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.

Suppose functions $f(x)$ and $g(x)$ are continuous with domain $X \subset \mathbb{R} $ which is nonempty, convex and compact, can we show that $$S \equiv (f(x), g(x)) $$ for all $x \in X$ is nonempty, compact and convex? If it is possible to show, please provide me hints on how to prove.

share|improve this question
1  
You haven't yet told us what's the image of the functions...! And is $\,X\,$ just a general compact convex topologuical space? –  DonAntonio Jan 8 '13 at 12:40
1  
$S$ non-empty and compact should be no problem. Convexity is: are $f$ and $g$ linear? If not, take $X = [0,2\pi]$ and $f(x) = \cos{x}$ and $g(x) = \sin{x}$ (for example). –  Martin Jan 8 '13 at 12:42
    
@DonAntonio I have improved my acceptance rate. What I can tell is that $X$ is interval -- closed and bounded. and $f$ and $g$ are continuous such that $$f(x)=\sqrt{x} - \max (x,0)$$. Would that suffices to show the claim? –  Daniel Lårs Jan 8 '13 at 12:46
    
Then you are working on $\,\Bbb R\,$ and the functions are real...good. –  DonAntonio Jan 8 '13 at 12:46
    
Think of $x \mapsto (f(x),g(x))$ as a path in the plane. You should see that the answer depends a lot on what your $g$ looks like. // I removed (proof-theory) from your question because that's a branch of mathematical logic. –  Martin Jan 8 '13 at 12:53

1 Answer 1

up vote 0 down vote accepted
  1. $S$ is non-empty - trivial since both $f(X)$ and $g(X)$ are nonempty subsets of $\mathbb{R}$
  2. $S$ is compact - since $f$ and $g$ are continuous on a compact set $X$, it follows that $f(X)$ and $g(X)$ are compact sets. This means, by Heine-Borel, they are closed and bounded. Boundedness implies that $S=(f(x),g(x))$ is bounded. To show $S$ is closed, take a sequence in $\{(a_n,b_n)\}_{n=0}^\infty\subset S$ that coverges to $(a,b)\in\mathbb{R}^2$. Since convergence in $\mathbb{R}^2$ occurs if and only if we have convergence in both components, it follows that $a_n\rightarrow a$ and $b_n\rightarrow b$. By construction $a_n=f(\tilde{a}_n)$ and $b_n=g(\tilde{b_n})$ for some tilde real sequences. Since $f(X)$ is compact, it is closed. This tells us that $a_n=f(\tilde{a}_n)\rightarrow a \in f(X)$. Since $g(X)$ is compact, it is closed. This tells us that $b_n=f(\tilde{b}_n)\rightarrow b \in g(X)$. Therefore $a=f(\tilde{a})$ and $b=g(\tilde{b})$ implying $(a,b)\in S$. This shows closedness. Since closed and bounded, by Heine-Borel, S is compact.
  3. Since $f(x)=g(x)$ for all $x\in X$ and since continuous functions from $\mathbb{R}$ to $\mathbb{R}$ take convex sets to convex sets, convexity follows.
share|improve this answer

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.