1
$\begingroup$

show that $(c_0,||\cdot||_\infty)$ be the space of real valued sequences converging to $0$ with the supremum norm is complete

Okay so I know that the normed space is complete if any Cauchy sequence in my normed space converges to a value which is inside my normed space.

but how would I use this to show that my normed space is complete? is it enough to find one example inside my normed space and show that it is Cauchy and that it converges to $0$??

$\endgroup$
  • $\begingroup$ Why would finding one example of a Cauchy sequence that converges imply that all Cauchy sequences converge? $\endgroup$ – zhw. Feb 22 '16 at 18:40
  • $\begingroup$ I guess that's incorrect then... it was just an attempt to show what's going through my mind instead of just posting a question on its own ... so I am required to show all Cauchy sequences within my normed space are convergent? $\endgroup$ – smith Feb 22 '16 at 18:43
  • $\begingroup$ Use the definition: Take a Cauchy sequence $\{ a_n \}_n$ of elements of $c_0$: every $a_n$ is itself a sequence $\{ a_{nm} \}_m$! Now, you have to find a sequence $a$ such that $a_n \to a$ in the sup norm. What would you do? How to define your limit sequence $a$? $\endgroup$ – Crostul Feb 22 '16 at 18:43
  • $\begingroup$ okay that makes sense but what sequence would I take in the first place? $\endgroup$ – smith Feb 22 '16 at 19:04
  • $\begingroup$ As in when you say take a Cauchy sequence $\{a_n\}_n$ $\endgroup$ – smith Feb 22 '16 at 19:09
1
$\begingroup$

No that won't be enough. You have to proceed that way:

Let's $(U_n)$ be a Cauchy sequence of $c_0$ (then $U_n$ is a sequence of sequence). Thus : $\forall \epsilon >0, \exists N \in \mathbb{N}, \forall p,q \geq N \parallel U_p-U_q\parallel _{\infty}<\epsilon$

Then you have to prove that $(U_n)$ converges to a certain $(U) \in C_0$.

EDIT : We have an equivalence between (in fact this is more the definition of completeness for a metric space) :

  1. $(c_0, \parallel \cdot \parallel )$ is complete,
  2. Every Cauchy sequence of $c_0$ converges (for the sup norm).

So if you take any Cauchy sequence $(U_n)$ of $c_0$ and proove that this sequence converges to $U$ in $c_0$ for the sup norm, then you have won.

Answer :

First we are going to prove that the sequence (of sequences) $(U_p)$ converges to a certain $(U)$. Let's get back to the definition of $(U_p)$ being a Cauchy sequence : $$\forall \epsilon >0, \exists N \in \mathbb{N}, \forall p,q \geq N \parallel U_p-U_q\parallel _{\infty}<\epsilon$$ By the definition of the sup norm, this implies $$\forall \epsilon >0, \exists N \in \mathbb{N}, \forall p,q \geq N,\forall n \in \mathbb{N}, |U_p^n-U_q^n|<\epsilon$$ Where $U_p^n$ is the $n$-th element of $U_p$ and the norm being the usual distance on $\mathbb{R}$. But $\mathbb{R}$ is complete, and $(U_p^n)_p$ is a real Cauchy sequence, so : $$\forall n \in \mathbb{N},U_p^n \underset{p \rightarrow \infty}\longrightarrow U^n$$ where $U^n$ is the $n$-th element of the sequence $(U)$. So now we have proven that $(U_p)$ converges to a sequence $(U)$, now let's prove that $(U) \in c_0$.

Because of the convergence of $(U_p)$, we have : $$\forall \epsilon >0, \exists N \in \mathbb{N}, \forall p \geq N,\forall n \in \mathbb{N}, | U_p^n-U^n |<\epsilon $$ But, $$\forall p,\forall \epsilon' >0 ,\exists N' \in \mathbb{N}, \forall n>N', |U_p^n-0|<\epsilon'$$ because $\forall p, (U_p) \in c_0$. Let's combine the two.Thus, $$\forall \epsilon,\epsilon' >0, \exists N,N' \in \mathbb{N}, \forall p \geq N,\forall n >N', |U^n-U_p^n|<\epsilon \text{ and } |U_p^n|<\epsilon'$$

By elementary manipulations on thoose inequalities, we get to $|U^n|<\epsilon + \epsilon'$. So the proof is complete.

$\endgroup$
  • $\begingroup$ Could you elaborate on this answer? ... Is $u_n$ a sequence of $U_n$? $\endgroup$ – smith Feb 22 '16 at 19:10
  • $\begingroup$ Sorry typo, $u_n$ was meant to be $U_n$ $\endgroup$ – nicomezi Feb 22 '16 at 20:10
  • $\begingroup$ so essentially I have to show that $(U_n)$, which is my sequence of sequences, converges to a sequence $(U)$ where the sequence $(U) \in c_0$ ?? $\endgroup$ – smith Feb 22 '16 at 20:42
  • $\begingroup$ could you explain why this would show completeness? I understand what needs to be done but why is this correct? $\endgroup$ – smith Feb 22 '16 at 20:43
  • $\begingroup$ ohhhh that makes more sense, so I wont be taking a specific Cauchy sequence, rather I will be taking any sequence and showing that converges to a value $U$ which is in $c_0$ $\endgroup$ – smith Feb 22 '16 at 21:12

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.