 k}=\frac{1}{2}>0 .$$ Hence the sequence converges in $d_{l^{\infty}}$, but does not converge to zero in $d_{l^{1}}$. \end{proof} \begin{example} Consider the following: $$a_{n}^{(k)}=\frac{k}{k^{2}+n^{2}} \quad k \geq 1, n \geq 0$$ First let us show that $\left\{a_{n}^{(k)} ight\} \in X$. $$ \sum_{n=0}^{\infty}\left|\frac{k}{k^{2}+n^{2}} ight|=\frac{k}{k^{2}} \sum_{n=0}^{\infty}\left|\frac{1}{1+\left(\frac{n}{k} ight)^{2}} ight|=\frac{1}{k} \sum_{n=0}^{\infty} \frac{1}{1+\left(\frac{n}{k} ight)^{2}} $$ Let $f_{k}(x)=\frac{1}{1+\left(\frac{x}{k} ight)^{2}}$. So for all $k, f_{k}(x)$ is positive and decreasing on $[0, \infty)$. Hence we can use the Integral test: $$ I=\int_{0}^{\infty} f_{k}(x)=\int_{0}^{\infty} \frac{1}{1+\left(\frac{x}{k} ight)^{2}} $$ Let us make the substitution $u=\frac{x}{k} \Rightarrow d x=k d u$, and $u(0)=0, u(\infty)=\infty$. So, $$ I=k \int_{0}^{\infty} \frac{1}{1+u^{2}} d u=\left.k \arctan u ight|_{u=0} ^{u=\infty}=k \frac{\pi}{2} $$ Thus by the integral test, the series $\frac{1}{k} \sum_{n=0}^{\infty} \frac{1}{1+\left(\frac{n}{k} ight)^{2}}$ converges. But, $$\frac{1}{k} \sum_{n=0}^{\infty} \frac{1}{1+\left(\frac{n}{k} ight)^{2}}=\sum_{n=0}^{\infty}\left|\frac{k}{k^{2}+n^{2}} ight|.$$ Thus $\left\{a_{n}^{(k)} ight\} \in X$. Now let us show that $d_{l \infty}\left(\left\{a_{n}^{(k)} ight\},\{0\} ight)=0$. Clearly $\frac{k}{k^{2}+n^{2}}$ is decreasing, so we get that $\sup _{n \in \mathbb{N}}\left|\frac{k}{k^{2}+n^{2}} ight|=\frac{1}{k}$. And, $\lim _{k ightarrow \infty} \frac{1}{k}=0$. Hence, $d_{l \infty}\left(\left\{a_{n}^{(k)} ight\},\{0\} ight)=0$. Now let us show that $$d_{l^{1}}\left(\left\{a_{n}^{(k)} ight\},\{0\} ight) eq 0.$$ First note that if $f$ is a decreasing function on $[0, \infty)$ then we have the following inequalities (think upper and lower step functions): $$ \sum_{n=1}^{\infty} f(n) \leq \int_{0}^{\infty} f(x) d x \leq \sum_{n=0}^{\infty} f(n) $$ And since $f_{k}(x)=\frac{1}{1+\left(\frac{x}{k} ight)^{2}}$ is decreasing on $[0, \infty)$ we get that: $$ \int_{0}^{\infty} f_{k}(x) d x=\frac{\pi}{2} $$ Hence, $\sum_{n=0}^{\infty} f(n) \geq \frac{1}{2}$. Thus, $d_{l^{1}}\left(\left\{a_{n}^{(k)} ight\},\{0\} ight) eq 0$. Note: You may have noticed that in both examples we showed that the subsequences did not converge to a specific point with respect to $d_{l^{1}}$, but the exercise says to show that it does not converge at all with respect to $d_{l^{1}}$. However, we were also asked to prove that if a sequence converges to $x$ with respect to $d_{l^{1}}$ then it converges to $x$ with respect to $d_{l^{\infty}}$. So if we know that a subsequence converges to $y$ with respect to $d_{l \infty}$ then if it converges with respect to $d_{l^{1}}$ then it must also converge to $y$ (otherwise what we proved earlier would not be a true statement). \end{example}

Exercise 1.1.15

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Exercise 1.1.15

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