Give a sequence of functions $f_n:\mathbb{R}\to\mathbb{R}$ such that $$f_n \to 0 \text{ in } L^3(\mathbb{R}), f_n\rightharpoonup 0 \text{ in } L^2(\mathbb{R}), f_n \nrightarrow 0 \text{ in } L^2(\mathbb{R}).$$

I took $f_n(x)=n^{1/2}I_{[0,1/n]}$, then $||f_n||_{L^2}=1$ and the third condition holds. For $\phi \in C_c^{\infty}(\mathbb{R})$ dense in $L^2(\mathbb{R})$ we have$$\int_0^{1/n} f_n\phi=\int_0^{1/n} n^{1/2}\phi =n^{-1/2}\phi(1/n)-\int_0^{1/n} n^{1/2}x\phi'\leq n^{-1/2}\phi(1/n)-\frac12n^{-3/2}\sup |\phi'|\to 0$$ which shows the second condition. But I I cannot the satisfy first one as well. Tweaking the exponents makes one of the other conditions fail…


Since $L^3 \hookrightarrow L^2$ on bounded domains, the support of the functions $f_n$ cannot be uniformly bounded. Otherwise (1) and (3) contradict.

Define $$ f_n = \chi_{[0,n]} n^{-1/2}. $$ Then $\|f_n\|_{L^3}^3=n^{-1/2}$, $\|f_n\|_{L^2}^2=1$. In addition, $\int_{\mathbb R} f_n \phi \to0$ for all $\phi \in C_c(\mathbb R)$. Such functions are dense in $L^2(\mathbb R)$, hence $f_n \rightharpoonup 0$ in $L^2(\mathbb R)$.

  • $\begingroup$ Do we have that $\phi(n)=0$? We get a boundary term $[n^{-1/2}x\phi(x)]_0^n$ from integrating by parts? $\endgroup$ – user30523 2 days ago
  • $\begingroup$ ?? there is no integration by parts: $\int |f_n\phi| \le n^{-1/2} \|\phi\|_{L^\infty}\to0$. $\endgroup$ – daw 2 days ago

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