Consider the functional $$\min f= \min \bigg(\sum_{n=1}^{\infty}(2^{-n}x_n-2^{-\frac{3n}{2}})^2+\lambda \sum_{n=1}^{\infty}x_n^2\bigg)$$ where $x_1, x_2, \cdots $ are infinite real variables and $\lambda$ is Lagrange multiplier. How to find minimizer of the above functional?
My Idea:I know that the function $f$ has a unique minimizer because it is strictly convex, bounded below. From theory, I know the first order necessary criteria according to which $$\frac{\partial f}{\partial x_n}=0\ \forall n.$$ Is this criteria sufficient also? Please help.
Calculating the stationary points
$$ \frac{\partial f}{\partial x_n} = 2\cdot 2^{-n}(2^{-n}x_n-2^{-\frac{3n}{2}})+2\lambda x_n = 0 $$
we have
$$ x_n^* = \frac{2^{-\frac{5n}{2}}}{2^{-2n}+\lambda} $$
then $x_n^*$ should obey the restriction
$$ \sum_{n}^{\infty}(x_n^*)^2 = \sum_{n}^{\infty}\frac{2^{-5n}}{(2^{-2n}+\lambda)^2}=0\Rightarrow \lambda \to\infty $$
This is possible only if $x_n^* = 0\forall n$ so the minimum is
$$ \min f = \sum_{n=1}^{\infty}2^{-3n}= \frac 17 $$