prove $\rho\in K\to ||\sqrt{\rho}'||_{L^2}^2$ is convex, where $K$ is a convex cone

Question

$$K = \{\rho\in H^1(0,1);\rho \ge 0,\sqrt{\rho}\in H^1(0,1)\}$$

I have proved that $K$ is a convex cone. Now I'm asked to prove that $\rho\in K\to ||\sqrt{\rho}'||_{L^2}^2$ is convex.

I tried several ways to compute, but I can only get $$||\sqrt{tu+(1-t)v}'||_{L^2}^2 \le 2t||\sqrt{u}'||_{L^2}^2 + 2(1-t)||\sqrt{v}'||_{L^2}^2 \text{ or }\le \sqrt{t}||\sqrt{u}'||_{L^2}^2 + \sqrt{1-t}||\sqrt{v}'||_{L^2}^2$$ So any ideas how to get convexity?

Answer 1

Hints:

• Compute $\sqrt{\rho}'$.
• Use this to get a nice (integral) expression for $\|\sqrt{\rho}'\|_{L^2}^2$.
• Prove that $\mathbb R^2 \ni (a,b) \mapsto a^2/b$ is convex.