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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


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.

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1 Answer 1

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.

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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

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