Absolute Minimums and Maximums — Question 6

Question 6

Let $$D=\{(x,y):x^2+y^2<1\}$$ be the open unit disk, and let $f(x,y)=x+y$.

Tasks

1. Determine whether $f$ has an absolute maximum or minimum on $D$.

2. Find its supremum and infimum and give sequences approaching them.

3. Explain exactly which hypothesis of the Extreme Value Theorem is absent.

Question 6 – Solution

Strategy. Bound the linear function with Cauchy–Schwarz, then determine whether equality points belong to the open disk.

Step 1: Bounds For $(x,y)\in D$, $$|x+y|\le\sqrt{x^2+y^2}\sqrt 2<\sqrt 2.$$ Therefore every function value lies strictly between $-\sqrt 2$ and $\sqrt 2$.

Step 2: Approaching the bounds For $0<r<1$, the point $$\left(\frac r{\sqrt 2},\frac r{\sqrt 2}\right)$$ lies in $D$ and gives $f=r\sqrt 2\to\sqrt 2$ as $r\to 1^-$. Its negative gives values tending to $-\sqrt 2$. Thus $$\boxed{\sup_D f=\sqrt 2},\qquad \boxed{\inf_D f=-\sqrt 2}.$$ Neither value is attained, so $f$ has no absolute maximum or minimum on $D$.

Step 3: The theorem The function is continuous and the disk is bounded, but $D$ is not closed: its boundary circle is omitted. Hence the domain is not compact, and the Extreme Value Theorem does not guarantee attained extrema. The missing equality points are precisely $$\left(\frac 1{\sqrt 2},\frac 1{\sqrt 2}\right) \quad\text{and}\quad \left(-\frac 1{\sqrt 2},-\frac 1{\sqrt 2}\right).$$