# Thread: Banach spaces or not?

1. ## 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

2. 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 $\displaystyle f_n(x)\to 0$

3. 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?
The zero function is in both of the spaces $\displaystyle \Omega$ and $\displaystyle \Psi$.

4. 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?

5. 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 $\displaystyle [a,b]$.

6. 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.

7. 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 $\displaystyle 1/2$.

8. 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.

9. That argument looks correct to me.