Thanks in advance for the help. I'm looking to find an example for the following, but I'm hitting a total wall in my understanding of what it means to find functions in the topologies generated by the metrics. Any help in understanding would be amazing. The problem follows:
Let $I = [0, 1]$ (with the subspace topology from $\mathbb{R}$), $Y = I^I = \{f | f: I \rightarrow I\}$, and $X = \{f \in Y|f\text{ continuous}\}$. Let $\mathcal{T}_1$ = topology on $X$ generated by metric (called the $L^1$ metric) $d_1(f, g) = \int_0^1 |f(t) - g(t)| dt$. Let $\mathcal{T}_p$ = subspace topology on $X$ from product topology on $Y$. Let $\mathcal{T}_\infty$ = topology on $X$ generated by $d_{\infty}(f, g) = \sup\{|f(t) - g(t)| \mid t \in I\}$.
Give an example of $f_1, f_2, f_3, ... \in X$ and $g_1, g_2, g_3, ... \in X$ such that $f_1, f_2, f_3, ...$ converges to the zero function $(0(x) = 0)$ in $\mathcal{T}_1$ but not in $\mathcal{T}_p$ and $g_1, g_2, g_3, ...$ converges to o in $\mathcal{T}_p$ but not in $\mathcal{T}_\infty$.
It helps to first rewrite what it means to converge to the zero function in each of the different topologies. Given a sequence $(h_n)$ in $X$, note that \begin{align*} h_n\to 0\ \text{in $\mathcal{T_1}$} &\iff \lim_{n\to\infty}\int_0^1h_n(t)\,dt=0; \tag{1}\\ h_n\to 0\ \text{in $\mathcal{T_p}$} &\iff \lim_{n\to\infty}h_n(t)=0\ \text{for each $t\in[0,1]$}; \tag{2}\\ h_n\to 0\ \text{in $\mathcal{T_\infty}$} &\iff \lim_{n\to\infty}\sup_{t\in[0,1]}h_n(t)=0. \tag{3}\\ \end{align*} (Note that there are no absolute values in (1) or (3) because all functions in $X$ take only positive values.)
Now we want $(f_n)$ to satisfy (1) but fail (2). Take $f_n$ to be the piecewise linear function whose graph connects the points $(0,1)$, $(1/n,0)$, and $(1,0)$. Then $\int_0^1f_n(t)\,dt=1/(2n)$, but $f_n(0)=1$ for all $n$.
We want $(g_n)$ to satisfy (2) but fail (3). Take $g_n$ to be the piecewise linear function whose graph connects the points $(0,0)$, $(1/(2n),1)$, $(1/n,0)$, and $(1,0)$. Then $h_n(0)=0$ for all $n$, and if $t>0$, there exists $N$ such that $h_n(t)=0$ for all $n>N$. However $\sup_{t\in[0,1]}f_n(t)=1$ for each $n$.