Banach spaces or not?

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

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

Thanks.
bkarpuz

What about $\displaystyle \displaystyle f_n(x)=\frac{\mathbf{1}_{[c,b)}}{n}$. Then it's clear that $\displaystyle \left\{f_n\right\}$ is a Cauchy sequence in $\displaystyle \Omega_c([a,b])$ yet $\displaystyle f_n(x)\to 0$
• Mar 30th 2011, 04:24 PM
bkarpuz
Quote:

Originally Posted by Drexel28
What about $\displaystyle \displaystyle f_n(x)=\frac{\mathbf{1}_{[c,b)}}{n}$. Then it's clear that $\displaystyle \left\{f_n\right\}$ is a Cauchy sequence in $\displaystyle \Omega_c([a,b])$ yet $\displaystyle f_n(x)\to 0$

Forgive me because I dont understand it, what does this mean exactly? (Speechless)
The zero function is in both of the spaces $\displaystyle \Omega$ and $\displaystyle \Psi$. (Itwasntme)
• Mar 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 $\displaystyle \Omega$ and $\displaystyle \Psi$.

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

Originally Posted by Drexel28
So $\displaystyle \Omega_c([a,b])$ is the space of all functions which may have a left-hand discontinuity at $\displaystyle c$?

Yes, and $\displaystyle c$ is a given fixed point in $\displaystyle [a,b]$.
• Mar 30th 2011, 05:14 PM
Drexel28
Quote:

Originally Posted by bkarpuz
Yes, and $\displaystyle c$ is a fixed point in $\displaystyle [a,b]$.

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

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

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

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

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

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

It would be very appreciated if you confirm that my arguments are correct.
• Apr 1st 2011, 02:41 PM
Opalg
That argument looks correct to me. (Hi)