Does scaling lead to weak convergence to the null function?

Let $f\in L^p(\mathbb{R}^d)$, with $1<p<\infty$. Is it true that $$\lambda^{\frac{d}{p}}f(\lambda x ) \rightharpoonup 0\quad \text{ weakly in }L^p\text{ as }\lambda\to+\infty?$$

One has the easy case for $f\in L^{p-\epsilon}\cap L^p$. This condition allows the use of Hölder's inequality as follows (here $\frac{1}{p}+\frac{1}{p'}=1$ and $\phi$ is a continuous function with compact support): $$\begin{split}\left\lvert \int_{\mathbb{R}^d} \lambda^{\frac{d}{p}} f(\lambda x)\phi(x)\, dx\right\rvert& \le \lambda^{\frac{d}{p}}\lVert f(\lambda x)\rVert_{L^{p-\epsilon}}\lVert \phi\rVert_{L^{(p-\epsilon)'}} \\ &= \lambda ^{\frac{d}{p}- \frac{d}{p-\epsilon} } \lVert f\rVert_{L^{p-\epsilon}}\lVert\phi\rVert_{L^{(p-\epsilon)'}}\to 0. \end{split} $$ But what happens if that condition is removed? I carried out an explicit check on the standard example of a function $f$ that belongs to $L^2(\mathbb{R})$ and does not belong to $L^{2-\epsilon}(\mathbb{R})$ for any $\epsilon > 0$, namely $$f(x)=\begin{cases} 0 , & x<2 \\ \frac{1}{\sqrt{x}\log x}, & x \ge 2\end{cases}.$$ Taking $\phi=\chi_{[a, b]}$ with $2<a<b$ one has $$\left \lvert \int_{\mathbb{R}} \lambda^{\frac{1}{2}}f(\lambda x)\phi(x)\, dx \right\rvert = \dfrac{ \int_{\lambda a }^{\lambda b} \frac{dy}{\sqrt{y}\log(y)} } {\lambda^{\frac{1}{2}}}, $$ and an application of l'Hôpital's rule shows that the right hand side tends to $0$ as $\lambda \to \infty$. This implies weak convergence to $0$ by a standard density argument.

This seems to point towards an affirmative answer to the question in the gray box.

Solution 1:

Fix $g$ in $L^q$ and $\epsilon>0$. There is $R>0$ such that $\|f\chi_{\{|x|>R\}}\|_p<\epsilon$. There is also $r>0$ such that $\|g\chi_{\{|x|<r\}}\|_q<\epsilon$.

Let $f_\lambda(x) = \lambda^{\frac{d}{p}}f(\lambda x )$. If $\lambda>R/r$, then $$\| f_\lambda \chi_{\{|x|>r\}}\|_p \le \|f\chi_{\{|x|>R\}}\|_p<\epsilon$$ Hence $$\int|f_\lambda g| = \int_{\{|x|<r\}} |f_\lambda g| + \int_{\{|x|> r\}} |f_\lambda g| \le \|f\|_p\,\epsilon + \epsilon\, \|g\|_q $$ by Hölder's inequality applied to each integral.

Since $\epsilon$ was arbitrarily small, $\int|f_\lambda g| \to 0$ as $\lambda\to\infty$.