# Compare the topology of pointwise convergence with the vector topology in the space of the continuous scalar functions $C(I)$

examples-counterexamplesfunctional-analysisgeneral-topologytopological-vector-spacesuniform-convergence

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
$$[f;x_1,\dots,x_n;r]:=\{g\in C(I):|f(x_i)-g(x_i)|
the collection
$$\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?

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?