# Show that this soft maximum (log-sum-exp) variant approaches the hard maximum

I am trying to prove something that I believe is related to the logsumexp soft maximum trick but I am getting stuck on formalising the bit where one can claim that one of the exponents dominates every other term in the summand. The question is as follows:

Suppose we have constants $$\delta_i$$ and $$\alpha_{im}$$, for $$1 \leq i \leq n$$, $$1 \leq m \leq M$$, with constraints

1. $$0 = \delta_1 < \delta_2 \leq \delta_3 \leq \dots \leq \delta_n$$.
2. $$0 < \alpha_{im}$$ for all $$i, m$$, and,
3. $$\alpha_{i1} + \alpha_{i2} + \dots + \alpha_{iM} = 1$$ for all $$i$$.

Show that there exists a constant $$C$$ such that:

$$C = \lim_{k \rightarrow \infty} \frac{1}{k}\log\sum_m\left(\sum_i\beta_{im}e^{C - \delta_i}\right)^k,$$ where $$\beta_{im} = (\alpha_{im})^{\frac{1}{k}}$$.

What I did was to use the generalised Pascal's triangle to expand out the inner exponent so that:

$$\begin{split} \text{RHS} &= \lim_{k \rightarrow \infty} \frac{1}{k}\log\left(\sum_i(\sum_m \alpha_{im})e^{k(C-\delta_i)} + \text{lower order terms})\right)\\ &= \lim_{k \rightarrow \infty} \frac{1}{k}\log\left(\sum_ie^{k(C-\delta_i)} + \text{lower order terms})\right) \hspace{1cm}\text{using constraint (3)} \end{split}$$

The bit that I am stuck with is showing that there is a value of $$C$$ so that $$\sum_ie^{k(C-\delta_i)}$$ dominates the 'lower order terms', since, after that, we can reduce the problem to the soft maximum as discussed by John D. Cook:

$$\max(x_1, x_2, ...) = \lim_{k \rightarrow \infty} \operatorname{softmax}(k, x_1, x_2, ...)$$

where

$$\operatorname{softmax}(k, x_1, x_2, ...) = \frac{1}{k}\log(\exp(kx_1) + \exp(kx_2) + \dots)$$