prove min $x$ of $f(x) = \sum_{j = 1}^{n}\sum_{i = 1}^{n} (x_i-a_{ij})^2 $ is the same as $g(x) = \sum_{j = 1}^{n}\sum_{i = 1}^{n} |x_i-a_{ij}|$

Question

Let $x'$ and $y'$ be the points at which $f(x)$ and $g(y)$ are globally minimized, respectively. Can we say that $x'$=$y'$? If so, how can I proof that? I guess the inequality $|x-a|\leq|x|+|a|$ might be useful.

$$f(x) = \sum_{j = 1}^{n}\sum_{i = 1}^{n} (x_i-a_{ij})^2, a_{ij}>0, x \in \mathbb{R}^n , f(x) \in \mathbb{R} $$ $$g(y) = \sum_{j = 1}^{n}\sum_{i = 1}^{n} |y_i-a_{ij}|, a_{ij}>0 ,y \in \mathbb{R}^n, g(y) \in \mathbb{R} $$

Answer 1

The proof is trivial. As suggested in a comment, the minimizer is $(a_1,\dots,a_n)$. In fact, if you observe that in both functions you have the sum of non-negative quantities then the best you can do to minimize them is to make vanishing each member of the sums.

Alternatively you can evaluate the gradient of the two functions and verify that in both cases it is zero in $(a_1,\dots,a_n)$.