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I have a question about extending the interval of an estimate using continuity.

So suppose that I have positive constants $c_1, c_2, D$, some real number $r >1$ and continuous function $f(x) : [0,\infty) \rightarrow \mathbb{R} $ such that the following estimate holds for all $x \geq 1$

\begin{equation} c_1 \exp( -D x^\frac{2}{r}) \leq f(x) \leq c_2 \exp( -D x^\frac{2}{r}) \end{equation}

How can I extend the above result (with possibly different constants $c_{1}^{*}$ and $c_{2}^{*}$) to all $x>0$ using the function $x \rightarrow \exp( -D x^\frac{2}{r}) $ is bounded on $[0,1]$ above and below by positive constants depending only on $r$?

Will the same argument work if the origional estimate held only for $x > n$ where $n>2$?

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    yes I did thank you2012-04-07

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You need to know more about the function $f$. For all you've said so far it might be negative at some points of $(0,1]$. But it would work if $f$ is bounded below by some positive constant on $[0,1]$ (we already know by continuity that it's bounded above there), since $ \exp(-D x^{2/r})$ is bounded above and below by positive constants on that interval (they depend on $D$, not on $r$, though). Namely if $0 < a \le f(x) \le b$ on $[0,1]$, and knowing that $\exp(-D) \le \exp(-D x^{2/r}) \le 1$ on $[0,1]$, you have $a \le f(x)/\exp(-D x^{2/r}) \le b/\exp(-D)$ there, so $a \exp(-D x^{2/r}) \le f(x) \le b \exp(D) \exp(-D x^{2/r})$. Thus on $[0,\infty)$ you have $\min(c_1, a) \exp(-D x^{2/r}) \le f(x) \le \max(c_2,\exp(D)) \exp(-D x^{2/r})$.

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    The only thing special is that the values of $\exp(-D x^{2/r})$ at $0$ and $1$ don't depend on $r$.2012-04-09