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I am currently reading a treatment on Fourier Series and the Heat trace, where the following estimate is stated:

Suppose $\lambda > 0$ and $t \in (0,\infty)$. We have

\begin{equation} e^{-t\lambda}\lambda^j \leq t^{-j} C(j) e^{-t\lambda/2} \end{equation} where $C(j)$ denotes a constant, dependent on the positive number $j$.

I'd like to understand how to derive this, but struggle to fill in the details - do I need to expand the exponential into its power series expression to see this ?

Many thanks for your help !

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up vote 0 down vote accepted

We have to show that there is a constant $C(j)$ such that for each $x>0$, $e^{-x/2}x^j\leq C(j)$, i.e. for each $s>0$, $e^{-s}2^js^j\leq C(j)$. Since $e^s\geq \frac{s^j}{j!}$, we get that $e^{-s}\leq \frac{j!}{s^j}$ hence $s^je^{-s}\leq j!$, so we get the result with $C(j)=\frac{j!}{2^j}$.

An alternative way: let $f_j(x)=x^je^{-x}$, then $f_j'(x)=e^{-x}(-x^j)-e^{-x}jx^{j-1}=e^{-x}x^{j-1}(j-x)$, so $f_j$ gets its maximum at $x=j$ and $f_j(j)=j^je^{-j}$.

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Many thanks !! Just one question, by working through your argument I get $C(j) = 2^j j!$ Have I done something wrong ? – harlekin May 1 '12 at 14:00

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