Can somebody please explain me what the difference is between totally bounded and uniformly bounded functions?
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asked Oct 16, 2012 at 19:37
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1$\begingroup$ Please formulate your question as a question. $\endgroup$– J.R.Commented Oct 16, 2012 at 19:40
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$\begingroup$ Could you give the definitions, and where are the functions defined? $\endgroup$– Davide GiraudoCommented Oct 16, 2012 at 19:42
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1 Answer
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To illustrate the concepts, I consider real functions in one real variable in the following. Of course this carries over to arbitrarly generalized contexts (domains in $\mathbb{R}^n$, metric spaces, Banach spaces, whatever).
A single function $f:\mathbb{R}\rightarrow\mathbb{R}$ is bounded, if there exists a constant $C\ge 0$ such that $|f(x)|\le C$ for all $x\in\mathbb{R}$.
The term uniformly bounded only makes sense if you are considering an object that depends on at least one additional parameter, e.g. a sequence of functions $(f_k)_k$ ($f_k(x)$ depends on the index $k$ and on $x$).
A sequence $(f_k:\mathbb{R}\rightarrow\mathbb{R})_k$ of functions is uniformly bounded if there exists a constant $C\ge 0$ s.t. for all $k$ we have $|f_k(x)|\le C$ for all $x\in\mathbb{R}$. The important thing here is that C does not depend on $x$. This is what the word uniformly means.
In contrast, such a sequence is (pointwise) bounded, if for all $x\in\mathbb{R}$ there exists a constant $C=C(x)\ge 0$ such that $|f_k(x)|\le C$ for all $k$. Here, $C$ depends on $x$.
Uniformly bounded means each of the functions in the set is bounded with respect to the sup metric if we are talking about a metric space.
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$\begingroup$ What if $C$ does not depend on $x$ but does depend on $k$? $\endgroup$– user_ACommented Nov 18, 2023 at 8:45
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$\begingroup$ Could the term uniformly bounded be used for a function $f:\mathbb{R}\rightarrow\mathbb{R}^n$, with $n$ playing the role of $k$ in your definition? In that case, this definition would be equivalent to $||f(x)||_\infty \le C$ for all $x \in \mathbb{R}$? $\endgroup$– SivaCommented Jun 8 at 3:22