Let $B_0(1)$ be the unit ball in $\mathbb R^n$, $n\geq2$. Let $f\in W_0^{1,2}(B_0(1))$, and $W^{1,2}(B_0(1))\ni f_i\to f$ in the sense of $L^2(B_0(1))$-norm, as $i\to \infty$.

Question 1: Can we find a sequence of positive numbers $\{\delta_i\}$ (may depends on $\{f_i\}$), going to $0$, s.t., $$\frac{1}{\delta_i^2}\int_{1-\delta_i\leq|x|\leq1}f_i^2\,dx\to0,$$ as $i\to\infty$ ?

Question 2: If Question 1 is not true, we then strengthen the condition that $W^{1,2}(B_0(1))\ni f_i\to f$ in the sense of $W^{1,2}(B_0(1))$-norm. Can we prove Question 1's conclusion ?


2 Answers 2


I think Question 1 has a positive answer. Denote $B=:B_0(1)$ and $A_r:=\{r<\|x\|<1\} $ for $0<r<1$. For functions $f\in W^{1,2}_0(B)$ we have a Poincaré inequality on $A_r$ : $$\int_{A_r} f^2dx\le \Big(\frac{1-r}{r}\Big)^2 \int_{A_r} |\nabla f|^2dx \ .$$

Now let $(f_i)$ a sequence in $W^{1,2}(B)$ converging to $f$, so $f^2_i$ converges to $f^2$ in $L^1(B)$. For any sequence $0<\delta_i<1$ we have, by the above inequality:

$$\frac{1}{\delta_i^2}\int_{A_{1-\delta_i}}f_i^2dx\le \frac{1}{\delta_i^2}\int_{A_{1-\delta_i}}f ^2dx + \frac{1}{\delta_i^2} \|f^2-f_i^2\|_{1} \le$$

$$\le \frac{1}{ (1-\delta_i)^2} \int_{A_{1-\delta_i}}\|\nabla f\| ^2dx + \frac{1}{\delta_i^2} \|f^2-f_i^2\|_1\ .$$

Now if $\delta_i=o(1)$, the first term on the RHS is $o(1)$, just because $\|\nabla f\| ^2$ is integrable and $\operatorname{meas}(A_{1-\delta_i})=o(1)$.

On the other hand, if we also choose the sequence $\delta_i$ such that $ \|f^2-f_i^2\|_1=o(\delta_i^2)$ as $i\to\infty$, the other term is also $o(1)$.


Details on the above Poincaré inequality. By density, one can assume $f\in C^\infty_c(B)$. Let $0<r\le \|x\|\le 1$. So $f(x/r)=0$, and we have $$f(x)= -\int_1^{1/r}\partial_t f(tx)dt=-\int_1^{1/r}\nabla f(tx)\cdot x\ dt\ . $$ Thus by Cauchy-Schwarz $$ f(x) ^2\le \Big(\frac{1}{r} -1\Big) \int_1^{1/r}\|\nabla f(tx)\|^2 dt\ , $$ and $$ \int_{A_r} f(x) ^2 \ dx\le \Big(\frac{1}{r} -1\Big) \int_{A_r}\int_1^{1/r}\|\nabla f(tx)\|^2 \ dt\ dx = $$ $$=\Big(\frac{1}{r} -1\Big) \int_1^{1/r} t^{-n} \int_{A_{tr}} \|\nabla f( x)\|^2 \ dx \ dt $$ $$\le \Big(\frac{1}{r} -1\Big)^2 \int_{A_{r}} \|\nabla f( x)\|^2 \ dx \ . $$


The answer to Q1 is yes. As pointed out by both Pietro and me in this very similar question, we will have that $$ \frac{1}{\delta^2}\int_{1-\delta\le |x|\le 1} f^2\, dx \to 0 , $$ since $f\in W^{1,2}_0(B)$.

Take $\sigma_k$ so small that this is $<1/k$ for $\delta\le\sigma_k$. Then also $$ \frac{1}{\sigma_k^2}\int_{1-\sigma_k\le |x|\le 1} f_n^2 < \frac{1}{k} $$ for all large $n\ge N=N_k$ by $L^2$ convergence. We can also assume that the $N_k$ increase and $\sigma_k\to 0$, and now we can define $\delta_n$ to be $\sigma_k$ for $N_k\le n<N_{k+1}$.

  • $\begingroup$ This also says that the sequence $\delta_k$ must converge to $0$ not too fast. For instance, keeping fixed $\sigma_k$, any other $N_k'\ge N_k$ works as well, and the corresponding $\delta_n'$ would converge to $0$ slowly (just to check the conclusion I got). $\endgroup$ Dec 25, 2015 at 19:47
  • $\begingroup$ @PietroMajer: Yes, I think we are saying the same thing in our answers. Before I can make $\delta_n$ smaller, I must wait for $\int_{1-\delta\le |x| \le 1} f_n^2$ to become sufficiently small, with the new $\delta$. $\endgroup$ Dec 25, 2015 at 21:40
  • $\begingroup$ Both answers are so useful, thank you so much. I find that I could not accept two answers at the same time. $\endgroup$
    – user84068
    Dec 27, 2015 at 4:50

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