1

I want to say: The $x$ that minimizes $f(x)$ while keeping $f(x)$ above $t$.

I know I can write "the $x$ that minimizes $f(x)$" like this: $x = \arg\min_x f(x)$.

How would I complete that to say "while keeping $f(x)$ above $t$"?

amWhy
  • 209,954
ak321
  • 21
  • How about $\forall y:(t\leq f(y))\wedge (t\leq f(x)\leq f(y))$ – Kevin Jan 02 '23 at 17:03
  • 4
    $\arg\min_{x, f(x) > t} f(x)$ – lhf Jan 02 '23 at 17:04
  • "arg" ?? ....Write $t(x)=\min {f(y):y>t}$ or $f(x)=\min_{y>t}f(t).$ – DanielWainfleet Jan 02 '23 at 17:07
  • 1
    @DanielWainfleet "$\min_x f(x)$" returns the minimum function value of $f(x)$, whereas $\operatorname{argmin}x f(x)$ returns the value(s) of $x$ that minimise $f(x)$ (as in the "argument" that "minimises" the function). For example, $\min{x \in \Bbb{R}} (3 + (x - 1)^2) = 3$, but $\operatorname{argmin}_{x \in \Bbb{R}} (3 + (x - 1)^2) = 1$ (or perhaps, technically, ${1}$). – Theo Bendit Jan 02 '23 at 17:13

1 Answers1

1

$\newcommand{amin}{\operatorname*{argmin}}$The best suggestion was made by lhf in the comments: $$\amin_{\substack{x \\ f(x) > t}} f(x).$$ You could alternatively use the inverse image notation, i.e. $$f^{-1}(t, \infty) := \{x : f(x) \in (t, \infty)\} = \{x : f(x) > t\}$$ to write $$\amin_{x \in f^{-1}(t, \infty)} f(x).$$

Theo Bendit
  • 50,900