I am wondering if anyone can give me a hint to evaluate the following limit:
$\lim_{\lambda \rightarrow -\infty} \lambda y_1 - \log( \sum_{i=1}^N a_i e^{\lambda y_i})$.
The book I'm reading says that it's equal to $-\log a_1$. Here, $a_i > 0$ however, I do not know the sign of the $y_i's$ -- it is not specified. Furthermore, it can be assumed that $y_1 < y_2 < \dots y_N$.
Clearly, if $y_2 > 0$ (implying that the remaining $y_i$'s except for $y_1$ are positive), I recover the above result. Otherwise, what tricks can I use to evaluate this limit?
I am assuming you meant $y_1<y_2<\dots <y_n$, as if $y_1=y_2$ this is not necessarily true. (You can check it for yourself with e.g. $N=2$, $a_1=a_2=1$ and $y_1=y_2 = 1$.)
Rewrite $$ \lambda y_1 - \log\sum_{i=1}^N a_i e^{\lambda y_i} = \log e^{\lambda y_1} - \log\sum_{i=1}^N a_i e^{\lambda y_i} = \log \frac{e^{\lambda y_1}}{\sum_{i=1}^N a_i e^{\lambda y_i}} = \log \frac{1}{a_1+\sum_{i=2}^N a_i e^{\lambda(y_i-y_1)}}\tag{1} $$ Since $y_i-y_1>0$ for every $2\leq i\leq N$, we have $$ \lim_{\lambda\to-\infty}\sum_{i=2}^N a_i e^{\lambda(y_i-y_1)} = 0\tag{2} $$ and therefore, from (1), $$ \lim_{\lambda\to-\infty} \lambda y_1 - \log\sum_{i=1}^N a_i e^{\lambda y_i} = \log \frac{1}{a_1 + \lim_{\lambda\to-\infty}\sum_{i=2}^N a_i e^{\lambda(y_i-y_1)}} = -\log a_1\,. \tag{3} $$