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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?

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  • $\begingroup$ This isn't true. The sphere in $\mathbb{R}^2$ is not convex. The open ball is though. Is that what you meant? (e.g. the origin is not in the unit ball even though it is on the line connecting any two points. $\endgroup$ Mar 26 '13 at 2:29
  • $\begingroup$ Yes that's what I meant...woops $\endgroup$
    – james
    Mar 26 '13 at 2:30
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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 $-1\le a\le 1$.

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

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  • $\begingroup$ But then I have to reduce the problem to what Rudin does. I am avoding this. Let me show you what I did $\endgroup$
    – james
    Mar 26 '13 at 2:36

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