Minimizing $ |t|^2 + a(|t + u| - |u|) $ over $ t \in \mathbb{C} $

Question

I'm not so experienced with Complex Analysis and I somewhere stumbled upon the following function $ f: \mathbb{C}^2 \rightarrow \mathbb{R} $ that I need to minimize with respect to the Complex variable $ t $ ($ t \neq 0 $):

$$ f(t, u) = |t|^2 + a(|t + u| - |u|), $$

where $ u $ is also Complex and $ a $ is a known Real constant (I can also assume $ a \geq 0 $). I wonder if I can find an explicit solution (even considering cases) for $ \hat{t} $ the minimizer, and if so I would appreciate if one can help me how to proceed?

Answer 1

Expanding on the hint in my comment, here is a way to solve it without calculus, using just basic algebra and the triangle inequality for complex numbers.

• For constant $u$ the last term in $f(t) = |t|^2 + a|t + u| - a|u|$ is a constant, so minimizing $f(t)$ is equivalent to minimizing $g(t) = |t|^2 + a|t + u|\,$.

• $|t+u| \ge \big||t| - |u|\big|$ by the triangle inequality, with equality iff $t = \lambda u$ with $\lambda \in \mathbb R^-$.

• Then $\,g(t) = |t|^2 + a|t + u| \;\ge\; |t|^2+a\big||t|-|u|\big|\,$ when $\,a \ge 0\,$.

• With $|t|=r \in \mathbb R^+$ the real function $h(r)=r^2 + a\big|r-|u|\big|$ has a minimum at either $\,r_0=|u|\,$ or $\,r_0=a/2\,$ depending on $\,a,u\,$.

• Piecing it all together, $\,f(t)+a|u|=g(t)\ge h\left(|t|\right) \ge h(r_0)\,$ with equality when $\,|t|=r_0\,$ and the triangle inequality becomes an equality, so $\,t_0=-\frac{u}{|u|}r_0\,$.

Answer 2

You can avoid complex variables and transfer all of them to the Euclidean space. Let us denote the real part and the imaginary part of a complex number $ u $ as $ R(u) $ and $ I(u) $, so that $ u = R(u) + i I(u) $. Using this notation, you have

$$ f(t, u) = (R(t))^2 + (I(t))^2 + a \left( \sqrt{(R(t) + R(u))^2 + (I(t) + I(u))^2} - \sqrt{(R(u))^2 + (I(u))^2} \right) . $$

Minimizing $ f(t, u) $ with respect to $ t $ is same as minimizing

$$ g(x, y) = x^2 + y^2 + a \left( \sqrt{(x + R(u))^2 + (y + I(u))^2} - \sqrt{(R(u))^2 + (I(u))^2} \right) $$

with respect to real valued variables $ x $ and $ y $ and then taking $ \hat{t} = \hat{x} + i \hat{y} $, where $ (\hat{x} , \hat{y}) $ minimizes $ g(x, y) $. Since $ g(x, y) $ is a differentiable function of $ x, y $, you can use derivative of $ g(x, y) $ with respect to $ x, y $ to find the minimizers.

Answer 3

This is just expanding on @dxiv: idea. Say we have a normed space $E$, $\phi_1$, $\phi_2$ increasing functions on $[0, \infty)$, $x_1$, $x_2$ points in $E$. Consider the function $$E\ni x \mapsto f(x) \colon= \phi_1 (d(x, x_1)) + \phi_2 (d(x, x_2))$$ If $f$ has a point of minimum, then it has a point of minimum on the segment $[x_1, x_2]$.

Proof: It is enough to show that for every $x\in E$ there exists $x'\in [x_1, x_2]$ with $f(x')\le f(x)$. We distinguish two cases:

1. $d(x_1, x) > d(x_1, x_2)$. Then we have $f(x_2)\le f(x)$.

2. $d(x_1, x)\le d(x_1, x_2)$. Consider $x'$ the (unique) point on the segment $[x_1, x_2]$ such that $ d(x_1, x') = d(x_1, x)$. From the triangle inequality we get $d(x', x_2) \le d(x, x_2)$. We conclude $f(x')\le f(x)$.