Sup of a function

1. Mar 4, 2004

Rick

I don't have background in analysis, but was looking for
a simple explanation(proof?) of this statement,

Over a compact set, a differentiable function V Rn-> R, with V<0 in that set, then sup(V)<0 in that set

Actually I'm not certain if I interpreted the statement right, so maybe the statement as it is might be wrong/incomplete.

2. Mar 5, 2004

NateTG

You mean
$$f:\Re^n\rightarrow \Re$$
and $$f(v) < 0$$
?

Consider the possibility that the image of the compact set $$S$$ might be $$(-1,0)$$ which certainly has sup $$0$$.

3. Mar 5, 2004

matt grime

The image of a compact set under a continuous (diffble) map is compact.
So considering (0,1) as the image won't get you very far.

Moreover it is bounded and attains its bounds, hence there is some element in the set with V(x)=sup{V(y)}

this is negative by hypothesis