I want to maximize the function $$u(a)=\sum_{i=0}^\infty r^i\log a_i$$ defined on some proper subset of $V=\{a\in R^\infty:|\sum r^ia_i|<\infty\}$ for some fixed $r\in (0,1)$.
In particular, let $D\subset V$ such that $$a\in D \iff |\sum _{i=0}^{\infty}r^i\log a_i|<\infty $$ and for fixed $M \in R$, define $S \subset V$ such that $$a\in S \iff \sum_{i=0}^{\infty}r^ia_i \leq M$$ I want to solve the following problem $$max_{a\in S \cap D} \sum_{i=0}^{\infty}r^i\log a_i$$
I want to apply the lagrangian method to this problem. But I am worried about whether it is legitimate to use the lagrangian method and also worried about the existence and uniqueness of the solution. That is, let $$ L=\sum_{i=0}^{\infty}r^i\log a_i+\lambda(M-\sum_{i=0}^{\infty}r^ia_i)$$ Carelessly applying the lagrangian method gives the FOC of $$\frac{\partial L}{\partial a_i}=r^i(a_i^{-1}-\lambda)=0$$ Hence we get $$a_0=a_1=...$$ Combined with the constraint $$\sum_{i=0}^{\infty}r^ia_i=M$$ the above equalities give $$a_i=(1-r)M$$ Can this solution justified? Any hint would be really appreciated!
OK. I solved it. It is obvious that the constraint $$\sum_{i=0}^{\infty}r^ia_i=M$$ should hold.
Claim: If $M>0,$ then $a=(1-r)M$ is the unique solution.
Proof: Let $b$ be the sequence such that $\sum_{i=0}^{\infty}r^ib_i=M$ and $b\neq a. $ Then, there must exist $n,m\in N$ such that $n \neq m$ and $b_n>(1-r)M, \ ~ b_m<(1-r)M$. Take $\epsilon _n, \epsilon _m>0$ such that $r^n\epsilon _n=r^m\epsilon _m$
Define $c_n=b_n-\epsilon_n, c_m=b_m+\epsilon _m, c_k=b_k$ for $k\neq n,m$. Then, $$\sum_{i=0}^{\infty}r^i\log c_i-\sum_{i=0}^{\infty}r^i\log b_i=r^n[\log(b_n-\epsilon_n)-\log(b_n)]+r^m[\log(b_m+\epsilon_m)-\log(b_m)]$$
If we devide both sides by $r^n\epsilon_n=r^m\epsilon_m$ and take $\epsilon_n\rightarrow0$, we have $-b_n^{-1}+b_m^{-1}$. Since we have chosen $n,m$ that satisfiy $b_n>b_m$, it follows that $-b_n^{-1}+b_m^{-1}>0$. This shows that $u(a)\geq u(c) > u(b)$.