1
$\begingroup$

Hello I have problems with exercise

Let $M$ be the space of all real sequences. Given $x = (x_i)_i \; , y = (y_i)_i \in M$, define

$d(x,y) = \displaystyle\sum_{i=1}^\infty{\displaystyle\frac{1}{i^2} \; min \{ |x_i-y_i| , 1 \}} $

Prove that $(M, d)$ is a metric space. Let $ \{ x^{(k)} \}_k$ be a sequence in $M$. Prove that $x^{(k)} \rightarrow x$ for $d$ if and only if ${x_i}^{(k)} \rightarrow x_i $ for all $i \in \mathbb{N}.$

My attempt

$(i) \; d(x,x)=0$

$d(x,x)= \displaystyle\sum_{i=1}^\infty{\displaystyle\frac{1}{i^2} \; min \{ |x_i-x_i| , 1 \}} = \displaystyle\sum_{i=1}^\infty{\displaystyle\frac{1}{i^2} \; min \{ |0| , 1 \}} = \displaystyle\sum_{i=1}^\infty{\displaystyle\frac{1}{i^2}} \cdot 0 = 0 $

$(ii) \; If \; x \neq y $ then $d(x,y) > 0$

$d(x,y) = \displaystyle\sum_{i=1}^\infty{\displaystyle\frac{1}{i^2} \; min \{ |x_i-y_i| , 1 \}}$ (This expression is always greater than zero)

$(iii) \; d(x,y)=d(y,x)$

$d(x,y) = \displaystyle\sum_{i=1}^\infty{\displaystyle\frac{1}{i^2} \; min \{ |x_i-y_i| , 1 \}} = \displaystyle\sum_{i=1}^\infty{\displaystyle\frac{1}{i^2} \; min \{ |y_i-x_i| , 1 \}} = d(y,x)$

$(iv) \; d(x,z) \leq{} d(x,y) + d(y,z) $ (I have problems with triangular inequality)

I don't know how to prove this part:

Prove that $x^{(k)} \rightarrow x$ for $d$ if and only if ${x_i}^{(k)} \rightarrow x_i $ for all $i \in \mathbb{N}.$

Thanks

$\endgroup$
1
1
$\begingroup$

Concerning the triangular inequality, note that, for each $i\in\Bbb N$,\begin{align}\min\{|x_i-z_i|,1\}&\leqslant\min\{|x_i-y_i|+|y_i-z_i|,1\}\\&\leqslant\min\{|x_i-y_i|,1\}+\min\{|y_i-z_i|,1\};\end{align}see this question.

If $\lim_{k\to\infty}x^{(k)}=x$ and if $i\in\Bbb N$, take $\varepsilon>0$. Then, if $k$ is large enough, $d\left(x^{(k)},x\right)<\frac\varepsilon{i^2}$. In particular, $\frac1{i^2}\min\left\{\left|x_i^{(k)}-x_i\right|,1\right\}<\frac\varepsilon{i^2}$, and therefore $\min\left\{\left|x_i^{(k)}-x_i\right|,1\right\}<\varepsilon$, which implies that $\left|x_i^{(k)}-x_i\right|<\varepsilon$.

Finally, if $\lim_{i\to\infty}x_i^{(k)}=x_i$ for each $i\in\Bbb N$ and if $\varepsilon>0$, take $M\in\Bbb N$ such that $\sum_{i=M+1}^\infty\frac1{i^2}<\frac\varepsilon2$. For each $i\leqslant M$, take $N_i\in\Bbb N$ such that $k\geqslant N_i\implies\frac1{i^2}\left|x_i^{(k)}-x_i\right|<\frac\varepsilon{2N}$. Then$$k\geqslant\max\{N_1,N_2,\ldots,N_M\}\implies d\left(x^{(k)},x\right)<\frac\varepsilon2+\frac\varepsilon2=\varepsilon.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.