Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $k\rightarrow \infty$.

share|improve this question
    
solved. No need to labor others anymore. –  Bombyx mori Jun 6 '12 at 5:25
2  
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

1 Answer 1

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 $k\to +\infty$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.