Let $a,b\in\mathbb R$ and $L^\infty([a,b])$ be a space of all bounded functions $f:[a,b]\to\mathbb R$. It is a metric space with a metric function given by $ d(f,g) = \sup\limits_{x\in[a,b]}|f(x) - g(x)|. $ Let us say that the function $f\in L^\infty([a,b])$ is in the class $S$, i.e. $f\in S$ if there is a finite partition $ \mathcal T = \{a = t_0
Closure of space of simple functions in $L^\infty([a,b])$
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0These are usually called step functions. Simple functions allow characteristic functions of sets that are not intervals. – 2012-02-27
2 Answers
The closure for the uniform norm of step functions is the space of functions which have left and right limit at each point of $(a,b)$ (right for $a$ and left for $b$). These functions are called regulated functions.
If $f$ is in the closure for the uniform norm of step functions, we fix $\varepsilon>0$. We can find a step function $f_1$ such that $||f-f_1||_{\infty}\leq \varepsilon$. Let $x_0\in (a,b)$. Then we can find $\eta>0$ such that $f_1$ is constant on $(x_0,x_0+\eta)$. If $x,y\in (x_0,x_0+\eta)$ then $|f(x)-f(y)|\leq |f(x)-f_1(x)|+|f_1(x)-f_1(y)|+|f_1(y)-f(y)|\leq 2\varepsilon,$ so by Cauchy's criterion $f$ has a right limit at $x_0$. A similar argument shows that $x_0$ has a left limit and that $a$ has a right limit, $b$ a left limit.
Conversely, if $f$ is regulated, we fix $\varepsilon>0$. Then for all $x\in [a,b]$, we can find $\eta(x)>0$ such that if $y,z\in (x-\eta(x),x)\cup (x,x+\eta(x))$ we have $|f(x)-f(y)|<\varepsilon$. Let $I_n:=[a,b]\cap (x-\eta(x),x+\eta(x))$. Then $(I_x)_{x\in [a,b]}$ is an open cover of $[a;b]$, so exists a $\eta>0$ such that for all open interval $I\subset [a,b]$ of diameter $<\eta$ is contained in a set $I_x$. Let $(t_0,\ldots,t_m)$ a subdivision of $[a,b]$ such that $\max_it_{i+1}-t_i<\eta$, $x_i$ such that $(t_i,t_{i+1})\subset I_{x_i}$. Let $S=(a_1,\ldots,a_p)$ a subdivision containing $t_i$ and $x_i$. We have for all $y,z\in (a_i,a_{i+1})$: $|f(y)-f(z)|<\varepsilon$. Fix $c_j:=f\left(\frac{a_i+a_{i+1}}2\right)$ and $f_1$ the step function defined by $f_1(x)=c_j$ if $x\in (a_i,a_{i+1})$ and $f_1(a_i)=f(a_i)$. Then $||f-f_1||_{\infty}\leq \varepsilon$.
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0@DavideGiraudo: Do you know what happens if we replace $[a,b]$ by an arbitrary topological space? From Wikipedia, it seems that regulated functions are defined only for one real variable, yet the closure of the space of step functions exists for any topological space. – 2017-07-12