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I'm working on the following question. I'm having trouble with the solution presented in the textbook - specifically weak convergence.

Let $f \in L^p(\mathbb{R}^N), 1<p<\infty$ with $\alpha>0$ real, define $f_n = n^\alpha f(nx)$ for $n = 1,2,3,\dots$. Does $\{f_n\}$ strongly converge? Does it weakly converge?

It can be shown that for $\alpha < N/p$ the sequence strongly converges, otherwise it does not strongly converge. Clearly, for $\alpha < N/p$ the sequence also weakly converges (I got this far). Now consider $\alpha = N/p$. Suppose $f$ vanishes for $|x| >K$...(some analysis to conclude weak convergence). For $f$ arbitrary, given $\epsilon > 0$ and $\phi \in L^q(\mathbb{R}^n)$, let $g$ be a function vanishing outside a compact set such that $\|f-g\|_p \leq \epsilon/2\|\phi\|_q$.

  1. Why do they consider $f$ compact first?
  2. How do we get that estimate between $f$ and the compactly supported $g$?
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I think you want $\mathbb{R}^N$ rather than $\mathbb{R}^d$, so your later $N/p$ makes sense.

  1. The analysis that you omitted should tell you why. Essentially when $f$ is compactly supported, eventually $f_n$ lives inside a very small ball centered 0 so you can estimate the integrals by turning the condition onto your $\phi\in L^q$, i.e., $\lvert\int\phi f_n\rvert\leq \lVert f_n\rVert_p\cdot\lVert \phi 1_{\lvert x\rvert<K/n}\rVert_q\to 0$.

  2. Consider $g_R=f\cdot 1_{|x|<R}$. As $R\to\infty$ we have $g_R\to f$ in $L^p$ so there is an $R$ where $\lVert f-g\rVert_p$ satisfies this estimate.

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  • $\begingroup$ yes I did mean $N$ not $d$ -- thanks! $\endgroup$ – yoshi May 23 at 15:54

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