The set of piecewise continuous functions form a Banach space?

[bkarpuz]

Dear MHB members,

I have a quick question as follows.
Let $I$ be a compact interval and $\Omega$ denote the set of functions $f:I\to\mathbb{R}$,
which has at most finite number of discontinuities on $I$ such that if $t$ is a discontinuity point of $f$, then $\lim_{\tau\to t^{+}}f(\tau)=f(t)$ and $\lim_{\tau\to t^{-}}f(\tau)$ exists.
Is $\Omega$ a Banach space?

Thanks.
bkarpuz

 

[ThePerfectHacker]

Let \(\Omega\) be the set as defined and let us put a norm on \(\Omega\) be defining \(|| f || = \sup_{x\in I} |f(x)|\).

Let \(q_1,q_2,q_3,....\) be an enumeration of all rational numbers in \( I \).

For each \(n\geq 1\) define the a function \(f_n\in \Omega\) as follows. We define \(f_n(q_1) = 1\), \(f_n(q_2)=1/2\), \(f_n(q_3)=1/3\), ... , \(f_n(q_n) = 1/n\), and \(f_n(x) = 0\) otherwise if \( x\not \in \{q_1,q_2,...,q_n \} \).

If \( m>n \) then \( || f_m - f_n || = \frac{1}{m+1} \). Thus, the sequence \(f_1,f_2,f_3,...\) is a Cauchy sequence in \( (\Omega, || \cdot || ) \).

Now what is the pointwise limit of this sequence of functions? Well, if \( x\in I \) is irrational then \( f_n(x) \to 0 \), if \( x \) is rational then \( f_n(x) \to \frac{1}{k} \) where \( k \) is the positive integer such that \( x = q_k \) by the enumeration. Let us define the function \( f:I \to \mathbb{R}\) to be \( f(x) = 0 \) if \( x \) is irrational and \( f(x) = \frac{1}{k} \) where \( k \) is the positive integer such \( x = q_k \).

This function \( f \) is a Dirichlet-type function so it is discontinous at every point of \( I \). Which means that \( f \not \in \Omega \) as \( \Omega \) was defined to be the set of all piecewise continous functions on \( I \). Thus, we have produced a Cauchy sequence above that does not have a limit in \( \Omega \), for if it did, then its limiting function \( f \) would need to belong to \( \Omega \) but it does not.

This means that your space of functions is not a Banach space (as by definition a Banach space is a normed vector space that is complete).

 

[bkarpuz]

Let \(\Omega\) be the set as defined and let us put a norm on \(\Omega\) be defining \(|| f || = \sup_{x\in I} |f(x)|\).

...

This means that your space of functions is not a Banach space (as by definition a Banach space is a normed vector space that is complete).

Thank you TPH, I was thinking about the same!

 

[sandy]

Thank you TPH, I was thinking about the same!

Hi, I realized that a long time has elapsed, but I hope it will be helpful.

The constructed sequence is not left continuous, so it is not a correct example.

Actually the space of piecewise left cont. funcs. is complete as long as the interval is compact. For reference you can follow the url:
Principles of Discontinuous Dynamical Systems - Marat Akhmet - Google Kitaplar

Best regards

 

[Opalg]

Let $I$ be a compact interval and $\Omega$ denote the set of functions $f:I\to\mathbb{R}$,
which has at most finite number of discontinuities on $I$ such that if $t$ is a discontinuity point of $f$, then $\lim_{\tau\to t^{+}}f(\tau)=f(t)$ and $\lim_{\tau\to t^{-}}f(\tau)$ exists.
Is $\Omega$ a Banach space?

Hi, I realized that a long time has elapsed, but I hope it will be helpful.

The constructed sequence is not left continuous, so it is not a correct example.

Actually the space of piecewise left cont. funcs. is complete as long as the interval is compact. For reference you can follow the url:
Principles of Discontinuous Dynamical Systems - Marat Akhmet - Google Kitaplar

Best regards

Hi Sandy, and welcome to MHB. Thanks for your contribution, but I disagree with your conclusion.

The question does not ask for the functions to be piecewise left continuous, but that they should have only finitely many discontinuities, at each of which there should be a left and a right one-sided limit. TPH's example satisfies these conditions, because each of his functions has left and right one-sided limits equal to $0$ at each discontinuity. His limit function has infinitely many discontinuities, so is not in the space $\Omega$. So he correctly shows that $\Omega$ is not a Banach space.

 

[sandy]

Hi Sandy, and welcome to MHB. Thanks for your contribution, but I disagree with your conclusion.

The question does not ask for the functions to be piecewise left continuous, but that they should have only finitely many discontinuities, at each of which there should be a left and a right one-sided limit. TPH's example satisfies these conditions, because each of his functions has left and right one-sided limits equal to $0$ at each discontinuity. His limit function has infinitely many discontinuities, so is not in the space $\Omega$. So he correctly shows that $\Omega$ is not a Banach space.

Thanks Opalg,
yes, it is a suitable example for any function having finitely many discontinuitites .
But $\Omega$ is the space of piecewise right continuous functions because of the condition $lim_{\tau\to\t+}f(\tau)=f(t)$. As it seems this condition has not noticed.

So I made a mistake by saying left continuous. But still $\Omega$ is complete, and the proof is very similar to the one in the book. (page 19)

Best regards

 

[Opalg]

Thanks Opalg,
yes, it is a suitable example for any function having finitely many discontinuitites .
But $\Omega$ is the space of piecewise right continuous functions because of the condition $lim_{\tau\to t+}f(\tau)=f(t)$. As it seems this condition has not noticed.

So I made a mistake by saying left continuous. But still $\Omega$ is complete, and the proof is very similar to the one in the book. (page 19)

Best regards

Yes, you are quite right. It seems none of us had noticed the right continuity condition. So a suitable counterexample would be to define $f_n(x)$ to be the function on the unit interval taking the value $\frac1k$ in the interval $\bigl[\frac1k,\frac1{k-1}\bigr)$, for $2\leqslant k\leqslant n$, and $f(x) = 0$ when $x\in \bigl[0,\frac1n\bigr)$. There is a limit function (in the sup norm), which is right continuous everywhere but has infinitely many discontinuities and so does not lie in $\Omega$ as defined in the OP.