If $φ:I→J$ is a homeomorphism then $f_n→f$ implies that $(f_n∘φ) →(f∘φ)$ with respect the uniform, pointwise and $L_2$ topology respectively?

Question

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 $C(I)$ whose topology $\mathcal T_\infty$ is said topology of uniform convergence or rather uniform topology. 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 $C(I)$ said topology of pointwise convergence or rather pointwise topology.

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So let be now $(f_n)_{n\in\Bbb N}$ a sequence of functions defined in an interval $I$ and thus let be $\varphi$ a homoeomprhism from an interval $J$ to $I$. So, I would like understand if the sequence $(f_n)_{n\in\Bbb N}$ converges pointwisely or uniformly in $I$ then also the sequence $(f_n\circ\varphi)_{n\in\Bbb N}$ converges respectively pointwisely or uniformly.

First of all I tried to prove that if $(f_n)_{n\in\Bbb N}$ converges uniformly then also $(f_n\circ\varphi)_{n\in\Bbb N}$ converges uniformly and precisely I tried to do this observing that if $X$ and $Y$ are homeomorphic topological spaces then a net $(x_\lambda)_{\lambda\in\Lambda}$ on $X$ converges to $x$ if and only if for any homeomorphism $\phi$ from $X$ to $Y$ the net $\big(\phi(x_\lambda)\big)_{\lambda\in\Lambda}$ converges on $Y$ to $\phi(x)$. Indeed, knowing this result I defined a map $\Phi$ from $C(I)$ to $C(J)$ through the equation $$ \Phi(f):=f\circ\varphi $$ for any $f\in C(I)$ and thus I attempted to prove that it is a homeomorphism and I did this as to follow. So first of all, I defined the application $\Phi^{-1}$ form $C(J)$ to $C(I)$ through the equation $$ \Phi^{-1}(g):=g\circ\varphi^{-1} $$ for any $g\in C(J)$ and thus I observed that for any $f\in C(I)$ and for any $g\in C(J)$ the equalities $$ \big(\Phi^{-1}\circ\Phi\big)(f)=\Phi^{-1}\big(\Phi(f)\big)=\Phi^{-1}(f\circ\varphi)=(f\circ\varphi)\circ\varphi^{-1}=f \\\text{and}\\ \big(\Phi\circ\Phi^{-1}\big)(g)=\Phi\big(\Phi^{-1}(g)\big)=\Phi(g\circ\varphi^{-1})=(g\circ\varphi^{-1})\circ\varphi=g $$ hold so that I concluded that $\Phi$ is bijective and in particular $\Phi^{-1}$ is exactly its inverse. Now, the application $\varphi$ and $\varphi^{-1}$ are surjective so that $$ \sup|(f\circ\varphi)-(h\circ\varphi)|=\sup|f-h|\,\,\,\text{and}\,\,\, \sup|(g\circ\varphi^{-1})-(k\circ\varphi^{-1})|=\sup|g-k| $$ for any $f,h\in C(I)$ and for any $(g,k)\in C(J)$ and thus I conclude that $\Phi$ and $\Phi^{-1}$ are isometry and so they are, in particular, continuous and this proves that $\Phi$ is properly a homeomorphism with respect the unfirom topolgy.

Well, we observe now that if $(f_n)_{n\in\Bbb N}$ is a sequence on $C(I)$ then for any finite sequence $(x_i)_{i=1}^n$ of elements of $I$ the numerical sequence $\Big((f_n\circ\varphi)\big(\varphi^{-1}(x_i)\big)\Big)_{n\in\Bbb N}$ for $i=1,\dots ,n$ converges to $(f\circ\varphi)(\varphi^{-1}(x_i)\big)$ for any $f\in C(I)$ wheter the numerical sequence $\big(f_n(x_i)\big)_{n\in\Bbb N}$ for $i=1,\dots, n$ converges to $f(x_i)$: indeed if $\big(f_n(x_i)\big)$ converges to $f(x_i)$ then for any $\epsilon>0$ ther exist $n_\epsilon\in\Bbb N$ such that $$ \Big|(f_n\circ\varphi)\big(\varphi^{-1}(x_i)\big)-(f\circ\varphi)\big(\varphi^{-1}(x_i)\big)\Big|=|f_n(x_i)-f(x_i)|<\epsilon $$ for any $n\ge n_\epsilon$ so that the statement follows trivially. So we conclude that if $(f_n)_{n\in\Bbb N}$ converges pointwisely at $x$ then $(f_n\circ\varphi)_{n\in\Bbb N}$ converges pointwisely at $\varphi^{-1}(x)$: however, I was not able to understand if the last argumentations show that the application $\Phi$ is also a homeomorphism with respect the pointwise topology and I would not dislike to understand this.

