The course this question is from is called Wavefield imaging, It gives the vector p-norm of an N-by-1 vector $x=[x_1,x_2,…,x_N]^T$ is defined as
$$
||x||_p=\left(\sum_{n=1}^N |x_n|^p \right)^{1/p}
$$
Suppose we are given a 3-by-1 data vector $d=[d_1,d_2,d_3]^T$ with $d_1\geqslant d_2 \geqslant d_3 $. Let the system matrix be given by $G=[1,1,1]^T$ and let $x$ be real-valued scalar. Verify that
a)$x_{opt}=\frac{d_1+d_2+d_3}{3}$ solves the minimization problem $x = \underset{x \in R}{\mathrm{argmin}} ||d-Gx||_2$. The solution is equal to the arithmetic mean of the data.
In my attempted solution I tried to replace $x$ in $||d-Gx||_2$ with $x_{opt}$ as follows…
$$
Gx=[1,1,1]^T \cdot \frac{d_1+d_2+d_3}{3}= \begin{bmatrix}
\frac{d_1+d_2+d_3}{3} \\
\frac{d_1+d_2+d_3}{3} \\
\frac{d_1+d_2+d_3}{3}
\end{bmatrix}
$$
Then I tried to put the resulting vector into $||d-Gx||_2$
$$
||x||_2=\sqrt{\left|d_1-\frac{d_1+d_2+d_3}{3}\right|^2+\left|d_2-\frac{d_1+d_2+d_3}{3}\right|^2+\left|d_3-\frac{d_1+d_2+d_3}{3}\right|^2}
$$
my final solution was like this
$$
||x||_2=\frac{d_1^2+d_2^2+d_3^2-4d_1d_2-4d_1d_3-4d_2d_3}{3}
$$
I couldn't arrive at the correct solution, I checked some textbooks and the internet trying to figure out what I am doing wrong but I couldn't figure out how to arrive at the correct solution so any help would be much appreciated.
Best Answer
Your minimization problem is equivalent to minimizing the function
$$\|Gx-d\|^2=(x-d_1)^2+(x-d_2)^2+(x-d_3)^2$$
This is a quadratic in the variable $x$ whose minimal value is attained at $x=\frac{d_1+d_2+d_3}{3}$, which can easily be verified by differentiation.