The closed unit ball on $\mathbb{R}^2$ is convex

Question

i realize there is a much much easier way to handle this problem in Rudin which generalizes an $n$ dimensional ball. But I am more interested in doing it this "hard" way.

Consider the closed unit ball $C = \{ (x,y): x^2+y^2\leq1\}$. I want to show that $z = (x_0,y_0)t+(1-t)(x_1,y_1)$ is in the ball. But I honestly do not know how to make the $x^2 + y^2=1 $ condition hold for my line. Should I write my line in cartesian like $y = mx + b$?

Using $z$, I know that $x_0^2 + y_0^2 \leq 1$ and $x_1^2+y_1^2\leq1 \implies t^2(x_0^2+y_0^2)\leq t^2 < t < 1$ and $(1 -t)^2(x_1^2+y_1^2)\leq(1-t)^2< 1$

So both parts are less than one and can we conclude that's done?

Answer 1

We have $0\le t\le 1$. The generic point on the line segment has coordinates $(x,y)$ where $x=tx_0+(1-t)x_1$ and $y=ty_0+(1-t)y_1$.

Calculate $x^2+y^2$. We get, after using the fact that $x_0^2+y_0^2=1$ and $x_1^2+y_1^2=1$, $$x^2+y^2=t^2 +(1-t)^2 +2t(1-t)(x_0x_1+y_0y_1).\tag{$1$}$$

To estimate $x_0x_1+y_0y_1$, note that $$(x_0x_1+y_0y_1)^2+(x_0y_1-x_1y_0)^2=(x_0^2+y_0^2)(x_1^2+y_1^2)=1,$$ and therefore $|x_0x_1+y_0y_1|\le 1$.

Now it is finished. For the expression $t^2+(1-t)^2 +2at(1-t)$ is $\le 1$ when $-1\le a\le 1$.

Answer 2

Hint (after clarifying you meant the unit ball): Use the triangle inequality.