So, I'd like finally understand if the $L_2$ convergence on $I=(a,b)$ implies the $L_2$ convergence on $J=(c,d)$. Unfortunately, I was not able to say anything about - I was not only able to prove the $L_2$ convergence on $I$ implies $L_2$ convergence on $J$ and vice versa but I was not able to find a counterexample showing the contrary. Anyway here is showed that the inclusion $$ \mathcal T_p,\mathcal T_{L_2}\subseteq\mathcal T_\infty $$ holds so that the uniform convergence implies the $L_2$ convergence and thus this let to state that if $(f_n)_{n\in\Bbb N}$ is a sequence on $I$ converging uniformly then it converges with respect $L_2$ topology and in particular by the previous arguments also $(f_n\circ\varphi)_{n\in\Bbb N}$ converges with respect $L_2$. Moreover, let be $P_0(I)$ and $N_0(I)$ the subsets of $C(I)$ defined through the equation $$ P_0(I):=\{f\in C(I):f(x)\ge0\,\forall x\in I\}\,\,\,\text{and}\,\,\,N_0(I):=\{f\in C(I):f(x)\le 0\,\forall x\in I\} $$ and thus we observe that they are linear subspace and in particular $$ fg\ge 0 $$ for any $f,g\in P_0\cup N_0$. Now we suppose that $\varphi$ is of class $C^1$ so that by the Weierstrass theorem $|D\varphi|$ is bounded and thus observing that $$ \Big(d_{L_2}(f\circ\varphi,g\circ\varphi)\Big)^2=\big\|(f\circ\varphi)-(g\circ\varphi)\big\|_{L_2}^2=\biggl|\int_J(fg)\circ\varphi\biggl|=\\ \biggl|\int_{\varphi[J]}\big((fg)\circ\varphi\big)\circ\varphi^{-1}\cdot\big|\det D\varphi\big|\biggl|= \biggl|\int_I(fg)\cdot|D\varphi|\biggl|=\\ \int_I(fg)\cdot |D\varphi|\le\int_I(fg)\le\max|D\varphi|=\max|D\varphi|\cdot\int_Ifg=\\ \max|D\varphi|\cdot\big\|f-g\big\|_{L_2}^2=\max|D\varphi|\cdot\Big(d_{L_2}(f,g)\Big)^2 $$ for any $f,g\in P_0(I)\cup N_0(I)$ we concluded that if $(f_n)_{n\in\Bbb N}$ is a sequence on $P_0\cup N_0$ converging to $f\in P_0(I)\cup N_0(I)$ then $(f_n\circ\varphi)_{n\in\Bbb N}$ converges to $f\circ\varphi\in P_0(J)\cup N_0(J)$ and moreover this surely happens if $P_0(I)$ and $N_0(I)$ was closed but unfortunately I was not able to clarify this. Anyway if $\varphi$ is of class $C^1$ then, as here showed, it is a lipschit function but unfortunately it seems that this does not help to understand if the $L_2$ topology on $I$ is compatible with the $L_2$ topology on $J$ through $\varphi$. Finally, we suppose that $$ \varphi(x):=\frac{b-a}{d-c}\cdot(x-c)+a $$ for any $x\in J$ and thus we observe that if $d-c\le b-a$ then $$ \Big(d_{L_2}(f\circ\varphi,g\circ\varphi)\Big)^2=\big\|(f\circ\varphi)-(g\circ\varphi)\big\|_{L_2}^2=\biggl|\int_J(fg)\circ\varphi\biggl|=\\ \biggl|\int_{\varphi[J]}\big((fg)\circ\varphi\big)\circ\varphi^{-1}\cdot\big|\det D\varphi\big|\biggl|= \biggl|\int_I(fg)\cdot\Big|\frac{b-a}{d-c}\Big|\biggl|=\\ \frac{b-a}{d-c}\cdot\biggl|\int_Ifg\biggl|\le\biggl|\int_Ifg\biggl|=\|f-g\|_{L_2}^2=\Big(d_{L_2}(f,g)\Big)^2 $$ for any $f,g\in C(I)$ so that we conclude that the application $\Phi$ above defined is an isometry and so it also continuous and thus if $(f_n)_{n\in\Bbb N}$ is a sequence on $C(I)$ converging to $f$ then the sequence $\big(\Phi(f_n)\big)_{n\in\Bbb N}$ on $J$ converges to $\Phi(f)$. However, what happens the inequality $(d-c)\le(b-a)$ does not holds? Anyway, this proves that if the inequality $(d-c)\le (b-a)$ holds then ther exist at least a homeomorphism from $ J$ to $I$ that carries the $L_2$ topology on $I$ to the $L_2$ topology on $J$.

