Showing $f(x)=||x-A||^2$ attains a global minimum in $\mathbb{R}^2$?

Question

Having slight difficulties in understanding how to display that

$f(x)=||x-A||^2$ attains a global minimum for $x,A \in \mathbb{R}^2$, $A$ being fixed.

I first thought that since there's the norm I could simply write

$$f(x) \ge 0 \space \forall x$$

and claim that therefore it attains the minimum value at precisely when $x-A=0$ since by properties of the norm $f(x)=0 \iff x-A=0$.

Another thing I've been thinking would be to look at the expansion of the norm

$$\sqrt{x-A}^2=x-A$$

and then look for critical points. But I don't know how the partial derivatives will leave any variables, since this one has only first order variables.

Answer 1

I first thought that since there's the norm I could simply write $$f(x) \ge 0 \space \forall x$$ and claim that therefore it attains the minimum value at precisely when $x-A=0$ since by properties of the norm $f(x)=0 \iff x-A=0$.

This is probably the easiest way of solving the problem, yes. The minimum is attained at $x=A$, and this is a global minimum.

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In your second part, you write the expression $$\sqrt{x-A}^2=x-A$$ which is nonsensical, since $\sqrt{}$ is a function from $[0,\infty)$ to $\mathbb R$, while $x-A$ is an element of $\mathbb R^2$.

The expression $\|x-A\|^2$ can be written as $(x_1-A_1)^2 + (x_2-A_2)^2$, and using derivation (on both variables), you can quckly show that there exists only one critical point, $x=A$. That, means that if $f$ has a minimum, then the minimum must be attained at $x=A$.