# If f is continuous on [a, b], then f is bounded on [a,b].

1. Jan 21, 2008

### Esran

Dear friends,

I just joined the forums, and I'm looking forward to being a part of this online community. This semester, I signed up for Analysis II. I'm a math major, so I should be able to understand pretty much everything you say (hopefully). However, I'd really appreciate it if you try not to be too arcane in your explanations.

So, here's the theorem I'm stuck on.

Suppose f is continuous on [a,b] and S is the set such that x is in S if and only if x is in [a,b] and, (1) x = a or (2) f is bounded on the subinterval [a,x]. Then S is [a,b].

This is what I have so far.

Let S = {x in [a,b] | x = a or f is bounded on [a,x]}
Thus, by definition, S is a subinterval of [a,b].
Because of this fact, S is bounded.
Because S is bounded, S has a least upper bound, call it p.
It follows that p > a must be true.
Because b is an upper bound of S, p <= b.

Now, we employ an indirect argument.
Thus, assume p < b.
By definition, b is the LUB of set [a,b].

From here on out, I'm confused about where to go and what to do. Does anyone have any pointers or suggestions? What approach do I need to take to complete this proof? The only thing I can think of is that it might have to do with f being continuous on an interval implying that f is bounded on that interval. However, I'm not sure how to prove this implication either.

Last edited: Jan 21, 2008
2. Jan 22, 2008

### zhentil

Have you taken analysis I? The proof that the image of a compact set under a continuous function is compact is straightforward, and should be in your book in any case. Once you have that, you're done.

3. Jan 22, 2008

### mathboy

I assume you are speaking of a compact subset of a metric space. In that case, then the theorem will support the OP's goal, since a compact subset of a metric space is necessarily closed and bounded.

4. Jan 23, 2008

Well this is kind of one of the theorems it's good to know a number of approaches to solve.. Such as [a,b] is compact, you wouldn't want to just apply the Heine-Borel theorem, and actually - the proof above that f([a,b]) is bounded is somewhat like the usual proof that [a,b] is compact from first principles..

So first clarify: what you are trying to show basically is that f([a,b]) = {y: y = f(x) for some x in [a,b]} is bounded.

To do this, I think the first thing to do is to establish that S contains more than one point {a}. You can apply continuity (at a) here: Let e = 1. Then there exists d > 0 such that f[a,a+d]) $\subseteq$ [f(a)-1,f(a)+1]. Thus f([a,a+d]) is bounded.

The next step is to show that if M = sup S, then M is in S, i.e. f([a,M]) is bounded: to do that you have to apply the continuity at M as you did at a in above. (I won't write everything out here..)

Then the next thing to show is that M = b. To do this, assume M < b, apply continuity at M, and get a contradiction to the fact that M = sup S...

This is very much like the proof that [a,b] is compact, by taking an open cover {U_i} of [a,b] and letting S = {x: [a,x] has a finite subcover}.

5. May 28, 2010

### Esran

Theorem: If $$f$$ is continuous on $$\left[a,b\right]$$, then $$f$$ is bounded on $$\left[a,b\right]$$.

Proof: Let $$\epsilon > 0$$ and define $$S=\left\{x: x=a\ or \left[a,x\right] \ is \ bounded\right\}$$. Clearly, $$S$$ is nonempty, bounded below by $$a$$, and bounded above by $$b$$. Thus, there exists $$p\in\left[a,b\right]$$ such that $$p=LUB\left(S\right)$$.

We shall proceed with a proof by contradiction. Assume $$p\in\left[a,b\right)$$. Because $$f$$ is continuous at $$p$$, there exists $$\delta > 0$$ such that $$x\in \left(p - \delta,p + \delta\right)\rightarrow \left|f\left(x\right) - f\left(p\right)\right|<\epsilon$$, whence it is easy to see that $$f$$ is bounded on $$\left(p - \delta,p + \delta\right)$$. Furthermore, we note there exists $$x\in S$$ such that $$x\in \left(p-\delta, p\right)$$, which implies $$f$$ is bounded on $$\left[a,x\right]$$. Now, pick $$y\in \left(p, p+\delta\right)$$; it follows that $$f$$ is bounded on $$\left[x,y\right]$$, and thus on $$\left[a,y\right]$$, wherefore $$y\in S$$ but $$y>p$$, which is a contradiction.

Hence, $$p=b$$. It remains to show $$b\in S$$. This is established by logic almost identical to that which was used previously to show $$y\in S$$.

This completes the proof.

Last edited: May 28, 2010
6. May 28, 2010

### DrGreg

The right symbols, but not in the right order. The correct deduction would be

for each $M\geq 0$ there exists an $x\in\left[a,b\right]$ such that $\left|f\left(x\right)\right|\geq M$.​

You need to make use of the fact that [a,b] is closed, because the theorem wouldn't be true on (a,b).

7. May 28, 2010

### Esran

Very true. I just thought I'd type something out to reply to myself two years ago just for the fun of it. I know the standard proof.

8. Oct 16, 2011

### albert1993

[STRIKE]I don't see why we just don't get f is bounded on [a,y] right away from the second part. Why does it only prove that it's bounded on [x,y]?[/STRIKE] See below for revised question
I don't see how what logic this is supposed to be... Can anyone elaborate?

Last edited: Oct 16, 2011
9. Oct 16, 2011

### HallsofIvy

That was a typo- he meant [a, b], not [x, y]. By the way, this thread was started almost two years ago and the last post before yours was a year and a half old.

10. Oct 16, 2011

### albert1993

Hmm I don't think that's a typo. Isn't he saying that [a,x] is bounded, [x,y] is bounded, and therefore [a,y] is bounded?

Oops that was a very poorly worded question as well on my part... I meant: So we have [a,x] is bounded above from the first part("we note there exists x∈S such that x∈(p−δ,p)"), but I guess I didn't understand why fully because I have trouble seeing why the second part ("Now, pick y∈(p,p+δ)") shows that f is bounded on [x,y] and not [a,y] right away... Could somebody explain why? Thanks

Hmm is digging old posts looked down upon? Should I just make new threads for minor questions like this? Sorry if it is.