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Suppose that $A$ and $B$ are compact subsets of $\mathbb{R}^{n}$. Consider the function $G(x;a,b)$ where $x\in\mathbb{R}^{n}$, $a\in A$ and $b\in B$. We also suppose that $G(x;a,\cdot)$ is uniformly continuous. If it holds that \begin{align} G(x;a,b)\geq \delta \end{align} is it possible for all $\lambda\in{\cal B}(b,r)\cap B$ (where $r=r(b)>0$), to hold
\begin{align} G(x;a,\lambda)\geq \frac{3\delta}{4} \end{align} where ${\cal B}$ stands for the open ball?

I think the answer is yes and has something to do with the uniform continuity. Any suggestion will be highly appreciated.

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No measure-theory in there. –  Did Dec 14 '12 at 15:13
    
The answer to the question is YES: choose $\lambda=b$. –  Did Dec 14 '12 at 15:13
    
A counterexample to the question you probably mean to ask could be $G(x;a,c)=\exp(-c^2x^2)$ around $b=0$. –  Did Dec 14 '12 at 15:16
    
Listen, you might want to seriously revise your post. I tried to indicate that the existence of some $\lambda$ in $B(b,r)$ is trivial since $\lambda=b$ fits the bill. If you want the inequality to hold for every $\lambda$ in $B(b,r)\cap B$, please modify the question--and then ponder the example in my other comment. –  Did Dec 14 '12 at 15:22
    
@did, actually i did not understand your example. Could you be more clear please ? –  Peter 5 Dec 14 '12 at 15:37

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up vote 1 down vote accepted

A counterexample to the question you probably mean to ask could be $$ G(x;a,c)=\frac1{1+|c\cdot x|}, $$ around $b=0$. Then, $G(x;a,0)=1$ but $G(x;a,b)\geqslant3/4$ only if $|x|⩽1/(3|b|)$ hence $G(x;a,λ)\geqslant3/4$ uniformly on $x$ is impossible for each $λ≠0$. A fortiori, $G(x;a,λ)\geqslant3/4$ is impossible uniformly on $x$ and on every $|λ|⩽r(0)$ as soon as $r(0)>0$.

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