let $a\in(0,1)$, let $f:M\times N\times [0,T)\rightarrow \mathbb{R}$ a $C^{\infty}$ function, with $M,N$ compact manifolds. Assume that we have a closed, convex set $F$ of such functions such that the subset $W=\{f\in F | $ there exist $x,y\in N$ with $f(p,x,t)> af(p,y,t)\}$ is bounded for every $p,t$, then why we can find a positive constant $C$, independent of $p,t$, such that $\min_N f(p,q,t)\geq a\max_N f(p,q,t) - C$
find a bound on a particular class of functions
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real-analysis
analysis
functions
functional-analysis
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1You might get more answers if you make the question look nicer and more readable. – 2011-05-08
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0How do you know that you can find such a constant $C$? Is this a homework problem? Nothing wrong with that, just that it might help if people knew where the problem was coming from. – 2011-05-09
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0@Gerry Myerson,@Rasmus It is a lemma. It arises in the following way: take a $C^{\infty}$ family of metrics $g(t)$ on a Riemannian maniford $M$. For every $t$ fix isometries $T_p M\cong_{g(t)} V$ where $V$ is a vector space of right dimension. consider the space of algebraic curvature tensors of $V$, $F^{'}$ is a pinching set in such space and $f(p,x,t)=K_{(p,t)}(x)$ is the sectional curvature, computed on the 2-plane $x$, associated to the curvature tensor of $g(t)$ in $p$, $F$ is the set associated to $F^{'}$. – 2011-05-09
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1When you say, "it is a lemma," do you mean that it is stated and proved as a lemma in some paper? If so, doesn't the proof in the paper tell you why you can find that positive constant $C$? – 2011-05-09
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0@Gerry Myerson this is stated in the last part of the proof as something that "should be obvious" and so not proved there :) – 2011-05-09
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1We're making progress. Is this proof published? Maybe if you edit your question to include a link or reference, someone will be able to work out an answer from the context. – 2011-05-10