Is $\mathcal{M}(X)$ a Banach space?

  • MHB
  • Thread starter Euge
  • Start date
  • Tags
    2017
In summary, a Banach space is a complete normed vector space, named after mathematician Stefan Banach, where every Cauchy sequence converges to a limit in the space. The significance of $\mathcal{M}(X)$ being a Banach space is that it allows for the development of powerful mathematical tools for analyzing functions on a set $X$. To show that $\mathcal{M}(X)$ is a Banach space, one must prove that every Cauchy sequence of measurable functions on $X$ converges to a limit in $\mathcal{M}(X)$, which can be done by using the completeness property and the space of bounded functions being closed under uniform convergence. Not every normed vector space
  • #1
Euge
Gold Member
MHB
POTW Director
2,073
243
Here is this week's POTW:

-----
Consider the normed space $\mathcal{M}(X)$ of all complex regular Borel measures on a locally compact Hausdorff space $X$, with total variation norm $\|\mu\| := \lvert \mu\rvert (X)$, for all $\mu\in \mathcal{M}(X)$. Prove that $\mathcal{M}(X)$ is a Banach space.

-----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
Physics news on Phys.org
  • #2
Hi MHB community,

My apologies for the delayed post, I was unwell. No one answered this week's problem, but you can read my solution below.
Let $(\mu_j)$ be an absolutely summable sequence in $\mathcal{M}(X)$. We want to show $\sum_j \mu_j$ converges in $\mathcal{M}(X)$. Let $\lambda := \sum_j \lvert \mu_j\lvert$. Then $\mu_j <<\lambda$ for every $j$, so there exists a sequence $f_j\in \mathcal{L}^1(\lambda)$ such that $d\mu_j = f_j\, d\lambda$ for all $j$. Since $\sum_j \|f_j\|_1 = \sum_j \|\mu_j\| < \infty$ and $\mathcal{L}^1(\lambda)$ is complete, $\sum_{j = 1}^\infty f_j$ converges to a function $f\in L^1(\lambda)$. Define a measure $\mu$ on $X$ by the assignment $A \mapsto \int_A f\, d\mu$. It is an element of $\mathcal{M}(X)$, and the series $\sum_{j} \mu_j$ converges to $\mu$.
 

FAQ: Is $\mathcal{M}(X)$ a Banach space?

What is a Banach space?

A Banach space is a complete normed vector space, meaning that every Cauchy sequence in the space converges to a limit in the space. It is named after the mathematician Stefan Banach.

What is the significance of $\mathcal{M}(X)$ being a Banach space?

If $\mathcal{M}(X)$ is a Banach space, it means that it is a complete normed vector space of bounded measurable functions on a set $X$. This property is important because it allows for the development of powerful mathematical tools and techniques for analyzing functions on $X$.

How do you show that $\mathcal{M}(X)$ is a Banach space?

To prove that $\mathcal{M}(X)$ is a Banach space, one must show that every Cauchy sequence of measurable functions on $X$ converges to a limit in $\mathcal{M}(X)$. This can be done by using the completeness property of the underlying normed vector space, along with the fact that the space of bounded functions is closed under uniform convergence.

Is every normed vector space a Banach space?

No, not every normed vector space is a Banach space. For a normed vector space to be a Banach space, it must also be complete, which means that every Cauchy sequence converges to a limit in the space. Some examples of normed vector spaces that are not Banach spaces include the space of continuous functions with the sup norm and the space of differentiable functions with the $L^1$ norm.

Can a Banach space have an infinite dimension?

Yes, a Banach space can have an infinite dimension. In fact, most commonly studied Banach spaces, such as $\mathcal{C}[a,b]$ (the space of continuous functions on $[a,b]$) and $L^p(\mathbb{R}^n)$ (the space of $p$-integrable functions on $\mathbb{R}^n$), have an infinite dimension.

Back
Top