Prove convexity of a particular set

Question

How would one prove the convexity of a set such as

$$\left \{(x, y) : (x^2 + y^2)^{n} < x + y \right \}$$

where $n$ is some positive integer? For $n = 1$ we have a circle and the result is trivial, but for any higher $n$ I am at a loss.

Answer 1

If $f$ is convex then $\{x | f(x) \le 0 \}$ is convex.

If $f$ is convex and $g$ is convex and non decreasing then $g \circ f$ is convex.

Take $f(x) = x_1^2+x_2^2$, $g(t)= t^n 1_{[0,\infty)}(t)$. Then $g\circ f$ is convex. If $l$ is linear then $g\circ f-l$ is also convex. Here we have $l(x) = x_1+x_2$.

Then $\{x |g(f(x))-l(x) \le 0 \}$ is convex.

Answer 2

You can show it directly by using that

• (1): $t^2$ is convex: $(pt_1 + qt_2)^2 \leq pt_1^2 + qt_2^2$ for $p \in [0,1], q = 1-p$
• (2): $t^n$ is convex for $t\geq 0$: $(pt_1 + qt_2)^n \leq pt_1^n + qt_2^n$ for $p \in [0,1], q = 1-p$

So, consider $p(x,y) + q(u,v)$ with $(x,y),(u,v) \in C = \left \{(x, y) : (x^2 + y^2)^{n} < x + y \right \}$ with $p \in [0,1], q = 1-p$. To show is

• $ \color{blue}{((px+qu)^2 + (py+qv)^2)^n <} px+qu + py+qv = \color{blue}{p(x+y) + q(u+v)}$

The calculation is straight forward: $$ \color{blue}{((px+qu)^2 + (py+qv)^2)^n} \stackrel{(1)}{\leq} (px^2 + qu^2 + py^2 + qv^2)^n = (p(x^2 +y^2) + q(u^2+ v^2))^n$$ $$\stackrel{(2)}{\leq} p(x^2+y^2)^n + q(u^2+v^2)^n \color{blue}{\stackrel{(x,y),(u,v)\in C}{<} p(x+y) + q(u+v)}$$