Considering the comparison of norms I have the following proposition from a certain book:
Let $X$ be a vector space and let $\Vert \cdot\Vert_{1}$ and $\Vert \cdot\Vert_{2}$ be two norms on this space.
$\Vert \cdot\Vert_{1}$ is weaker than $\Vert \cdot\Vert_{2}$ iff every open $\Vert \cdot\Vert_{1}$ ball contains and open $\Vert \cdot\Vert_{2}$ open ball.
I have been trying to prove the 'if' direction, that is if every open $\Vert \cdot\Vert_{1}$ ball contains an open $\Vert \cdot\Vert_{2}$ ball, then $\Vert \cdot\Vert_{1}$ is weaker than $\Vert \cdot\Vert_{2}$.
I am inclined to believe that concerning the statement of the theorem, the author intended to further say that the two ball have the same centre. Without this assumption, I am pretty sure the identity map $I: (X,\Vert \cdot\Vert_2)\rightarrow (X, \Vert \cdot\Vert_{1})$ will not be uniformly continuous and we know that another characterisation of $\Vert \cdot\Vert_{1}$ being weaker than $\Vert \cdot\Vert_{2}$ is that this map is uniformly continuous.
What's your take on this? Do you reckon the author meant that the two balls in the proposition are concentric?
I've had a think about this and I came this far:
Let $y\in X$. It suffices to show that if every open $\Vert\cdot\Vert_{1}$ ball contains an open $\Vert\cdot\Vert_{2}$ ball then the identity map, $I:(X,\Vert\cdot\Vert_{2})\rightarrow(X, \Vert\cdot\Vert_{1})$ is continuous at $y$.
Consider the open $\Vert\cdot\Vert_1$ ball, $B(y, \epsilon)$. Then we know there is an open $\Vert\cdot\Vert_{2}$ ball, $B(z,\delta)$ such that \begin{equation} B(z, \delta)\subset B(y, \epsilon). \end{equation}
Now suppose that $\Vert y-z\Vert_{2}<\frac{\delta}{2}$ and define
\begin{equation} R\equiv d(y, \partial B(z, \frac{\delta}{2})) \text{ (with respect to }\ \Vert_\cdot\Vert_{2}). \end{equation}
Note $R<\frac{\delta}{2}$. Let $\delta^{\ast}
\begin{equation} \Vert x-z\Vert_{2}\leq \Vert x-y\Vert_{2}+\Vert y-z\Vert_{2}<\frac{\delta}{2}+\frac{\delta}{2}=\delta \end{equation}
and hence for such $x$ we conclude:
\begin{equation} \Vert x-y\Vert_{1}<\epsilon. \end{equation}
Thus the identity map is continuous.
I am not sure how to go about this problem if $\Vert y-z\Vert_{2}\geq\frac{\delta}{2}$.
Any ideas friends?