# find a bound on a particular class of functions

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 $\min_N f(p,q,t)\geq a\max_N f(p,q,t) - C$

-
You might get more answers if you make the question look nicer and more readable. – Rasmus May 8 '11 at 21:04
How 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. – Gerry Myerson May 9 '11 at 0:23
@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 $F^{'}$. – unk2 May 9 '11 at 12:34
When 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 $C$? – Gerry Myerson May 9 '11 at 12:45
We'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. – Gerry Myerson May 10 '11 at 0:16
show 1 more comment