Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $x$?

share|improve this question
13  
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. –  André Nicolas May 17 '12 at 4:03
2  
As we can see from the contradictory answers (I would choose George's answer), the OP must provide some context to get something useful. –  GEdgar Feb 10 '13 at 17:42

5 Answers 5

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)

share|improve this answer

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 $x$."

share|improve this answer

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.

share|improve this answer
2  
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. –  Rahul Feb 10 '13 at 17:43
    
It's not a problem since the point of this example is to show that not all finite functions are bounded. –  Student Feb 10 '13 at 20:49

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).

share|improve this answer

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 $|x_i-x_j|<\epsilon$

share|improve this answer
1  
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) –  Hurkyl Jun 23 '12 at 1:48

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.