Banach spaces or not?

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March 30th 2011, 03:55 PM
bkarpuz

Banach spaces or not?
Dear MHF members,

I have two questions on Banach spaces,
and I would really appreciate if you help me with a complete answer.
Let be a interval and .
1. Denote by $\Omega_{c}([a,b],\mathbb{R})$ the set of all real valued bounded functions on $[a,b]$ ,
  which may have jump type discontinuity from the left-hand side at the point $c$ ,
  i.e., $\lim_{t\to c^{-}}f(t)\neq f(c)$ and $\lim_{t\to c^{+}}f(t)=f(c)$ .
  Is $\Omega_{c}([a,b],\mathbb{R})$ a Banach space with the $\sup$ norm?
2. Denote by $\Psi_{c}([a,b],\mathbb{R})$ the set of all real valued bounded functions on $[a,b]$ ,
  which may have removable type discontinuity at the point $c$ .
  Is $\Psi_{c}([a,b],\mathbb{R})$ a Banach space with the $\sup$ norm?
Thanks.
bkarpuz
March 30th 2011, 04:01 PM
Drexel28
Quote:
Originally Posted by bkarpuz

Dear MHF members,

I have two questions on Babach spaces,
and I would really appreciate if you help me with a rigorous answer.
Let $[a,b]$ be a interval and $c\in[a,b]$ .

Denote by $\Omega_{c}([a,b],\mathbb{R})$ the set of all real valued bounded functions on $[a,b]$ , which might have jump type discontinuity from the left-hand side at the point $c$ , i.e., $\lim_{t\to c^{-}}f(t)\neq f(c)$ and $\lim_{t\to c^{+}}f(t)=f(c)$ .
Is $\Omega_{c}([a,b],\mathbb{R})$ a Banach space with the $\sup$ norm?
Denote by $\Psi_{c}([a,b],\mathbb{R})$ the set of all real valued bounded functions on $[a,b]$ , which might have removable type discontinuity at the point $c$ .
Is $\Psi_{c}([a,b],\mathbb{R})$ a Banach space with the $\sup$ norm?

Thanks.
bkarpuz
What about . Then it's clear that is a Cauchy sequence in yet
March 30th 2011, 04:24 PM
bkarpuz

Quote:

Originally Posted by Drexel28

What about $\displaystyle f_n(x)=\frac{\mathbf{1}_{[c,b)}}{n}$ . Then it's clear that $\left\{f_n\right\}$ is a Cauchy sequence in $\Omega_c([a,b])$ yet $f_n(x)\to 0$

Forgive me because I dont understand it, what does this mean exactly? (Speechless)
The zero function is in both of the spaces $\Omega$ and $\Psi$ . (Itwasntme)
March 30th 2011, 04:26 PM
Drexel28

Quote:

Originally Posted by bkarpuz

Forgive me because I dont understand it, but what does this mean exactly?
The zero function is in both of the spaces $\Omega$ and $\Psi$ .

So $\Omega_n([a,b])$ is the space of all functions which may have a left-hand discontinuity at $c$ ??? So, in other words it's all bounded functions?
March 30th 2011, 04:29 PM
bkarpuz

Quote:

Originally Posted by Drexel28

So $\Omega_c([a,b])$ is the space of all functions which may have a left-hand discontinuity at $c$ ?

Yes, and $c$ is a given fixed point in $[a,b]$ .
March 30th 2011, 05:14 PM
Drexel28

Quote:

Originally Posted by bkarpuz

Yes, and $c$ is a fixed point in $[a,b]$ .

I'm just misunderstanding the point of the extra condition. Namely, if the function needn''t have a left-hand discontinuity there, why even mention it? That's like saying $L(\mathbb{R})$ is the set of all functions which may have infinite integral....i.e. all functions.
March 30th 2011, 05:22 PM
bkarpuz

Quote:

Originally Posted by Drexel28

I'm just misunderstanding the point of the extra condition. Namely, if the function needn''t have a left-hand discontinuity there,
why even mention it? That's like saying $L(\mathbb{R})$ is the set of all functions which may have infinite integral....i.e. all functions.

