Let $(X,\rVert{\cdot}\lVert)$ be a normed vector space. Clearly there is a canonical way to induce a metric (and so a topology) on $X$ by defining $d(x,y)=\rVert x-y\rVert$.

Are there other metrics induced by $\rVert{\cdot}\lVert$? If does exist another metric $d_1\neq d$ induced by $\rVert{\cdot}\lVert$, are $d$ and $d_1$ equivalent?

  • 1
    $\begingroup$ What does "equivalent" mean here? $\endgroup$
    – 23rd
    May 2 '13 at 8:49
  • $\begingroup$ I mean that the induced topology (with $\epsilon$-balls) is the same. $\endgroup$
    – Dubious
    May 2 '13 at 8:50
  • $\begingroup$ I think metric induced by $||.||$ is defined as $||x-y||$, can you explain what do you mean by 'metric induced by $||. ||$' ? $\endgroup$
    – pritam
    May 2 '13 at 8:51
  • $\begingroup$ For example a metric $d_1=f(\rVert x-y\lVert)$ where $f$ is a generic function. $\endgroup$
    – Dubious
    May 2 '13 at 8:53
  • 2
    $\begingroup$ Then there are lots of such metrics. For example, $d_1=2d$. $\endgroup$
    – 23rd
    May 2 '13 at 8:54

Even if the question is loosely formulated, I would say that the answer is no. I'm thinking at the following examples: in the vector space $\mathbb{R}$, consider the usual absolute value $\lvert\cdot\rvert$, which indeed is a norm and induces the ordinary topology of the real line. We can define a metric on $\mathbb{R}$ in terms of $\lvert\cdot\rvert$ as follows: $$d_1(x, y)=\lvert \arctan(x)-\arctan(y)\rvert,$$ or as follows: $$d_2(x, y)=\frac{\lvert x-y\rvert}{1+\lvert x-y\rvert}$$.

Both are "metrics induced by $\lvert \cdot\rvert$" in some vague sense, and neither of them is metrically equivalent to $\lvert \cdot\rvert$, because both make $\mathbb{R}$ into a bounded space.

P.S. : I had not noticed that you consider "equivalent" two metrics which induce the same topology. Then $d_1$ and $d_2$ above do not qualify as counterexamples because they do induce the ordinary topology of the line. To obtain a counterexample in this case you can take the $\text{signum}$ function $$\text{signum}(s)=\begin{cases} 1& s>0 \\ 0 & s=0 \\ -1& s<0\end{cases}$$ and define $$d_3(x, y)=\text{signum}(\lvert x-y\rvert).$$ You get the discrete metric on $\mathbb{R}$, which obviously does not induce the same topology on $\mathbb{R}$ as $\lvert\cdot\rvert$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.