Let $X_1,...X_n$ be iid $N(0,\sigma^2)$ where $\sigma$ is unknown. My task is to find the BLUE for $\sigma$ within the set of linear functions of $|X_i|$ for $i=1,...,n$. Here is my work thus far:
First, let $T=\sum_{i=1}^nc_i|X_i|$ be an estimator for $\sigma$. Since $T$ is assumed to be unbiased, it must be true that $E[T]=\sigma$, and by the linearity of $E[.]$, it must also be true that: $$E[|X|]\sum_{i=1}^nc_i=\sigma.$$
By performing the appropriate transformation of $X$, it can be shown via the integration method that $E[| X|]=\sigma\sqrt\frac{2}{\pi} $.
At this point, I don't really know what to do. I can solve for $\sum_{i=1}^nc_i$, but I'm not even sure what this quantity represents, or if it's what I'm supposed to be looking for. Any help appreciated.
We have
\begin{align} E_{\sigma}(T)&=\sum_{i=1}^n c_iE_{\sigma}(|X_i|) \\&=\sigma\sqrt{\frac{2}{\pi}}\sum_{i=1}^n c_i \end{align}
For $T$ to be an unbiased estimator of $\sigma$, we must have
$$\sqrt{\frac{2}{\pi}}\sum_{i=1}^n c_i=1$$
Or, $$\sum_{i=1}^n c_i=\sqrt{\frac{\pi}{2}}$$
Now,
\begin{align} \operatorname{Var}_{\sigma}(T)&=\sum_{i=1}^n c_i^2\operatorname{Var}_{\sigma}(|X_i|) \\&=\sum_{i=1}^n c_i^2 \left[E_{\sigma}(X_i^2)-(E_{\sigma}(|X_i|))^2\right] \\&=\sum_{i=1}^n c_i^2 \left[\sigma^2+\frac{2}{\pi}\sigma^2\right] \\&=\sigma^2\left(1+\frac{2}{\pi}\right)\sum_{i=1}^n c_i^2 \end{align}
For $T$ to be the BLUE of $\sigma$, the value of $\operatorname{Var}_{\sigma}(T)$ has to the minimum among the variances of all linear unbiased estimators of $\sigma$.
So your problem boils down to
$$\text{ Minimize }\qquad\sum_{i=1}^n c_i^2 \\\quad\qquad\text{ subject to }\qquad\sum_{i=1}^n c_i=\sqrt{\frac{\pi}{2}}$$
Can you proceed now, using some known inequalities, or by using Lagrange multipliers?
By Cauchy-Schwarz,
$$\sum_{i=1}^n c_i^2\sum_{i=1}^n \left(\frac{1}{\sqrt{n}}\right)^2\ge \left(\sum_{i=1}^n \frac{c_i}{\sqrt{n}}\right)^2$$
Or,
$$\sum_{i=1}^n c_i^2 \ge \frac{1}{n}\left(\sum_{i=1}^n c_i\right)^2$$
Equality holds if and only if
\begin{align} c_i & \propto \frac{1}{\sqrt n}\quad ,\,i=1,2,\ldots,n \\ \iff c_i&=\frac{k}{\sqrt n}\quad\text{ for some }k \end{align}
From the constraint $\sum_{i=1}^n c_i=\sqrt{\frac{\pi}{2}}$, you will get $k$.