cauchy sequence

**surjective** · January 13th 2011, 02:23 AM

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.

**FernandoRevilla** · January 13th 2011, 02:56 AM

Have you found $\left\|{f_m-f_n}\right\|_1$ ?

Fernando Revilla

**HallsofIvy** · January 13th 2011, 05:57 AM

If m> n then 1/m< 1/n so

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

What is the norm of that? Does it go to 0 as m and n go to infinity.

**Drexel28** · January 13th 2011, 09:40 AM

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?

**surjective** · January 13th 2011, 10:15 AM

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?

**surjective** · January 13th 2011, 10:19 AM

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.

**Drexel28** · January 13th 2011, 12:16 PM

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.

**mus** · January 13th 2011, 03:16 PM

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.

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