1
$\begingroup$

Intuitively, if a function $f$ is defined on $[a, b]$, then it must be bounded. Is there a theorem for this? I remember reading something related to this, but not the details.

$\endgroup$
  • $\begingroup$ f is defined on [a,b] means f is bounded at each point of [a,b]? $\endgroup$ – Rupsa Aug 31 '15 at 13:24
  • $\begingroup$ @Rupsa When is a function bounded at a single point? $\endgroup$ – Siminore Aug 31 '15 at 13:50
  • $\begingroup$ @Siminore when f can be defined at that point i mean f does not take infinite value at that particular point $\endgroup$ – Rupsa Aug 31 '15 at 13:53
  • $\begingroup$ @Rupsa Well, all this can be made rigorous, but I do not think that Qed was working with functions taking values on the extended real line. $\endgroup$ – Siminore Aug 31 '15 at 14:00
  • $\begingroup$ @Siminore if f is bounded at each point of [a,b] then f must be bounded. bt i cant understand what Qed means by the term "defined on[a,b]" $\endgroup$ – Rupsa Aug 31 '15 at 14:05
5
$\begingroup$

Define $f \colon [0,1] \to \mathbb{R}$ by $f(x)=1/x$ for $x \in (0,1]$ and $f(0)=0$.

You probably remember a theorem (Weierstrass theorem) about continuous functions.

$\endgroup$
2
$\begingroup$

Theorem: If $f:[a,b]\rightarrow \mathbb{R}$ is a continuous funcion then $f([a,b])$ is a compact set. In particular, there is $x_m\in [a,b]$ and $x_M\in [a,b]$ such that $f(x_m)\leq f(x)\leq f(x_M)$ for every $x\in [a.b]$.

Well, this theorem said if you have a continuous function defined on compact interval then it is bounded. But $f:[0,+\infty)\rightarrow \mathbb{R}$ defined by $f(x)=x$ is defined on closed set but does not bounded.

$\endgroup$
2
$\begingroup$

My approach is, if $f(x)$ is bounded at each point of $D=[a,b]$ there is a neighbourhood $N(x)$ of $x$ s.t $f$ is bounded on $N(x)\cap D$. Consider $G=\{N(x):x\in D\}$ s.t $f$ is bounded on $N(x)\cap D$.Clearly $G$ is open cover of $D$ since $D$ is closed and bounded set in $\mathbb{R}$ by Heine-borel theorem there exists a finite subcollection $G'$ of $G$ s.t $G'$ also covers D. Let $G'=\{N(x_1),..,N(x_m)\}$ then $D\in N(x_1)\cup ... \cup N(x_m)$ & $f$ is bounded on $D\cap N(x_i)$ for $i=1,...,m$ .So there exists a +ve $M_i$ s.t $|f(x)|\leq M_i$ for all $x\in D\cap N(x_i)$ , $i=1,...,m$. Let $m=\max\{M_i,i=1,...,m\}$ & let $x\in D$ so $|f(x)|\leq M$ hence the proof.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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