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Let $(X,d)$ be a metric space, we define the following:

  • $D_b(X)$ is the set of all bounded functions $f:X\rightarrow \mathbb{C}$
  • $f\in $$D_b(X)$ is said to vanish at infinity if for each $\epsilon$>0 there is a bounded set $K\subseteq$ X such that $\left \| f|_{X|K} \right \|<\epsilon$, i.e. the set \begin{Bmatrix} x\in X|\left | f(x) \right |\geq \epsilon \end{Bmatrix} is bounded in $X$. The collection of all functions that vanish at infinity is denoted by $D_0(X)$.
  • $E\subseteq X\times X$ is called controlled set if $\exists r>0$ such that $d(x,y)<r$ for all $(x,y)\in E$.
  • Let $f \in D_b(X)$, and $E\subseteq X\times X$ is controlled set, the $E$-variation of f is the function $Var_E(f)$ defined by $$ Var_E(f)(x )=sup\left \{ \left | f(x)-f(y) \right |:(x,y) \in E \right \} $$
  • $f \in D_b(X)$ is said to have vanishing variation if $Var_E(f)\in D_0(X)$, for all controlled sets $E\subseteq X\times X$.The collection of all functions that have vanishing variation is denoted by $D_h(X)$. I need to verify that $D_0(X)\subseteq D_h(X)$.
Alex Ravsky
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Hussain Rashed
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1 Answers1

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The proof is a straightforward application of the provided definitions. Let $f\in D_0(X)$ be any function, $E\subseteq X\times X$ be any controlled set, and $\epsilon>0$ be any number. There exists $r>0$ such that $d(x,y)<r$ for all $(x,y)\in E$. There exists a bounded set $K$ of $X$ such that $|f(x)|<\epsilon$ for all $x\in X\setminus K$. Then the set $K’=\{x\in X: \exists y\in K (d(x,y)<r)\}$ is bounded too. If $x\in X\setminus K’$ is any element and $(x,y)\in E$ then $d(x,y)<r$ so $y\in X\setminus K$ and thus $|f(y)|<\epsilon$ and $|f(x)-f(y)|<2\epsilon$. Hence $Var_E(f)(x)\le 2\varepsilon$ (remark that if there are no $y\in X$ such that $(x,y)\in E$ then $Var_E(f)(x)$ is undefined).

Alex Ravsky
  • 90,434