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.

Let $\phi:\mathbb{R}^d\rightarrow\mathbb{R}$ be a continuous map w.r.t. Euclidean norm. Let $C\subseteq\mathbb{R}^d$ be a convex subset. Assume that there exists $y\in\overline{C}$ such that $\phi(y)>0$. Does there exist $y'\in C$ such that $\phi(y')>0$? Of course, if $y\in C$ then we are done. But what if $y\in\overline{C}\setminus C$? Maybe convexity is not needed here, but in my case, convexity is given.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

Since $\phi(y)>0$, there is an interval around it, say $(\delta,\varepsilon)$, with $0<\delta<\varepsilon$. Now, since $\phi$ is continuous, $\phi^{-1}(\delta,\varepsilon)$ will be open, and in particular, we can take the component containing $y$ and see that it intersects $C$. This implies there is an element $y'\in C$ such that $\phi(y')\in(\delta,\varepsilon)$. Equivalently, $\phi(y')>0$.

share|improve this answer
    
why can't it be that $\phi^{-1}(\delta,\varepsilon)\cap C=\emptyset$? –  Andy Teich Mar 11 '13 at 13:23
1  
One of the equivalent definitions of $\overline{C}$ is this: if $y\in\overline{C}$, then any open set containing $y$ contains a point $y'$ in $C$ distinct from $y$. Note that this is where the convexity is required (perhaps less, I think you can do it with connected). If $C$ was the unit ball unioned with $(2,2,2)$, then the fact that $\phi(2,2,2)>0$ has no bearing on the unit ball. –  Clayton Mar 11 '13 at 13:24
    
@AndyTeich: Which part has you confused? The characterization I gave is indeed for boundaries, and so we somehow have to exclude bad points (like the $(2,2,2)$ above). In order to do this, we impose some niceness condition on the set, say convexity or connectedness. –  Clayton Mar 11 '13 at 13:46

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.