Showing that $f(\lambda x)\to f(x)$ in $L^4$

Question

I would like to show that for $f\in L^4(\mathbb{R})$, we have that $$\lim_{\lambda\to 1} \int |f(\lambda x)-f|^4\,dx=0. $$ I first used the fact that $C_0^\infty$ is dense in $L^4$ so that, letting $\varepsilon>0$, we can find $\psi\in C_0^\infty$ such that $\|f-\psi\|_{L^4}<\varepsilon$. We now estimate using the triangle inequality, $$\|f(\lambda x)-f(x)\|_{L^4}\le \|f(\lambda x)-\psi(\lambda x)\|_{L^4} + \|f(x)-\psi(x)\|_{L^4} + \|\psi(\lambda x)-\psi(x)\|_{L^4}<\varepsilon + \varepsilon/\lambda^{1/4} +g(\lambda) $$ with $$g(\lambda)=\|\psi(\lambda x)-\psi(x)\|_{L^4}.$$ We would like to show that given $\varepsilon>0$ from earlier, there exists $\delta>0$ such that for $\lambda\in B(1, \delta)=\{\lambda\in\mathbb{R}: |\lambda-1|<\delta\}$, $ g(\lambda)<\varepsilon$. Without loss of generality, we assume that $\lambda>1/2$ for simplicity. We know that there exists $R>0$ such that $\mathrm{supp}(\psi)\subset B(0,R)$. Thus, $\mathrm{supp}(\psi(\lambda x))\subset B(0,R/\lambda)\subset B(0,2R)$. Hence, $\mathrm{supp}\left(\psi(\lambda x)-\psi(x)\right)\subset B(0,2R)$ and $$\int |\psi(\lambda x)-\psi(x)|^4\,dx = \int |\psi(\lambda x)-\psi(x)|^4 \chi_{B(0,2R)}(x)\,dx.$$ The integrand in the last integral satisfies $$|\psi(\lambda x)-\psi(x)|^4 \chi_{B(0,2R)}(x)\le 16\chi_{B(0,2R)}(x) \sup_{B(0,2R)} |\psi|^4:=h(x)$$ where $h(x)$ is independent of $\lambda$ and is integrable. Thus, by the dominated convergence theorem, we have $$\lim_{\lambda\to 1} \int |\psi(\lambda x)-\psi(x)|^4\,dx =\int \lim_{\lambda\to 1} |\psi(\lambda x)-\psi(x)|^4\,dx=0.$$ Thus, given our previous $\varepsilon>0$, we can pick $\delta>0$ such that whenever $|\lambda-1|<\delta$, $\|\psi(\lambda x)-\psi(x)\|_{L^4}<\varepsilon$. Hence, we have $$\|f(\lambda x)-f(x)\|_{L^4} < \varepsilon + 2 \varepsilon + \varepsilon = 4\varepsilon. $$ Since $\varepsilon$ is arbitrary we have shown that $$\lim_{\lambda\to 1} \int |f(\lambda x)-f|^4\,dx=0. $$ Is the proof correct, and if so, how may it be improved?

Answer 1

Your proof looks good! If you want to simplify it, you technically can by using the density of $C_c$ (continuous compactly supported), instead of the smooth and compactly supported functions. You'll note that nowhere in the proof do we need to use any derivatives of the approximant $\psi$. It looks like the only essential properties you needed were that

• $\mathrm{supp}\,\psi$ is bounded
• $\psi$ is continuous.

Also, I would dissent with the comment you received on your post. Using notation like $\|f(\lambda x)-f(x)\|_{L^p(dx)}$ is fine, as long as you specify somewhere in the notation what is the variable of integration (so don't omit the $L^p(dx)$ in the subscript!). It actually makes it clearer what you mean when there are more than one parameters (such as $\lambda$ and $x$, or $x, y, h$, etc.), but it has to be used correctly.