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Suppose $${a_n^{s_n} + b_n^{s_n} =1}$$ $$\lim_{n \to \infty} a_n = a >0$$ $$\lim_{n \to \infty} b_n = b >0$$ and $${a^s + b^s =1}$$ where $a+b<1$. Then how can we prove that $\displaystyle\lim_{n \to \infty} s_n = s$?

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There is a $\lim$ on the first relation? –  enzotib Aug 14 '12 at 21:07
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This is not true if eg $a=a_n=1$ and $b=b_n=0$ so you are missing some conditions. Once we get those special cases out of the way, have you tried thinking what happens if the limit is greater than, or less than $s$ - treat the two cases separately. –  Mark Bennet Aug 14 '12 at 21:10
    
Satisfied with an answer below? –  Did Sep 19 '12 at 17:31

3 Answers 3

up vote 3 down vote accepted

Here’s a more routine approach.

If $\lim_{n\to\infty}s_n\ne s$, then either $\limsup_n s_n>s$, or $\liminf_n s_n<s$.

If $\limsup_n s_n>s$, we may assume (by passing to a subsequence if necessary) that there is some $u>s$ such that $s_n\ge u$ and $1>a_n,b_n>0$ for all $n$. Then for all $n$ we have $a_n^{s_n}+b_n^{s_n}\le a_n^u+b_n^u$, so

$$1=\lim_{n\to\infty}(a_n^{s_n}+b_n^{s_n})\le\lim_{n\to\infty}(a_n^u+b_n^u)=a^u+b^u<1\;,$$

which is absurd.

Similarly, if $\liminf_n s_n<s$, we may assume that there is some $u<s$ such that $s_n\le u$ and $a_n,b_n>0$ for all $n$, and we then have

$$1=\lim_{n\to\infty}(a_n^{s_n}+b_n^{s_n})\ge\lim_{n\to\infty}(a_n^u+b_n^u)=a^u+b^u>1\;,$$

which again is absurd. Thus, $\lim_{n\to\infty}s_n=s$.

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Please reverse some inequalities: if $u\gt s$, then $a^u\lt a^s$ for $a$ in $(0,1)$. –  Did Aug 16 '12 at 10:35
    
@did: Argh. This not my day. Thanks. –  Brian M. Scott Aug 16 '12 at 10:40
    
@did: WLOG they are in $(0,1)$, since $a,b>0$ and $a+b<1$. I chose not to go through the details of passing to a subsequence, but I did explicitly say that I was doing so. –  Brian M. Scott Aug 16 '12 at 10:46

An elementary proof is based on the following not-so-usual way to express the convergence to any positive limit not equal to $1$.

Consider for example the fact that $\lim\limits_{n\to\infty}a_n=a$. Since $a\gt0$, $b\gt0$ and $a+b\lt1$, one knows that $0\lt a\lt1$, hence the function $\alpha:x\mapsto\log(x)/\log(a)$ is continuous and decreasing on $x\gt0$. In particular, the fact that $\lim\limits_{n\to\infty}\alpha(a_n)=\alpha(a)=1$ implies that, for every $0\lt\varepsilon\lt1$, there exists some finite $n_\varepsilon$ such that, for every $n\geqslant n_\varepsilon$, $1-\varepsilon\leqslant\alpha(a_n)\leqslant1+\varepsilon$, that is, $a^{1+\varepsilon}\leqslant a_n\leqslant a^{1-\varepsilon}$.

Likewise, one can, and we will, assume that $b^{1+\varepsilon}\leqslant b_n\leqslant b^{1-\varepsilon}$ for every $n\geqslant n_\varepsilon$. Since $a_n^{s_n}+b_n^{s_n}=1$ for every $n$, one gets, for every $n\geqslant n_\varepsilon$, $$ a^{(1+\varepsilon)s_n}+b^{(1+\varepsilon)s_n}\leqslant1\leqslant a^{(1-\varepsilon)s_n}+b^{(1-\varepsilon)s_n}. $$ Since $a^s+b^s=1$ and the function $r\mapsto a^r+b^r$ is decreasing, $(1-\varepsilon)s_n\leqslant s\leqslant(1+\varepsilon)s_n$ for every $n\geqslant n_\varepsilon$, that is, $\frac{s}{1+\varepsilon}\leqslant s_n\leqslant\frac{s}{1-\varepsilon}$. Finally, $\lim\limits_{n\to\infty}s_n=s$.

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Consider the function $$f(x,y,t):=x^t+y^t-1\quad(0<x<1,\ 0<y<1,\ t\in{\mathbb R})\ .$$ By assumption $(a,b,s)\in{\rm dom}(f)$ and $f(a,b,s)=0$. Furthermore ${\partial f\over\partial t}=\log x\ x^t+\log y\ y^t$, whence $$\left.{\partial f\over\partial t}\right|_{(a,b,s)}=\log a\ a^s+\log b\ b^s<0\ .$$ It follows by the implicit function theorem that there is an open box $W=U\times V\subset{\mathbb R}^3$ with center $(a,b,s)$ and a $C^1$-function $$\psi:\quad U\to V,\quad (x,y)\to t=\psi(x,y)\ ,$$ such that within this box the equation $f(x,y,t)=0$ is equivalent to $t=\psi(x,y)$.

Coming back to the original question we now have $$\lim_{n\to\infty} s_n=\lim_{n\to\infty} \psi(a_n,b_n)=\psi(a,b)=s\ .$$

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