Two non-comparable metrics on $X=C[0,\pi]$

Question

Problem:

Show that the two metrics$$d_1(x,y)=\sup_{[0,\pi]}t^2\lvert {x(t)-y(t)}\rvert\text{ and }d_2(x,y)=\sup_{[0,\pi]}\sin(t)\lvert {x(t)-y(t)}\rvert$$are not comparable on the set $X=C[0,\pi]$ of real valued continuous function defined on $[0,\pi]$. In particular, find a set $A\subset{X}$ that is open with the metric $d_1$ but not open with $d_2$.

Here, comparable means convergence of a sequence in one metric implies the convergence of it in the other with the same limit. Since this convergence relation can be characterized by means of being open, problem asks me to show it using open sets.

Now, I find it very difficult to construct such a set $A\subset{X}$. I believe it must be about the relation between the functions $sin(t)$ and $t^2$. For instance, I tried $A=\{x(t)\in{X} \lvert x(\pi)\gt{0}\}$ and showed that this set is open with respect to $d_1$. However, I do not know whether it is also open with respect to $d_2$ or not, and even if it is not open, it confuses me a lot, how to show it? Any comment/idea is highly appreciated.

Answer 1

Your guess seems correct, the set $A$ indeed is not open w.r.t. the metric $d_2$. To prove this, consider the constant function $f \equiv C$ for some constant $C > 0$. Then $f(\pi) = C > 0$ so $f \in A$. We will show that there is no open ball in the $d_2$ metric centered at $f$ which is contained in $A$.

Let $\varepsilon > 0$ be arbitrary. Since $\sin$ is continuous at $\pi$ and $\sin \pi = 0$, there exists $\delta \in \langle 0,\pi\rangle$ such that $\sin([\pi-\delta,\pi]) \subseteq \left[0,\frac\varepsilon{2C}\right].$ Define $g_\varepsilon \in C[0,\pi]$ as $$g_\varepsilon(t) = \begin{cases} C, &\text{ if }t \in [0,\pi-\delta], \\ \frac{C}{\delta}(\pi-t), &\text{ if }t \in [\pi-\delta,\pi]. \end{cases}$$ Then since $g_\varepsilon =f$ on $[0,\pi-\delta]$, we have \begin{align} d_2(f,g_\varepsilon) &= \sup_{t \in [0,\pi]} (\sin t)|f(t)-g_\varepsilon(t)| = \sup_{t \in [\pi-\delta, \pi]} (\sin t)\left|C-\frac{C}{\delta}(\pi-t)\right| \\ &= C \sup_{t \in [\pi-\delta, \pi]} \underbrace{(\sin t)}_{\le \frac\varepsilon{2C}}\underbrace{\left|1-\frac{1}{\delta}(\pi-t)\right|}_{\le 1}\le C \cdot \frac{\varepsilon}{2C}= \frac{\varepsilon}2< \varepsilon \end{align} so $g_\varepsilon \in B_{d_2}(f,\varepsilon)$. However, $g_\varepsilon(\pi) = 0$ so $g_\varepsilon\notin A$ which implies $B_{d_2}(f,\varepsilon) \not\subseteq A$.

Answer 2

Actually the inequivalence follows also from the fact that at $t=0$ the behaviours of the metrics $d_1$ and $d_2$ are different.

A result below is complementary to the one given in the answer of @mechanodroid.

The set $$A=\{x\in C[0,\pi]\,:\, d_2(x,0)<1\}$$ is open with respect to the metric $d_2$ (that's a general fact for every metric). On the other hand the set $A$ is not open with respect to the metric $d_1.$ To this end we will show that the function $0\in A$ can be approximated with respect to the metric $d_1$ by functions $x_n\notin A.$ For any fixed positive integer $n\ge 4$ let $$x_n(t) =\begin{cases} 2n^2t(2-nt) & 0\le t\le {2\over n}\\ 0 & {2\over n}<t\le \pi \end{cases} $$ The maximal value of $x_n$ is attained at $t={1\over n}$ and is equal $2n.$ Therefore $$d_2(x_n,0)\ge \max_{0\le t\le {1\over 2}}\sin( t )x_n(t)\ge {2\over \pi}\max_{0\le t\le {1\over 2}} t x_n(t)\ge {2\over \pi}[tx(t)]_{t={1\over n}}={4\over \pi}>1$$ On the other hand $$d_1(x_n,0)=\max_{0\le t\le \pi}t^2x_n(t)\le {4\over n^2}\max_{0\le t\le \pi}x_n(t)={4\over n^2}2n={8\over n}$$ Thus $x_n\notin A$ but $d_1(x_n,0)\underset{n\to \infty}{\longrightarrow} 0.$

Remark Concerning the original question in OP it can be shown that the set $$B=\{x\,:\, d_1(x,0)<1\}$$ which is open with respect to the metric $d_1,$ is not open with respect to the metric $d_2.$