Ytukyg

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?

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

b. Notice $int_0^1 |x^n|dx to 0$ as $n to infty$ BUT $sup_{xin [0,1]} |x^n|=1$ 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 $f(x):=lim_{nto infty} f_n(x)$ 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 $F(x):=left(1-frac1xright)^x$ is nondecreasing on $[1, infty)$ with $lim_{x to infty} F(x)=frac1e$. So with $varepsilon=frac1{4}(1-e^{-1})$ and for any $ngeq 2$, we may select $m=2n$ and have
$$sup_{xin[0,1]} |x^n-x^m| geq left(1-frac1nright)^nleft(1-left(1-frac1nright)^nright)geq varepsilon$$
which shows that the sequence of functions ${f_n}_{n=1}^infty$ is not Cauchy in $(X_1, d_1)$ by definition.

We know that a continuous map between metric spaces is one which preserves limits: $f$ is continuous if and only if $limlimits_{ntoinfty} f(x_n)=f(limlimits_{ntoinfty} x_n)$ 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 $limlimits_{ntoinfty} f_n=0$ 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 $xin X_i$, $d_j(x,0)le Mcdot d_i(x,0)$ -- 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.