# What does it mean when a function is finite?

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 $x$?

• 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. May 17, 2012 at 4:03
• As we can see from the contradictory answers (I would choose George's answer), the OP must provide some context to get something useful. Feb 10, 2013 at 17:42
• 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. Aug 7, 2015 at 7:12
• 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 $f(0)=+\infty$.
– Surb
Oct 5, 2015 at 15:28
• Possible duplicate of : What does 'finite-valued' mean? Aug 27, 2016 at 9:02

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 $x$."

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.

• 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.
– user856
Feb 10, 2013 at 17:43
• It's not a problem since the point of this example is to show that not all finite functions are bounded. Feb 10, 2013 at 20:49

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)

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)

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).

• 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). Aug 21, 2020 at 11:18

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 $|x_i-x_j|<\epsilon$

• 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)
– user14972
Jun 23, 2012 at 1:48