Just to have a space including the continuous functions together with some new functions which have discontinuity at a given point.
Lets say the continuous functions on $[0,1]$ and those which have left-hand dc at $1/2$ .
March 30th 2011, 08:24 PM
bkarpuz

Here are my arguments for the first question.
I will try to give an affirmative answer.

Proof of 1. It is clear that $\|\cdot\|:=\sup$ is a norm on $\Omega_{c}([a,b],\mathbb{R})$ .
To complete the proof, we need to show that any Cauchy sequence $\{f_{k}\}_{k\in\mathbb{N}}\in\Omega_{c}([a,b],\mathbb{R})$
converges a function in this space.
Then, for every, $\varepsilon>0$ , there exists $k_{0}\in\mathbb{N}$ such that $\|f_{n}-f_{m}\|<\varepsilon/2$ for all $m,n\in\mathbb{N}$ with $m,n\geq k_{0}$ .
In particular, for each $x\in[a,b]$ , we have $|f_{n}(x)-f_{m}(x)|<\varepsilon/2$ for all $n,m\in\mathbb{N}$ with $m,n\geq k_{0}$ ,
i.e., $\{f_{k}(x)\}_{k\in\mathbb{N}}\subset\mathbb{R}$ is a Cauchy sequence, and thus for each $x\in[a,b]$ ,
the limit $f(x):=\lim_{k\to\infty}f_{k}(x)$ exists, to complete the proof, we have to now show that
$\{f_{k}\}_{k\in\mathbb{N}}$ converges uniformly to $f$ and $f\in\Omega_{c}([a,b],\mathbb{R})$ .

We have $|f_{n}(x)-f_{m}(x)|<\varepsilon/2$ for all $x\in[a,b]$ and all $m,n\in\mathbb{N}$ with $m,n\geq k_{0}$ ,
which yields by letting $m\to\infty$ that $|f_{n}(x)-f(x)|<\varepsilon$ for all $x\in[a,b]$ and all $n\in\mathbb{N}$ with $n\geq k_{0}$ ,
and proves that $\{f_{k}\}_{k\in\mathbb{N}}$ converges uniformly to $f$ .

Let $x_{0}\in[a,b]$ , then for every $\varepsilon>0$ , there exists $n_{0}\in\mathbb{N}$ such that $|f(x_{0})-f_{k}(x_{0})|<\varepsilon/3$ for all $k\geq n_{0}$ .

Let $x_{0}\neq c$ . As the functions are continuous at $x_{0}$ , there exists a neighborhood $U_{n_{0}}$ of $x_{0}$
such that $|f_{n_{0}}(x)-f_{n_{0}}(x_{0})|<\varepsilon/3$ for all $x\in U_{n_{0}}$ .
Then, for any $x\in U_{n_{0}}$ , we have
$|f(x)-f(x_{0})|=|f(x)-f_{n_{0}}(x)+f_{n_{0}}(x)-f_{n_{0}}(x_{0})+f_{n_{0}}(x_{0})-f(x_{0})|$
......................... $\leq|f(x)-f_{n_{0}}(x)|+|f_{n_{0}}(x)-f_{n_{0}}(x_{0})|+|f_{n_{0}}(x_{0})-f(x_{0})|$
......................... $\leq\dfrac{\varepsilon}{3}+\dfrac{\varepsilon}{3}+ \dfrac{\varepsilon}{3}=\varepsilon$
for all $x\in U_{n_{0}}$ proving that $f$ is continuous at $x_{0}\neq c$ .

Let $x_{0}=c$ , then we follow similar steps above with a right neighborhood of $x_{0}$ ,
and show that $f$ is right-continuous at $x_{0}=c$ .

This shows that $f\in\Omega_{c}([a,b],\mathbb{R})$ , and thus it is complete.
Hence, $\Omega_{c}([a,b],\mathbb{R})$ is a Banach space. $\rule{0.2cm}{0.2cm}$

It would be very appreciated if you confirm that my arguments are correct.
April 1st 2011, 02:41 PM
Opalg

That argument looks correct to me. (Hi)