It is currently Tue Mar 26, 2019 6:49 am

 All times are UTC [ DST ]

 Print view Previous topic | Next topic
Author Message
 Post subject: Not Complete Metric Space Posted: Mon Jul 11, 2016 9:06 am
 Team Member

Joined: Tue Nov 10, 2015 8:25 pm
Posts: 314
Consider the set $\displaystyle C \left( [a,b] \right) = \left\{ f : [a,b] \longrightarrow \mathbb{R} \; \Big| \; f : \text{continuous} \right\}$ Show that the function $\displaystyle d_{1} : C \left( [a,b] \right) \times C \left( [a,b] \right) \longrightarrow \mathbb{R} \; , \; d_{1} (f,g) = \int_{a}^{b} \big| f(t) - g(t) \big|\, \mathrm{d}t$ is a metric on $\displaystyle C \left( [a,b] \right)$.
Afterwards, show that the metric space $\displaystyle \left( \; C \left( [a,b] \right) \, , \, d_{1} \; \right)$ is not complete.

Top   Post subject: Re: Not Complete Metric Space Posted: Mon Jul 11, 2016 10:20 am
 Team Member

Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
Hi Nickos.

Let $\displaystyle{f\,,g\,,h\in C(\left[a,b\right])}$.

Obviously, $\displaystyle{d_1(f,g)\in\mathbb{R}\cap \left[0,+\infty\right)\,\,,\forall\,f\,,g\in C(\left[a,b\right])}$ and

\displaystyle{\begin{aligned} d_1(f,g)=0&\iff \int_{a}^{b}\left|f(t)-g(t)\right|\,\mathrm{d}t=0\\&\stackrel{|f-g|\geq 0}{\iff}\left|f-g\right|=0\\&\iff f=g\end{aligned}}

Also,

$\displaystyle{d_1(f,g)=\int_{a}^{b}\left|f(t)-g(t)\right|\,\mathrm{d}t=\int_{a}^{b}\left|g(t)-f(t)\right|\,\mathrm{d}t=d_1(g,f)}$.

Finally,

\displaystyle{\begin{aligned} d_1(f,h)&=\int_{a}^{b}\left|f(t)-h(t)\right|\,\mathrm{d}t\\&=\int_{a}^{b}\left|(f(t)-g(t))+(g(t)-h(t))\right|\,\mathrm{d}t\\&\leq \int_{a}^{b}\left[\left|f(t)-g(t)\right|+\left|g(t)-h(t)\right|\right]\,\mathrm{d}t\\&=d_1(f,g)+d_1(g,h)\end{aligned}}.

So, $\displaystyle{d_1}$ is a metric on $\displaystyle{C(\left[a,b\right])}$.

The metric space $\displaystyle{\left(C(\left[a,b\right],d_1\right)}$ is not complete.

Indeed, consider $\displaystyle{f_n:\left[a,b\right]\to \mathbb{R}\,,n\in\mathbb{N}\,,n\geq 3}$ by

$\displaystyle{f_{n}(t)=\begin{cases} 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,a\leq t\leq (a+b)/2\\ n\,\left(t-\dfrac{a+b}{2}\right)\,\,\,\,\,\,,(a+b)/2<t<(a+b)/2+(1/n)\\ 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,(a+b)/2+(1/n)\leq t\leq b \end{cases}}$

The sequence $\displaystyle{\left(f_n\right)_{n\geq 3}}$ is a $\displaystyle{\rm{Cauchy}}$ -sequence

since for $\displaystyle{n>m}$ we get

\displaystyle{\begin{aligned} d_1(f_n,f_m)&=\int_{a}^{(a+b)/2}|f_n(t)-f_m(t)|dt+\int_{(a+b)/2}^{(a+b)/2+(1/m)}|f_n(t)-f_m(t)|dt+\int_{(a+b)/2+(1/m)}^{b}|f_n(t)-f_m(t)|dt\\&=\int_{(a+b)/2}^{(a+b)/2+(1/m)}|f_{n}(t)-f_{m}(t)|dt\\&\leq \dfrac{1}{m}\end{aligned}}

Suppose now that there exists $\displaystyle{f\in C(\left[a,b\right])}$ such that $\displaystyle{f_n\to f}$, that is

$\displaystyle{\int_{a}^{b}|f_{n}(t)-f(t)|\mathrm{d}t\to 0}$. Then,

$\displaystyle{0\leq \int_{a}^{(a+b)/2}|f(t)|\,\mathrm{d}=\int_{a}^{(a+b)/2}|f_n(t)-f(t)|\,\mathrm{d}t\leq \int_{a}^{b}|f_n(t)-f(t)|\,\mathrm{d}t}$

and since $\displaystyle{f\in C(\left[a,b\right])}$, we have that $\displaystyle{f(t)=0\,,a\leq t\leq \dfrac{a+b}{2}}$.

Let $\displaystyle{\delta\in\left(\dfrac{a+b}{2},b\right)}$. There exists $\displaystyle{n_0\in\mathbb{N}}$

such that $\displaystyle{\dfrac{a+b}{2}+\dfrac{1}{n}<\delta\,,\forall\,n\geq n_0}$ and

$\displaystyle{f_n(t)=1\,,t\in\left[\delta,b\right]}$. But,

$\displaystyle{0\leq \int_{\delta}^{b}|f_{n}(t)-f(t)|dt\leq \int_{a}^{b}|f_{n}(t)-f(t)|dt\to 0}$ and

then $\displaystyle{\int_{\delta}^{b}|1-f(t)|dt=0}$.

Since $\displaystyle{f}$ is continuous, we get

$\displaystyle{f(t)=1\,,\forall\,t\in\left[\delta,b\right]\,\,,\forall\,\delta\in\left(\dfrac{a+b}{2},b\right)}$

and $\displaystyle{f(t)=1\,,\forall\,t\in\left(\dfrac{a+b}{2},b\right]}$.

But now, $\displaystyle{f}$ is not continuous at $\displaystyle{t=\dfrac{a+b}{2}}$, a contradiction.

Therefore, $\displaystyle{\left(C(\left[a,b\right]),d_1\right)}$ is not complete.

Top   Display posts from previous: All posts1 day7 days2 weeks1 month3 months6 months1 year Sort by AuthorPost timeSubject AscendingDescending

 All times are UTC [ DST ]

#### Mathimatikoi Online

Users browsing this forum: No registered users and 1 guest

 You cannot post new topics in this forumYou cannot reply to topics in this forumYou cannot edit your posts in this forumYou cannot delete your posts in this forumYou cannot post attachments in this forum

Search for:
 Jump to:  Select a forum ------------------ Algebra    Linear Algebra    Algebraic Structures    Homological Algebra Analysis    Real Analysis    Complex Analysis    Calculus    Multivariate Calculus    Functional Analysis    Measure and Integration Theory Geometry    Euclidean Geometry    Analytic Geometry    Projective Geometry, Solid Geometry    Differential Geometry Topology    General Topology    Algebraic Topology Category theory Algebraic Geometry Number theory Differential Equations    ODE    PDE Probability & Statistics Combinatorics General Mathematics Foundation Competitions Archives LaTeX    LaTeX & Mathjax    LaTeX code testings Meta Powered by phpBB® Forum Software © phpBB Group Color scheme created with Colorize It.
Theme created StylerBB.net