So first of all I ask to know if the argumentations I gave about the uniform and pointwise convergence and the observations I did about the convergence on $P_0\cup N_0$ with respect the $L_2$ topology are correct and then I ask to clarify if the $L_2$ convergence on $I$ is equivalent to the $L_2$ convergence on $J$ through $\varphi$; moreover, I repeat I do not dislike to understand if the application $\Phi$ is a homeomorphism with respect the pointwise topology.

So could someone help me, please?

Answer 1

The argument for $\mathcal{T}_p$ can be simplified considerably; in that case $C([0,1])$ just has the subspace topology it inherits from $\Bbb R^I$ (or $\Bbb R^J$ resp.) in the product topology and it's quite clear that $\Phi: \Bbb R^I \to \Bbb R^J$ (which is a bijection when $\phi$ is) restricts to a bijection on $C(I)$ when $\phi$ is a homeomorphism of $I$: if $f \in C(I)$, we have $\Phi(f)=f \circ \phi \in C(J)$ as a composition of continuous functions and if $f \in C(J)$, it's clear too that $f'=f \circ \phi^{-1} \in C(I)$ and $\Phi(f')=f$ etc.

That $\Phi$ is a homeomorphism on the full products is clear by the universal property on products as $\pi^J_x \circ \Phi(f) = \pi^J_x \circ f \circ \phi = \pi^I_{\phi(x)}(f)$, for all $x \in J$ and $f \in \Bbb R^I$, so that $\pi^J_x \circ \Phi = \pi^I_{\phi(x)}$ is continuous for all projections etc. The same holds mutatis mutandis for $\Phi^{-1}(f)=f \circ \phi^{-1}$, which has the same form as $\Phi$. A map like $\Phi$ is often called the pull-back by $\phi$ and denoted $\phi_{\ast}$. So $(\phi_\ast)^{-1}=\phi^{-1}_\ast$ is what I was saying before, essentially.

So $\Phi$ is a homeomorphism between $C(I)$ and $C(J)$ in the pointwise topologies. You had quite similar considerations. But you don't have to consider any basic open sets: just know it's the product topology.

It's also quite clear that $\|\Phi(f)\|_\infty = \|f\|_\infty$ so that on the sup-norm topology we get a simple isometry and all is well.

I think too that standard integral facts tell us

Edited: $\int_I f(x)dx = \alpha \int_J f(\phi(x)) dx$ for some $\alpha >0$ as we choose $\phi$ to be a suitable linear homeomorphism of intervals. So $\Phi$, or a scaled version of it, preserves $L^2$ inner products and so norm too etc. (adapted after comments discussion).