Let $f: \mathbb{R} \to \mathbb{R}$ continuos with compact support. ($supp f=K$)
$ a_n \rightarrow a>0, b_n \rightarrow b , f_n(x) = f(a_nx+b_n)$
And let $g=f(ax+b)$, I want to show that $||f_n-g||_p \rightarrow 0$ for $p\in[1,\infty)$.
It is obvious that we have pointwise convergence, and I was able to show $f_n\rightarrow g$ in measure and that we have $f \rightarrow g$ uniformly a.e. thanks to the compact support of f.
My problem is when trying to evaluate $\int_{\mathbb{R}} |f_n-g|^p dm$. Here are my calculations:
$\int_{\mathbb{R}} |f_n-g|^p dm$ =$\int_{K} |f_n-g|^p dm + \int_{K^c} |f_n-g|^p dm$.
$\int_{K} |f_n-g|^p dm \leq \int_{K\cap\delta(\epsilon^{1/p})} |f_n-g|^p dm + \int_{K\cap \delta(\epsilon^{1/p})^c} |f_n-g|^p dm \leq m(K)\epsilon+||f_n-g||_2m(\delta(\epsilon^{1/p}))\leq m(K)\epsilon+2||g||_2\epsilon$
where $\delta(\epsilon) = \{x : |f_n(x)-g(x)| < \epsilon\}$ which complement goes to 0 from convergence in measure, and $||f_n||_2=||g||_2$ (linear change of variables with $det \neq 0)$
Now the problematic part is bounding whatever happens in $K^c$:
$\int_{K^c} |f_n-g|^p dm \leq \int_{K^c\cap E^c} |f_n-g|^p dm + \int_{K^c\cap E} |f_n-g|^p dm \leq m(E^c)||f_n-g||_2 + \int_{K^c\cap E} |f_n-g|^p dm \leq 2||g||_2\epsilon + \int_{K^c\cap E} |f_n-g|^p dm$
Where $f_n\rightarrow g $ uniformly on $E$ and $m(E^c)<\epsilon$
Now what I was planning to do is to use the uniform convergence on $E$ to claim that there is $L_p$ convergence, however, $m(K^c\cap E)$ is not necessarily finite, so I'm stuck. I feel like I might be overcomplicating this question
Any help would be appriciated.
Here we prove a more general result:
For $\varepsilon>0$, let $g$ is a continuous function of compact support such that $\|f-g\|_p<\tfrac13\tfrac{\varepsilon}{2^{1/p}|a|^{-1/p}}$. Let $N$ be large enough so that $|a_n-a|<\frac{|a|}{2}$. This way, $|a_n|> \frac{|a|}{2}>0$ for all $n\geq N$
$$\begin{align} \Big(\int&|f(a_nx+b_n)-f(ax+b)|^p\,dx\Big)^{1/p}\leq \Big(\int|f(a_nx+b_n)-g(a_nx+b_n)|^p\,dx\Big)^{1/p} +\\ &\quad\Big(\int|g(a_nx+b_n)-g(ax+b)|^p\,dx\Big)^{1/p}+\Big(\int|g(ax+b)-f(ax+b)|^p\,dx\Big)^{1/p}\tag{1}\label{main} \end{align}$$ We may get rid of the terms $b_n$ and $b$ in the first and third integrals in above by using the translation invariance property of the Lebesgue measure. In the first integral we translate the origin to $\frac{b_n}{a_n}$ and in the second integral we translate the origin to $\frac{b}{a}$. All this yields $$\begin{align} \int|f(a_nx+b_n)-g(a_nx+b_n)|^p\,dx&=\int|f(a_nx)-g(a_nx)|^p\,dx\\ \int|f(ax+b)-g(ax+b)|^p\,dx&=\int|f(ax)-g(ax)|^p\,dx \end{align}$$ Using the positive homogeneity of the Lebesgue measure, i.e., $\lambda(cA)=|c|\lambda(A)$ for all measurable set $A$, we obtain that $$ \begin{align} \int|f(a_nx)-g(a_nx)|^p\,dx&=\frac{1}{|a_n|}\int|f(x)-g(x)|^p\,dx\\ &<\frac{2}{|a|}\int|f(x)-g(x)|^p\,dx<\frac{\varepsilon^p}{3^p}\\ \int|f(ax)-g(ax)|^p\,dx&=\frac{1}{|a|}\int|f(x)-g(x)|^p\,dx<\frac{\varepsilon^p}{3^p} \end{align} $$ for all $n\geq N$.
It remains to control the size of the second integral in the right-hand-side of \eqref{main}. Since $\operatorname{supp}(g)$ is compact, so is $K=\frac{1}{a}\big(\operatorname{supp}(g)-b\big)$. Notice that $$\begin{align} \operatorname{supp}(g(a_n\cdot+b_n))&=\frac{1}{a_n}(\operatorname{supp}(g)-b_n)\\ &=\frac{a}{a_n}K+\frac{b}{a_n}-\frac{b_n}{a_n}\end{align}$$ The continuity of linear functions implies that there is $N'\geq N$ such that for $n\geq N'$, $$\frac{a}{a_n}K+\frac{b}{a_n}-\frac{b_n}{a_n}\subset (-A,A).$$
Since $g\in\mathcal{C}_{00}(\mathbb{R})$, it is uniformly continuous and so, there is $\delta>0$ such that whenever $|x-x'|<\delta$, $|g(x)-g(x')|^p<\frac{\varepsilon^p}{3^{p}|2A|}$. Choose $N''\geq N'$ such that $|a_n-a||A|+|b_n-b|<\delta$ for all $n\geq N''$. Hence $$\int_{\mathbb{R}}|g(a_nx+b_n)-g(ax+b)|^p\,dx=\int_{(-A,A)}|g(a_nx+b_n)-g(ax+b)|^p\,dx\leq\frac{\varepsilon^p}{3^p}$$ for all $n\geq N''$.
Putting things together, we obtain that for $n\geq N''$ $$\Big(\int|f(a_nx+b_n)-f(ax+b)|^p\,dx\Big)^{1/p}<\varepsilon$$ This completes the proof.