Show that $\lim_{r\to s} \|D_rf-D_sf\|_p = 0,\,r,\,s > 0$

Question

Full Question: Let $D_r$ be the dilation operator $D_rf(x) = f(rx)$ on $L^p(\mathbb{R}^d),\, 1 \leq p < \infty$. Show that $\lim_{r\to s} \|D_rf-D_sf\|_p = 0,\,r,\,s > 0$

I was told to use $\int f(rx)d\lambda^d(x) = |r|^{-d}\int f(x)d\lambda^d(x)$.

So I was thinking that set $s=1$, and prove $\lim_{r\to 1}\|D_rf-D_1f\|_p = \lim_{r\to 1}\|f(rx)- f(x)\| = 0$, then show this scales for all $s$. To show this I was thinking that since we know $\|f-g\|_p < \epsilon$, where $g$ vanishes to $0$ outside a bound, then do something along the lines of $\|D_rf - f\|_p \leq \|D_rf-D_rg\|_p+\|D_rg-g\|_p+\|g-f\|_p$, and $\|D_rf-D_rg\| = \int(f(rx)-g(rx)^pd\lambda^{1/p} \to |r|^{-d/p}\|f-g\|_p < \epsilon$, but I've been stuck here.

Answer 1

Given $\epsilon >0$ there exists $g \in C_c(\mathbb R^{d})$ such that $\|f-g\|<\epsilon$. DCT tells you that the result is true with $g$ in place of $f$. Now $\|f(rx)-f(x)\| \leq \|g(rx)-g(x)\|+\|f(x)-g(x)\|+\|f(rx)-g(rx)\|$ and $\|f(rx)-g(rx)\|=r^{-d} \|f(x)-g(x)\|$.