# Log-sum inequality

Let $$(a^n)_{n \in \mathbb{N}} \subset \mathbb{R}^k$$ be a sequence of vectors with non-negative components and with $$\sum_{i=1}^k a^n_i=1$$ for all $$n\in\mathbb{N}$$. Then we have for each $$n\in\mathbb{N}$$ by the log-sum inequality:

$$\sum_{i=1}^k a_i^{n+1} \log\dfrac{a_i^{n+1}}{a_i^n} \geq \left(\sum_{i=1}^k a_i^{n+1}\right) \log\dfrac{\sum_{i=1}^k a_i^{n+1}}{\sum_{i=1}^k a_i^{n}}=0,$$

where the equality holds if and only if $$a_i^{n+1}/a_i^n$$ is constant for any $$i=1,...,k$$.

Now assume that

$$\lim\limits_{n\rightarrow \infty}\sum_{i=1}^k a_i^{n+1} \log\dfrac{a_i^{n+1}}{a_i^n} = 0.$$

Can we conclude that $$\lim\limits_{n\rightarrow \infty} \dfrac{a_i^{n+1}}{a_i^n} = 1$$? Intuitively yes, but how can this be proven rigorously?

• I find your notation $a_i^n$ confusing and keep wanting to see the $n$ as an exponent rather than an index. I would write this as $a_{i, n}$ or $a_{n, i}$. Dec 17, 2018 at 18:18
Not necessarily. For instance, put $$k=2$$ and for each $$n\in\Bbb N$$ put $$a^n_1=\frac 1{2^n}$$ and $$a^n_2=1-\frac 1{2^n}$$. Then
$$\lim_{n\to\infty}\sum_{i=1}^k a_i^{n+1} \log\dfrac{a_i^{n+1}}{a_i^n}= \lim_{n\to\infty}\frac 1{2^{n+1}}\log\dfrac{\frac 1{2^{n+1}}}{\frac 1{2^{n}}}+ \left(1-\frac 1{2^{n+1}}\right)\log\dfrac{1-\frac 1{2^{n+1}}}{1-\frac 1{2^{n}}}=0,$$
but $$\lim\limits_{n\rightarrow \infty} \dfrac{a_1^{n+1}}{a_1^n} = \frac 12.$$