Normed Spaces and Balls

Question

Let $(X, \Vert \cdot \Vert)$ be a normed space. Show that $B_{\Vert \cdot \Vert} (x, r) = x + r B_{\Vert \cdot \Vert} (0, 1)$ for $x \in X$ and $r > 0$.

$B_{\Vert \cdot \Vert} (x, r)$ contains all elements that are "$\Vert \cdot \Vert$" away from $x$ within a diameter of $r$, right? This contains $x$ itself. Now, $B_{\Vert \cdot \Vert} (0, 1)$ contains the same elements (excluding $x$) "scaled down" by $r$ (since its a diameter of $1$ instead of $r$), so by multiplying the ball by $r$ we "scale it back up," giving us the same elements of $B_{\Vert \cdot \Vert} (x, r)$, excluding $x$. By adding $x$, we get the original ball.

I'm honestly not even sure if I'm understanding the problem correctly. Should I instead be trying to show that given element $a \in B_{\Vert \cdot \Vert} (x, r)$, I can also get $a$ by using the RHS addition? How should I go about proving the problem? Any help would be great. Thank you.

Answer 1

Note that $$\begin{aligned} y\in B(x,r)\quad&\Leftrightarrow\quad\|y-x\|<r\\ &\Leftrightarrow\quad y-x\in B(0,r)=rB(0,1)\\ &\Leftrightarrow\quad y\in x+rB(0,1). \end{aligned}$$ And we are done.

Answer 2

Let $a \in B_{\|\cdot\|} (x, r)$. By definition of open ball it is $\|a-x\| < r$ so $a - x \in B_{\|\cdot\|} (0, r)$. Furthermore:

$$\left\|\frac1r (a-x)\right\| = \frac1r \|a-x\| < \frac{r}r = 1 \implies \frac1r (a-x) \in B_{\|\cdot\|} (0, 1)$$

Now we have:

$$a = x + r \cdot \underbrace{\frac1r(a - x)}_{\in B_{\|\cdot\|} (0, r)} \in x + rB_{\|\cdot\|} (0, r)$$

Conversely, let $a \in x + rB_{\|\cdot\|} (0, 1)$. By defintion, there exists $y \in B_{\|\cdot\|} (0, 1)$ such that $a= x + ry$.

Therefore, $$\|a - x\| = \|ry\| = r \underbrace{\|y\|}_{<1}< r \implies a \in B_{\|\cdot\|} (x, r)$$

We conclude $B_{\|\cdot\|} (x, r) = x + rB_{\|\cdot\|} (0, 1)$.