When someone says a real valued function $f(x)$ on $\mathbb{R}$ is finite, does it mean that $|f(x)| \leq M$ for all $x \in \mathbb{R}$ with some $M$ independent of $x$?
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15$\begingroup$ What you are describing is what is usually called a bounded function. I have not seen the English word finite used in this context. Perhaps you could mention the context in which the term was used. $\endgroup$– André NicolasMay 17, 2012 at 4:03
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2$\begingroup$ As we can see from the contradictory answers (I would choose George's answer), the OP must provide some context to get something useful. $\endgroup$– GEdgarFeb 10, 2013 at 17:42
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$\begingroup$ Just pointing out that "finite" sometimes also means nonzero (with a finite measure, usually $>0$). In the sense that $dx$ can be understood as infinitesimal, but still finite interval. Most likely not in this case, but if we are discussing semantics, we should include all the cases. $\endgroup$– orionAug 7, 2015 at 7:12
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$\begingroup$ For example $x\mapsto \frac{1}{x}$ is finite valued on $]0,\infty [$ since for all $x\in ]0,\infty [$, $-\infty <f(x)<\infty $, but it's not valued on $[0,\infty [$ since $f(0)=+\infty $. $\endgroup$– SurbOct 5, 2015 at 15:28
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$\begingroup$ Possible duplicate of : What does 'finite-valued' mean? $\endgroup$– GigiliAug 27, 2016 at 9:02
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6 Answers
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In Elias Stein's Real Analysis, at the beginning of Chapter 4.1, it reads "We shall say that $f$ is finite-valued if $-\infty<f(x)<\infty$ for all $x$."
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A function is finite if it never asigns infinity to any element in its domain. Note that this is different than bounded as $f(x):\mathbb R \to \mathbb R \cup\{\infty\}: f(x)=x^2$ is not bounded since $\lim_{x \to \infty}=\infty$. However, $f$ is finite since it does not assign $\infty$ to any real number.
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answered Feb 10, 2013 at 16:35
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4$\begingroup$ This doesn't sound right. No $\mathbb R\to\mathbb R$ function can assign infinity to any element in its domain, because infinity is not a real number. $\endgroup$– user856Feb 10, 2013 at 17:43
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1$\begingroup$ It's not a problem since the point of this example is to show that not all finite functions are bounded. $\endgroup$ Feb 10, 2013 at 20:49
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Since a valued function may have $\mathbb R \cup \{\infty\}$ as target, it's possible that finite function $f$ corresponds to cases where $\forall x \in \mathbb R \quad f(x) \neq \infty$ like $f(x)=x$ or $f(x)= \frac x {x^2+6}$ while $f(x)=\frac 1x$ , for example, is not finite according to this meaning, because $f(0)=\infty$ (thing that can be taken by defintion or convention)
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A convex function $f$ is said to be proper if its epigraph is non-empty and contains no vertical lines, i.e., if $f(x)<+\infty$ for at least one $x$ and $f(x)>-\infty$ for every x. (Section 4, Chapter 1, Convex analysis, Rockafellar, 1997)
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I am not familiar with the term finite in this context. One possible definition would be this.
A function $f: A \to B$ is finite if and only if $f(A) \subseteq B$ is finite.
However, I would not use this definition because the relation $f \subseteq A \times B$ is still an infinite set (if $A$ is infinite).
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$\begingroup$ The Icelandic term endanlegt fall is defined this way in a dictionary, and endanlegt means finite and fall means function. But that dictionary states that the English translation of the term is logic function (which I've never seen used in that meaning). $\endgroup$ Aug 21, 2020 at 11:18
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No, It means there are only finitely many $y$ such that $f(x)=y$, for example $f(x)=0,1,2,\dots,n$ where $n\in\mathbb{N}$ is finitely valued, although it is also bounded, but not all bounded functions are finitely valued for example $f(x)=x$ on $\mathbb{R}$ or $\mathbb{Q}$ takes uncountably many values within any $|x_i-x_j|<\epsilon$
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1$\begingroup$ Amusingly, one might also use the term for the opposite condition -- for each $y$, $f(x)=y$ only has finitely many solutions for $x$. (c.f. a finite morphism of schemes) $\endgroup$– user14972Jun 23, 2012 at 1:48