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 $\delta^{\ast}<R$. Then for all $x: \Vert x-y\Vert_{2}<\delta^{\ast}$, we have:

\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?

  • 3
    $\begingroup$ Saying «a certain book» is almost never better than saying explicitly what book you are talking about... $\endgroup$ – Mariano Suárez-Álvarez Aug 24 '12 at 19:33
  • $\begingroup$ What do you mean exactly by weaker? Determines a weaker topology? $\endgroup$ – Sean Eberhard Aug 24 '12 at 19:33
  • $\begingroup$ It's Promislow's book "A First Course in Functional Analysis" $\endgroup$ – Nirav Aug 24 '12 at 20:24
  • $\begingroup$ Below is the definition of weaker I am using:$\Vert \cdot\Vert_{1}$ is weaker than $\Vert\cdot\Vert_{2}$ if and only if $\Vert x_n-x\Vert_{2}\rightarrow 0\Rightarrow \Vert x_n-x\Vert_{1}\rightarrow 0$ $\endgroup$ – Nirav Aug 24 '12 at 20:27

By assumption the unit ball $B_1(0,1)$ with respect to $\|\cdot\|_1$ contains some ball $B_2(x,r)$ with respect to $\|\cdot\|_2$. But then $B_2(0,r) =B_2(x,r) -x \subseteq B_1(0,2)$ (since $x\in B_1(0,1)$). I think you know how to proceed.


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