Prove that $(x-1)^2 + (y-1)^2 \leq 1$ is convex with constraints

Question

Let $D = \{(x, y, z) \in \mathbb{R}^3: (x-1)^2 + (y-1)^2 \leq 1, \space x - z \geq 1,\space y + z = 0\}$ be a set. How do I show that it's convex or concave? It should be easily solved by definition, but instead can we re-formulate it, setting $x\geq1-y$ as condition, so that the elements of $D$ are defined as $(x, y) \in \mathbb{R}^2$?

Answer 1

$D$ is the intersection of the interior of a cylinder with a half-space and a plane. All 3 sets being convex, their intersection is convex.

Edit: If you want to $\mathbf {manually}$ check this, then let $A=(x_1, y_1, z_1)$ and $B=(x_2, y_2, _z2)$ be two points in $D$. Let $C=\lambda A + (1-\lambda) B=(x_3, y_3, z_3)$, with $0\leq \lambda \leq 1$, be a point on the $[A, B]$ line segment. We need to prove that $C\in D$:

Since $y_1+z_1=0$ and $y_2+z_2=0$, it's easy to show that $y_3+z_3=\lambda y_1+(1-\lambda)y_2+\lambda z_1+(1-\lambda)z_2=\lambda(y_1+z_1)+(1-\lambda)(y_2+z_2)=0$.

Likewise, it's easily verified that $x_3-z_3\geq 1$.

Finally, using the fact that $1=\lambda + (1-\lambda)$ $$\begin{split} (x_3-1)^2 &= \lambda^2(x_1-1)^2 +2\lambda(1-\lambda)(x_1-1)(x_2-1)+(1-\lambda)^2(x_2-1)^2\\ (y_3-1)^2 &= \lambda^2(y_1-1)^2 +2\lambda(1-\lambda)(y_1-1)(y_2-1)+(1-\lambda)^2(y_2-1)^2 \end{split}$$ Summing the two and using the fact that $(x_1-1)^2+(y_1-1)^2\leq 1$ $$ (x_3-1)^2+(y_3-1)^2\leq\lambda^2+2\lambda(1-\lambda)\left [(x_1-1)(x_2-1) + (y_1-1)(y_2-1)\right]+(1-\lambda)^2$$

Now, notice that for any real numbers $a, b$, you have $2ab\leq a^2+b^2$, so $$\begin{split} 2(x_1-1)(x_2-1)&\leq (x_1-1)^2+(x_2-1)^2 &\leq 1\\ 2(y_1-1)(y_2-1)&\leq (y_1-1)^2+(y_2-1)^2 &\leq 1\\ \end{split}$$ Plugging this above, $$(x_3-1)^2+(y_3-1)^2\leq \lambda^2+2\lambda(1-\lambda)+(1-\lambda)^2 = 1$$