I am reading a paper and I have a function $f(a)$ defined by: $f(a):= \sum_{x\in L} e^{-\pi x^2/a} \\$, with $a \ge 1$. the $x$'s are just points in a lattice we are summing over, but it's irrelevant.
We know that $f'(a) \le \frac{n}{2a}f(a),$ with $a \ge 1$.
They conclude that [log$f(a)$]' $\le \frac{n}{2a}$ which gives:
$f(a) \le a^{n/2}f(1)$, for $a \ge 1$.
I understand how they get the log part since taking the derivative yields the fraction but how do they get the the last line?
Integrate $(\log f)'(a)$, from $1$ to $x$; you get (assuming hereafter $f(1) > 0$) $$ \log \frac{f(x)}{f(1)} = \int_1^x (\log f)'(a) \,da \leq \frac{n}{2}\int_1^x \frac{da}{a} = \frac{n}{2}\log x \tag{1} $$ the inequality being preserved by monotonicity of the integral. Now, take the exponential of both sides of (1): $$ \frac{f(x)}{f(1)} \leq e^{\frac{n}{2}\log x} = x^{\frac{n}{2}}\tag{2} $$ by monotonicity of $\exp$. Since $f(1) > 0$, you can rearrange the inequality to get the desired result.