Definition: Let $f_n:X\to Y$ be a sequence of functions from a set $X$ to the metric space $Y$. Let $d$ be the metric for $Y$. The sequence $(f_n)$ d-converges uniformly to the function $f:X\to Y$ if given $\varepsilon>0$, there exists an integer $N$ such that $d(f_n(x),f(x))<\varepsilon$ for all $n\geq N$ and all $x\in X$.

I wonder if this definition depends on the metric $d$. More explicitly: consider $d_1,d_2$ metrics on $Y$ such that the corresponding induced topologies are the same, and let $f:X \to Y$ a function.

Is the following statement true?: If $f_n:X\to Y$ is a sequence of functions that $d_1$-converges uniformly to $f$, then it also $d_2$-converges to $f$.

If the answer is no, it would be quite interesting to see a counter-example.

  • 1
    $\begingroup$ Certainly it remains true for all strongly equivalent metrics to $d$: en.m.wikipedia.org/wiki/Equivalence_of_metrics $\endgroup$ – Alex R. Sep 8 '14 at 4:55
  • $\begingroup$ Can you think of a pair of metrics that are topologically, but not strongly equivalent? $\endgroup$ – Alex R. Sep 8 '14 at 4:56
  • $\begingroup$ If $d$ is a metric then $\frac{d}{1+d}$ is a topologically equivalent one but generally they are not strongly equivalent. For example the usual metric in $\mathbb{R}$, $|x-y|$ is not bounded and its standard bounded version $\frac{|x-y|}{1+|x-y|}$ is bounded, so they cannot be strongly equivalent. $\endgroup$ – Chilote Sep 8 '14 at 5:03

No, it's false.

Let $Y = (0,+\infty)$ with the metrics $d_1(x,y) = |y - x|$ and $d_2(x,y) = |\log y - \log x|$. The two metrics clearly define the same topology, since the second is isometric to $\mathbf{R}$ with the usual distance, and the exponential function is a homeomorphism of $\mathbf{R}$ onto $(0,+\infty)$.

Also let $X = (0, + \infty)$. Now define $f_n(x) = x + 1/n$. The sequence converges uniformly to $f(x) = x$ for $d_1$. However, $\sup_{x > 0} d_2(f_n(x),f(x)) = +\infty$, so $f_n$ doesn't converge uniformly for $d_2$.

  • $\begingroup$ ...and hence $d_1$ and $d_2$ cannot be strongly equivalent. $\endgroup$ – Chilote Sep 8 '14 at 6:13
  • $\begingroup$ Yes. But your example of $d/(1 + d)$ wouldn't work. In fact, if it's possible to simply calculate the second distance from the first, and the two define the same topology, then uniform convergence in one is equivalent to uniform convergence in the other. $\endgroup$ – Dave Sep 8 '14 at 6:20

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.