Consider two equivalent norms on the real vector space X. Denote the two norms as $||\cdot||_1$ and $||\cdot||_2$. We have for $b>a>0$, $$a||x||_2\leq ||x||_1\leq b||x||_2.$$
Let $B_r(x_0)$ be an open ball in the norm $||\cdot||_1$ and $B'_r(x_0)$ be an open ball in the norm $||\cdot||_2$.
We want to show that $$B'_{r/b}(x_0)\subset B_r(x_0)\subset B'_{r/a}(x_0)\quad (1).$$
Do we use the definition of open balls to prove the relation $(1)$? Do we have to show that a point $x$ in the open ball $B'_{r/a}(x_0)$ also belongs to the open ball $B_r(x_0)$?
For a point $x\in B'_{r/a}(x_0)$, we have $||x-x_0||_2<r/a\implies a||x-x_0||_2<r$ but $a||x-x_0||_2\leq||x-x_0||_1$ and $||x-x_0||_1<r$ so that leads to the conclusion that $x\in B_r(x_0)$.
For the reverse implication, we have $x\in B(x_0)$, we have $||x-x_0||_1<r\implies\frac{1}{b}||x-x_0||_1<r/b$ but $\frac{1}{b}||x-x_0||_1\leq||x-x_0||_2$ and $||x-x_0||_2<r$ so that leads to the conclusion that $x\in B'_{r/b}(x_0)$.
Are my arguments correct? Can I improve the rigor of the proof? Thanks.
Best Answer
In the first part of your proof, you are supposing that $x\in B^\prime_{r/a}(x_0)$ and attempting to prove that $x\in B_r(x_0)$. But this would prove the inclusion $$ B^\prime_{r/a}(x_0)\subseteq B_r(x_0) $$ which is the opposite of what you are trying to prove.
The problem is that the ineqalities $$ a \lVert x-x_0\rVert_2<r \quad\text{and}\quad a \lVert x-x_0\rVert_2 \leq \lVert x-x_0 \rVert_1 $$ do not imply that $\lVert x-x_0 \rVert_1 < r$.
For similar reasons, the second part of your proof is also incorrect.
I'll go over how you could prove the first part; the second part can be solved similarly.
To prove the inclusion $$ B_r(x_0) \subseteq B^\prime_{r/a}(x_0), $$ we will make use of the inequality $$ a\lVert x \rVert_2 \leq \lVert x\rVert_1 \iff \lVert x \rVert_2 \leq \frac{1}{a}\lVert x\rVert_1 \qquad\forall x\in X. $$ Now, suppose that $x\in B_r(x_0)$. Then $$ \lVert x-x_0 \rVert_2 \leq \frac{1}{a}\underbrace{\lVert x-x_0 \rVert_1}_{<r \text{ since }x\in B_r(x_0)} < \frac{r}{a} $$ so $x\in B^\prime_{r/a}(x_0)$.