# Banach space of discontinuous functions(Killing continuous functions)

Edit: According to the comment of Prof. Majer, I revise the question:

For a metric space $X$, we put $A=\{f:X\to \mathbb{C}\mid \text{f is bounded}\}$. We define two semi norm on $A$

$$\parallel f \parallel_{1}= \parallel \omega_{f}\parallel_{\infty}$$ or $$\parallel f \parallel_{2}= \int_{[0,\;1]} \omega_{f}(x)dx$$

where $\omega_{f}$ is the standard oscillation function..(Edit: In the later norm, we put $X=[0\;1]$)

We obtain two normed space on an appropriate quotient space. Are these resulting space, complete? After a possible completion how can we compare them with each other and how can we compare them with $A/C$ where $C$ is the space continuous functions and $A$ is equipped with sup norm. (By comparison I mean comparison as two Banach space). What type of topological or metric properties of $X$ is encoded in each of the above three Banach space. To what extent the endomorphism of these Banach space is classified.(Motivated by the fact that the endomorphisms of $C(X)$, as a banach space, are classified)

• "Appropriate quotient space" Do you have an idea how this space looks like (or how the space which you divide by)? I guess the quotient contains all characteristic functions of all points and then it wouldn't be separable in the first norm. The second norm looks pretty weired to me… – Dirk Dec 7 '14 at 18:54
• The second semi-norm is considered in the particular case of $X=[0,1]$, it seems. It vanishes exactly on the Riemann integrable functions. – Pietro Majer Dec 7 '14 at 21:42
• @Dirk thanks for your comment on separability.regarding the second norm, as it is commented above, the quotient is dividing by riemann integrable functions. – Ali Taghavi Dec 9 '14 at 19:54

Let $f\in A$ and $r>\|\omega_f\|_\infty/2$. By definition of $\omega_f$, for any $x\in X$ there is a nbd $U$ of $x$ such that for all $y\in U$ one has $|f(x)-f(y)|\le2r$, so that $B(f(y),r) \cap B(f(x),r)\neq\emptyset .$
As a consequence, the closed convex valued multi map $F:X\to 2^\mathbb{C}$ defined by $F(x):=\overline {B(f(x),r)}$ is lower hemicontinuous. By Michael selection theorem there exists a continuous selection $g(x)\in F(x)$, therefore verifying $\|f-g\|_\infty \le r.$
$$\inf_{g\in C}\|f-g\|_\infty \le \|\omega_f\|_\infty/2.$$
that is, the quotient norm on $A/C$ induced by $\|\cdot\|_\infty$ is not larger than $\|\cdot\|_1/2$. On the other hand, for all $f\in A$, one has $\|f\|_1 \le 2\|f\|_\infty,$ so that the norm induced on $A/C$ by $\|\cdot\|_1$ is exactly twice the quotient norm of $\|\cdot\|_\infty$. In particular, it is complete.