So given the set
$$
C(I):=\{f\in\Bbb R^I:f\,\text{continuous}\}
$$
where $I=[a,b]$ is a closed interval of the real line it is not hard to show that for any $f,g\in C(I)$ and for any $\lambda\in\Bbb R$ the equations
$$
(f+g)(x):=f(x)+g(x)\,\,\,\text{and}\,\,\,(\lambda\cdot f)(x):=\lambda\cdot f(x)
$$
for any $x\in I$ make $C(I)$ a vector space. Now it is also a well know result that the equation
$$
L_2(f,g):=\int_If\cdot g
$$
for any $f,g\in C(I)$ defines an inner product over $C(I)$ so that this set is a topological vector space. However, it is well known that the equation
$$
d_\infty(f,g):=\sup\{|f(x)-g(x)|:x\in I\}
$$
defines a metric in $C(I)$. Finally for any $x_1,\dots,x_n\in I$ putting
$$
[f;x_1,\dots,x_n;r]:=\{g\in C(I):|f(x_i)-g(x_i)|<r\,\,\,\forall i=1,\dots,n\}
$$
the collection
$$
\mathcal B_p:=\{[f;x_1,\dots,x_n;r]:f\in C(I)\wedge r\in\Bbb R^+\}
$$
defines a topology on $C(I)$.
So observing that
$$
d_{L_2}(f,g)=\|f-g\|_{L_2}=\Biggl|\int_I f-g\Biggl|\le\int_I|f-g|\le\int_I\sup|f-g|=\\
\sup|f-g|\cdot\int_I 1\le\frac r{b-a}(b -a)=r
$$
for any $g\in\mathcal B_\infty\Big(f,\frac r{b-a}\Big)$ we conclude that
$$
\mathcal B_{\infty}\Big(f,\frac r{b-a}\Big)\subseteq\mathcal B_{L_2}(f,r)
$$
so that the topology $\mathcal T_\infty$ generated by $d_\infty$ contains the topology $\mathcal T_{L_2}$ generated by $L_2$, that is
$$
\mathcal T_{L_2}\subseteq\mathcal T_\infty
$$
After all the functions $f,g$ are continuous in $I$ so that for any $r>0$ there exist $\delta_{f_i},\delta_{g_i}$ for any $i=1,\dots,n$ such that
$$
|f(x)-f(x_i)|<\frac r 3\,\,\,\text{and}\,\,\,|g(x)-g(x_i)|<\frac r 3
$$
for any $x\in I$ such that
$$
|x-x_i|<\min\big\{\delta_{f_i},\delta_{g_i}\big\}
$$
so that let be
$$
\delta_i:=\min\big\{\delta_{f_i},\delta_{g_i}\big\}
$$
for any $i=1,\dots,n$.
So given $x\in I$ such that $|x-x_i|<\delta_i$ for any $i=1,\dots,n$ we observe that
$$
|f(x_i)-g(x_i)|=\Big|\big(f(x_i)-f(x)\big)+\big(g(x)-g(x_i)\big)+(f(x)-g(x)\big)\Big|\le|f(x_i)-f(x)|+|g(x)-g(x_i)|+|f(x)-g(x)|\le\frac r 3+\frac r 3+\sup|f-g|\le\frac 2 3 r+\frac r 3=r
$$
for any $g\in\mathcal B_\infty\Big(f,\frac r 3\Big)$ and thus we conclude that
$$
\mathcal B_\infty\Big(f,\frac r 3\Big)\subseteq[f;x_1,\dots,x_n;r]
$$
and this allows to conclude that the topology $\mathcal T_p$ generated by the collection $\mathcal B_p$ is contained in the topology $\mathcal T_\infty$ generated by the metric $d_\infty$, that is
$$
\mathcal T_p\subseteq\mathcal T_\infty
$$
So first of all I ask if the arguments I gave to compare the topologies $\mathcal T_{L_2}$, $\mathcal T_p$ and $\mathcal T_\infty$ are correct and then I ask to compare the topology $\mathcal T_{L_2}$ and $\mathcal T_p$.
So could someone help me, please?
Best Answer
Yes, indeed $\mathcal{T}_p, \mathcal{T}_{L_2} \subseteq \mathcal{T}_\infty$ is correct (and the proofs work too).
You do make the $\mathcal{T}_p \subseteq \mathcal{T}_\infty$ argument too hard: it's clear from the definitions that $$B_\infty(f,r) \subseteq [f;x_1, \ldots,x_n; r]$$ for any finite subset $\{x_1,\ldots x_n\}$ of $I$ and $r>0$. You don't even need continuity of $f$ for that.
$\mathcal{T}_p \subseteq \mathcal{T}$ for a topology $\mathcal{T}$ on $C(I)$ iff for each $x \in I$ we have that the map $f \to f(x)$ from $(C(I),\mathcal{T})$ to $\Bbb R$ is continuous. Is this the case here?