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Define $f:[0,\infty) \times [0,\infty) \to [0,\infty)$ by $f(x.y)=(x-1)^2+(y-1)^2$.

Question: Do there exist continuous functions $g,h:[0,\infty) \times [0,\infty) \to [0,\infty)$, satisfying

  1. $g(x,y)=0$ if and only if $xy=1$.
  2. $h(x,y)=0$ if and only if $x=y$.
  3. $f(x,y) \le g(x,y)+h(x,y)$.

Comment: The motivation comes from the case were $x,y$ are interpreted as singular values of a $2 \times 2$ matrix. Then $f(x,y)$ is the distance of the matrix from $\operatorname{SO}(2)$. $g$ and $h$ are interpreted as measures for the deviation of the matrix from being area-preserving and conformal, respectively.

Asaf Shachar
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1 Answers1

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Let $z = x + i y, \, F(z) = (z-(1+i))^2$. Then $|F(z)| = |(z-(1+i))|^2 = f(x,y)$.

Now set $G(z) = z^2 - 2i$. Then $\Re G(z) = (x-y)(x+y), \, \Im G(z) = 2(xy-1)$.

Compute $\frac{F(z)}{G(z)} = \frac{z-(1+i)}{z+(1+i)}$ and thus $$\big|\frac{F(z)}{G(z)}\big| = \big|\frac{z-(1+i)}{z+(1+i)}\big| \le 1$$ if and only if $\Re z + \Im z \ge 0$, which is certainly true if $x \ge 0, \, y \ge 0$.

Therefore you now have for $x, \, y \ge 0$ $$ f(x,y) = |F(z)| \le |G(z)| \le |\Re G(z)| + |\Im G(z)|= |x-y||x+y| + 2|xy-1| $$ and you can read off $g$ and $h$.

Hans Engler
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  • Thanks! this is a very nice answer. I wonder how did you come up with this construction? I can understand that thinking of $F(z)$ is quite natural, but "conjuring" the right candidate for $G$ seems mysterious to me. Was there any particular inspiration or guiding principle behind your construction? (I understand this is an imprecise question). – Asaf Shachar Aug 19 '20 at 18:50
  • Thank you :) Complexifying to come up with $F$, then there is a natural choice for $G$ s.th. $xy$ is the imaginary part, then looking at the ratio which simplifies nicely. Btw this shows that actually $f \le \sqrt{g^2 + h^2}$. – Hans Engler Aug 19 '20 at 20:23