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Let $(f_k)_{1\le k\le \infty}\in L_{1}^\mathrm{loc}(\mathbb{R}^n)$ be a sequence of real valued functions such that $\operatorname{supp} f_k \subset \{|x|\le k^{-1}\}$, $$\int f_k (x)\,dx=1,k\in \{1,2,\ldots,\infty\}$$ Show that the sequence $(f_k^2)_{1\le k\le\infty}$ does not converge in $\mathcal{D}'(\mathbb{R}^{n})$ as $k\rightarrow \infty$.

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solved. No need to labor others anymore. – Bombyx mori Jun 6 '12 at 5:25
Why don't you post your solution as answer? It's not unlikely that others might be interested... – draks ... Jun 6 '12 at 5:54
@draks: As the solution posted by others showed, the problem is quite elementary. – Bombyx mori Jun 6 '12 at 14:56
up vote 0 down vote accepted

Take $\psi$ a bump function: smooth function, with support contained in $B(0,2)$, non-negative, and $\psi=1$ on $B(0,1)$. If $\{f_k^2\}$ where convergent in $\mathcal D'(\Bbb R^n)$, in particular the sequence $\{a_k\}:=\{\int_{\Bbb R^n}f_k^2(x)\psi(x)dx\}$ would be bounded. Since the support of $f_k^2$ is contained in $B(0,k^{-1})$, we have $a_k=\int_{|x|\leq k^{-1}}f_k^2(x)dx$. Denoting $M:=\sup_{j\in\Bbb N}\int_{|x|\leq j^{-1}}f_j^2(x)dx$ $$1=\int_{B(0,k^{-1})}f_k(x)dx\leq \sqrt{|B(0,k^{-1})|}\sqrt{\int_{|x|\leq k^{-1}}f_k^2(x)dx}\leq \sqrt{|B(0,k^{-1})|}\sqrt M$$ hence $\frac 1M\leq |B(0,k^{-1}|$, which is a contradiction, because the volume of the ball of radius $k^{-1}$ converges to $0$ as $k\to +\infty$.

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