Why is $x_0+\frac{r_0}{2}x \in B(x_0,r_0)$, $\|x\| \leq 1$?

Question

Why is $x_0+\frac{r_0}{2}x \in B(x_0,r_0)$?

Where $B(x_0,r_0) \subset E$ and $x \in E$, $\|x\| \leq 1$. $E$ is Banach (except that this probably doesn't matter here).

Answer 1

Because$$\left\lVert x_0+\frac{r_0}2x-x_0\right\rVert=\frac{r_0}2\lVert x\rVert\leqslant\frac{r_0}2<r_0.$$

Answer 2

In general if you have a normed space, then

$$ B(x,r) = \{x+ry : y \in B(0,1)\} $$

Let's see both inclusions: if $z \in B(x,r)$, then $\|z-x\| < r$. Thus $\frac{z-x}{r}$ is in $B(0,1)$, and we can always write

$$ z = x + r\frac{z-x}{r}. $$

Reciprocally, if we have $y$ in the unit ball, then

$$ \|x+ry -x\| = \|ry\| = r\|y\| < r $$

and so $x+ry \in B(x,r)$.

In your example, $\frac{1}{2}x$ is in the unit ball, and so $x_0 + r\frac{1}{2}x \in B(x_0,r)$.