0
$\begingroup$

How to counter this problem? Is it enough to show pre image of unit ball in some metric is open under another to show the continuity? I am not at all getting the path to proceed. And how to contradict when it's not continuous?

enter image description here

$\endgroup$

closed as off-topic by user21820, Adrian Keister, max_zorn, YiFan, Eevee Trainer Mar 29 at 2:15

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Adrian Keister, max_zorn, YiFan, Eevee Trainer
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ You can use epsilon-delta methods to prove continuity. Proceed as you would with the standard metric but substitute the appropriate metric to make arguments like $d_2(f,g)\lt\epsilon$. $\endgroup$ – John Douma Dec 21 '18 at 4:03
  • 2
    $\begingroup$ Tip: Questions are not well-received on here whenever the question asker doesn't mention any attempted efforts towards a solution. So always try to mention any attempted work even if it seems embarrassingly incorrect, otherwise other users are less motivated to help. $\endgroup$ – Matt A Pelto Dec 21 '18 at 4:29
0
$\begingroup$

a. Let $\varepsilon>0$ be given. Since $\sup_{x\in [0,1]} | \, f(x)-g(x)| \geq \int_0^1 |\, f(x)-g(x)|dx$, we may choose $\delta=\varepsilon$ so that $\int_0^1 |\, f(x)-g(x)|dx<\varepsilon$ whenever $\sup_{x\in [0,1]} |\, f(x)-g(x)|<\delta$. Therefore $id:X_1 \longrightarrow X_2$ is continuous.

b. Notice $\int_0^1 |x^n|dx \to 0$ as $n \to \infty$ BUT $\sup_{x\in [0,1]} |x^n|=1$ for every $n \in \mathbb N$. Therefore $id:X_2 \longrightarrow X_1$ is NOT continuous at the constant function $g \equiv 0$ because it is not sequentially continuous there. There are other counterexamples but this is the preferred (by at least 2 people in this thread).

c. I leave to you. The notion of sequential continuity seems good for demonstrating a counterexample to such statements, while the standard notion of continuity seems good for proving such statements true which is not exactly strict advice.


Technically for b. I used the fact that the functions $f_n(x):=x^n$ from the sequence of functions $\{f_n\}_{n=1}^\infty$ converge to the constant function $g \equiv 0$ in $(X_2, d_2)$ but the same is not true in $(X_1, d_1)$. And so $id: X_2 \longrightarrow X_1$ is not sequentially continuous at $g$ as opposed to identifying the pointwise limit $f(x):=\lim_{n\to \infty} f_n(x)$ which is not in the set $\mathcal C[0,1]$ as $f$ is not continuous at $x=1$.

If this seems like an issue, then showing that the chosen sequence of functions is not uniformly Cauchy on $[0,1]$ (Cauchy in $(X_1, d_1)$) avoids this. For this argument we make the following two observations, $|x^n-x^m|=|x^n||1-x^{m-n}|$ and the function $F(x):=\left(1-\frac1x\right)^x$ is nondecreasing on $[1, \infty)$ with $\lim_{x \to \infty} F(x)=\frac1e$. So with $\varepsilon=\frac1{4}(1-e^{-1})$ and for any $n\geq 2$, we may select $m=2n$ and have $$\sup_{x\in[0,1]} |x^n-x^m| \geq \left(1-\frac1n\right)^n\left(1-\left(1-\frac1n\right)^n\right)\geq \varepsilon$$ which shows that the sequence of functions $\{f_n\}_{n=1}^\infty$ is not Cauchy in $(X_1, d_1)$ by definition.

$\endgroup$
  • 1
    $\begingroup$ For the part c, Cauchy Schwartz says, integral 0 to 1 of h < (int 0 to 1 of h²)^1/2. Hence like part a, choosing delta same as given epsilon we get d3 < delta implies d2< epsilon. Therefore id: X3 to X2 is continuous. $\endgroup$ – ChakSayantan Dec 21 '18 at 6:09
  • $\begingroup$ You got it, btw I didn't downvote your question. Someone else did and so I shared my tip. My upvote was preemptive but I would say you proved worthy. Most people seem eager to rush towards negative judgement these days (not to say some don't deserve it...rhymes with rump). $\endgroup$ – Matt A Pelto Dec 21 '18 at 9:43
  • $\begingroup$ Oh but learn to write in latex too. Tip: right click>"show math as" AND for symbols there are pdf documents littering the web with thorough latex code catalogs. $\endgroup$ – Matt A Pelto Dec 21 '18 at 10:15
  • 1
    $\begingroup$ Regarding $a.$ If $d_1,d_2$ are metrics on a set $X$ and if there exists $K>0$ such that $d_2(f,g)\leq K\cdot d_1(f,g)$ for all $f,g\in X$ then the topology generated by $d_2$ is a subset of the topology generated by $d_1,$ so $ id_X:(X,d_1)\to (X,d_2)$ is continuous because the inverse of a $d_2$-open set, which is itself, is also $d_1$-open.... In $a.$ we have $K=1.$ $\endgroup$ – DanielWainfleet Dec 22 '18 at 3:06
  • $\begingroup$ Comparing topologies on the same set certainly can come in handy sometimes: imgur.com/cfOxqfs (an old assignment I did that plays off the same notion). I guess considering a and b together, we might say that an open bijection is not always continuous -_- $\endgroup$ – Matt A Pelto Dec 22 '18 at 3:24
2
$\begingroup$

We know that a continuous map between metric spaces is one which preserves limits: $f$ is continuous if and only if $\lim\limits_{n\to\infty} f(x_n)=f(\lim\limits_{n\to\infty} x_n)$ for each convergent sequence $x_n$. To prove that a map is not continuous, we can try to find a sequence that is convergent in one metric but not the other.

For example, I don't think that b) is true. Consider the sequence $$f_n(x)=x^n.$$

Then $\lim\limits_{n\to\infty} f_n=0$ in the metric $d_2$, but not in $d_1$: it even fails to be Cauchy in $d_1$.

Of course, in general it is fine to show that the preimage of any open ball is open. There is some general theory that tells you that if the identity map is bounded - that is, if for every $x\in X_i$, $d_j(x,0)\le M\cdot d_i(x,0)$ -- then the identity map from $X_i$ to $X_j$ is continuous. This uses the fact that all these metrics come from norms, though, and takes some work to establish.

$\endgroup$
  • $\begingroup$ nice counterexample $\endgroup$ – Matt A Pelto Dec 21 '18 at 4:25

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