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.

This afternoon, while working out this answer by Nate Eldredge, I made some vain attempts at building cutoff functions of various kinds. Especially I was looking for the following.

Can a sequence $\zeta_n \in C^\infty(-1, 1)$ s.t.

  1. $0 \le \zeta_n \le 1$ and $\zeta_n(0)=1, \zeta(-1)=\zeta(1)=0$;
  2. $\lVert \zeta_n-1\rVert_2 \to 0$;
  3. $\lVert \zeta'_n \rVert_2$ is bounded

exist?

Graphically I was looking for something like this: L2 nice cutoff

(in red the graph of $\zeta_n(x)=\exp(\frac{1/n}{x^2-1})$, in black the scaled graph of its first derivative). The task was to arrange things so that those black peaks stayed $L^2$-bounded.

After many unsuccessful trials I've come to the conclusion that such a $\zeta_n$ cannot exist. In fact, it should be $H^1$-bounded and so, up to a subsequence, $H^1$-weakly convergent. This implies $L^2$-weak convergence and so we should have $\zeta_n \stackrel{H^1}{\rightharpoonup}1$. But now we observe that $\zeta_n \in H^1_0(-1, 1)$. $H^1_0(-1, 1)$ is a norm-closed subspace of $H^1(-1,1)$, and so - because of its convexity - it is also weakly closed. We have thus gotten the contradiction $1 \in H^1_0(-1, 1)$.

Two questions:

  1. Is this reasoning correct?
  2. If 1. is affirmative, is there a more elementary way to see this? I've got the strong feeling of using a sledge-hammer to crack a nut.

Thank you for your attention.

share|improve this question
add comment

1 Answer

up vote 3 down vote accepted

I will try to answer question 2 by explicitly stating a solution, which I feel is as elementary as possible.

It's not so hard to see that it suffices to restrict ourselves to functions that are monotone on both $[-1,0]$ and $[0,1]$. In this case we can define $a_n$ as the unique value in $[0,1]$ with $\zeta_n(a_n) = \frac{1}{2}$. Then $\int_{-1}^1 (\zeta_n(x)-1)^2 dx \geq \int_{a_n}^1 (\zeta_n(x)-1)^2 dx \geq \int_{a_n}^1 \left(\frac{1}{2}\right)^2 dx = \frac{1}{4}(1-a_n)$, so since the original integral converges to $0$, $a_n$ must converge to $1$.

Now define $M_n = \frac{1}{1-a_n}\int_{a_n}^1 |\zeta_n^\prime(x)| dx \geq \frac{1}{1-a_n}\int_{a_n}^1 -\zeta_n^\prime(x) dx = \frac{1}{1-a_n}\left[-\zeta_n(x)\right]_{a_n}^1 = \frac{1}{2(1-a_n)}$. Then $0 \leq \int_{a_n}^1 (|\zeta^\prime(x)|-M_n)^2 dx = \int_{a_n}^1 |\zeta^\prime(x)|^2 dx - 2M_n\int_{a_n}^1 |\zeta^\prime(x)| dx + M_n^2(1-a_n)$. By our definition of $M_n$, $M_n^2(1-a_n) - 2M_n\int_{a_n}^1 |\zeta_n^\prime(x)| dx = -(1-a_n)M_n^2$, so we conclude $\int_{a_n}^1 |\zeta^\prime(x)|^2 dx \geq (1-a_n)M_n^2 \geq \frac{1}{4(1-a_n)}$ and therefore $\|\zeta_n^\prime\|_2 \geq \frac{1}{2}(1-a_n)^{-\frac{1}{2}}$, which since $a_n \rightarrow 1$, as we saw, is not bounded.

I hope this answer is elementary by your standards.

share|improve this answer
    
Tomorrow I will read this thoroughly (now it is late in my country), but it looks like the perfect "elementary" answer I was interested in. –  Giuseppe Negro Jun 4 '11 at 23:47
    
Your answer is a mine of nice techniques! Understanding it has been a good exercise for me. I would like to propose an alternate proof. Let $a_n$ be as in your proof, and assume we have already proven that $a_n \to 1$. Since $$\int_{a_n}^1 \lvert \zeta'_n(x) \rvert \, dx=\frac{1}{2}, $$ applying the Cauchy-Schwarz inequality we have $$\frac{1}{4} \le \int_{a_n}^1 \lvert \zeta'_n(x) \rvert^2 \, dx \underbrace{(1-a_n)^2}_{\to 0}, $$ from which we infer that $\int_{a_n}^1 \lvert \zeta'_n(x)\rvert^2\, dx \to +\infty$. –  Giuseppe Negro Jun 5 '11 at 18:14
    
Agreed, your alternative is definitely neater than the excessive algebra in my approach. –  gfes Jun 5 '11 at 18:22
    
I've liked very much what you call excessive algebra. Substantially you have used the probabilistic inequality $$\mathbb{E}(X^2) \ge (\mathbb{E}(X))^2. $$ I find it very beautiful. I've also liked very much the main idea of your proof, that is taking into account the terminal part of the interval $[0, 1]$, where we must expect the most of the derivative will be. –  Giuseppe Negro Jun 5 '11 at 18:37
add comment

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.