infinite dimensional constrained minimization problem I have come up with the following constrained minimization problem:
\begin{eqnarray}
\min\ \sum_{i=1}^\infty x_i^2\\
\sum_{i=1}^\infty a_ix_i=1
\end{eqnarray}
If it were a finite-dimensional case it would be easily solved via Lagrange multipliers; in this case I ask your help since I don't know where to begin.
 A: Assume wlog that there are no zero terms in the sequence by 'ommiting' them.
Assuming that the sequence  $a_i $ is square summable,  then one can easily use cauchy schwartz inequality to show that 
$$x_i := \frac {a_i}  {\sum_{i=1}^{\infty}a_i^2} $$
minimizes the quantity $\sum_{i=1}^{\infty}x_i^2$. 

For the case where $a_i$ is not square summable i.e. $\sum_{i=1}^{\infty}a_i^2=\infty$, I will show that $$\inf\{\sum_{i=1}^{\infty}x_i^2: \sum_{i=1}^{\infty}a_ix_i=1\}=0$$
Consider the 'sequence' of sequences $\{x^{(n)}_i\}_{i\geq 1}$, where  $x^{(n)}$ is defined as follows:
1) $x^{(n)}_i=\frac{a_i}{\sum_{j=1}^{n}a_j^2}\,\,\,$  if $i\leq n$
2)  $x^{(n)}_i=0\,\,\,$  if $i> n$
It is easy to see the following that all the sequences $x^{(n)}$ satisfy the constraint i.e. $\sum_{i=1}^{\infty}a_ix^{(n)}_i=1$. Moreover :
$$ \lim_{n\rightarrow \infty} \sum_{i=1}^{\infty}[x^{(n)}_i]^2=\lim_{n\rightarrow \infty}\frac{1}{\sum_{j=1}^{n}a_j^2}=0$$
$\square$
A: Assuming the Cauchy's inequality still holds for infinite series,
$$\left(\sum_{i=1}^{\infty}x_i^2\right)\left(\sum_{i=1}^{\infty}a_i^2\right)\geq \sum_{i=1}^{\infty}x_ia_i=1$$
$$\sum_{i=1}^{\infty}x_i^2\geq \frac{1}{\sum_{i=1}^{\infty}a_i^2}$$
