# cauchy sequence

• January 13th 2011, 03:23 AM
surjective
cauchy sequence
I am considering the sequences $(f_{n})$ of continuous functions $f_{n}:[-1,1]\to \mathbb{R}$ defined by:

$
\begin{displaymath}
f_{n}(x) = \left\{
\begin{array}{lr}
1 & : -1\leq x \leq 0\\
1-nx & : 0 \leq x \leq \tfrac{1}{n}\\
0 & : \tfrac{1}{n} \leq x \leq 1
\end{array}
\right.
\end{displaymath}
$

How do I show that $(f_{n})$ is a cauchy-sequence in the normed vector space $(C([-1,1],\mathbb{R}), \Vert \cdot \Vert_{1})$ and how do I examine if the mentioned normed vector-space is a banach space (i know that every finite dimensional normed vector space is a banach space).

$(C([-1,1],\mathbb{R}), \Vert \cdot \Vert_{1})$ is the normed vector space of continuous real-valued functions defined in the closed interval $[-1,1]$.

Thanks.
• January 13th 2011, 03:56 AM
FernandoRevilla
Have you found $\left\|{f_m-f_n}\right\|_1$ ?

Fernando Revilla
• January 13th 2011, 06:57 AM
HallsofIvy
If m> n then 1/m< 1/n so

$f_n- f_m (x)= \left\{\begin{array}{cc}0 & -1< x< 0 \\ (m-n)x & 0

What is the norm of that? Does it go to 0 as m and n go to infinity.
• January 13th 2011, 10:40 AM
Drexel28
Quote:

Originally Posted by surjective
I am considering the sequences $(f_{n})$ of continuous functions $f_{n}:[-1,1]\to \mathbb{R}$ defined by:

$
\begin{displaymath}
f_{n}(x) = \left\{
\begin{array}{lr}
1 & : -1\leq x \leq 0\\
1-nx & : 0 \leq x \leq \tfrac{1}{n}\\
0 & : \tfrac{1}{n} \leq x \leq 1
\end{array}
\right.
\end{displaymath}
$

How do I show that $(f_{n})$ is a cauchy-sequence in the normed vector space $(C([-1,1],\mathbb{R}), \Vert \cdot \Vert_{1})$ and how do I examine if the mentioned normed vector-space is a banach space (i know that every finite dimensional normed vector space is a banach space).

$(C([-1,1],\mathbb{R}), \Vert \cdot \Vert_{1})$ is the normed vector space of continuous real-valued functions defined in the closed interval $[-1,1]$.

Thanks.

Every finite dimensional vector space is a Banach space just by the "accident" that on finite dimensional vector spaces all norms are equivalent. For infinite dimensional vector spaces such as $\mathcal{C}[-1,1]$ this isn't always the case. You need to show that the norm induces a complete metric space structure on your vector space. Does it here?
• January 13th 2011, 11:15 AM
surjective
cauchy sequence
Stupid question: How did you get $\Vert f_{n}-f_{m} \Vert_{1}$?

Since I have shown that:

$\Vert f \Vert _{1}=\int_{-1}^{1} \left| f(x) \right| dx$

the norm you are asking for becomes:

$\vert f_{n}-f_{m} \Vert_{1}=
\int_{-1}^{1} \left| f_{n}-f_{m} \right| dx=
\int_{0}^{\frac{1}{m}} \left| (m-n)x \right| dx + \int_{\frac{1}{m}}^{\frac{1}{n}} \left| 1-nx \right| dx =
\frac{1}{2m} \left| \frac{m-n}{m} \right|\cdot\left( 1-\frac{1}{n}(m-n) \right)$

I would say that this expression goes to 0 as $n$ and $m$ tend towards infinity.

Per definition i know that for a sequence to a be a cauchy-sequence it must hold that: $n,m\geq n_{0} \Rightarrow \Vert f_{n}-f_{m} \Vert_{1} < \epsilon$

Could I conclude that since the expression above tends towards 0 (which is smaller than $\epsilon$) for $n,m$ going towards infinity, that $f_{n}$ is a cauchy-sequence?
• January 13th 2011, 11:19 AM
surjective
Hmmmm, not sure how to do that. To be honest I don't understand what it means for a norm to induce a complete metric structure. Elaboration would be appreciated.
• January 13th 2011, 01:16 PM
Drexel28
Quote:

Originally Posted by surjective
Hmmmm, not sure how to do that. To be honest I don't understand what it means for a norm to induce a complete metric structure. Elaboration would be appreciated.

You need to show that every Cauchy sequence converges, or find a counterexample.
• January 13th 2011, 04:16 PM
mus
I can come up with the following counter example: If $n \to \infty$ then:

$f_{n} \to \begin{displaymath}
\left\{
\begin{array}{lr}
1 & : -1\leq x \leq 0\\
- & : 0 \leq x \leq \tfrac{1}{n}\\
0 & : \tfrac{1}{n} \leq x \leq 1
\end{array}
\right.
\end{displaymath}$

The "-" means that the values is not defined.