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I'm trying to read this paper (Schoen-Simon-Yau '74) and I'm struggling to understand the comment to (1.34).

(See also here for another question on this paper.)

SSY are showing an estimate for the second fundamental form of an immersed $n$-dimensional Riemannian manifold $M$ in an ambient $(n+1)$-dimensional Riemanninan manifold $N$.

The inequality of interest is \begin{align} \tag{*} |A| \, \Delta |A| + |A|^4 &\geq \frac 2 {(1+\varepsilon) \, n} \, |\nabla |A||^2 - \frac {n \, (n-1)} {2 \, \varepsilon} \, (K_1 - K_2)^2 \\ &\quad - 2 \, c \, |A| + n \, (2 \, K_2 - K_1) \, |A|^2, \end{align}

where $|A|^2 = \sum_{i,j} h^2_{ij}$ with $h_{ij}$ the second fundamental form of $M$, $\nabla$ the covariant derivative on $M$, $K_1 \leq K_2 \in \mathbb R$ and $c \in \mathbb R$ constants.

Now my question is due to the comment:

"[$(*)$ holds] at all points where $|A| \neq 0$. Actually since $|A| \, \Delta |A| = \frac 1 2 \Delta |A|^2 - |\nabla|A||^2$ we can in fact see that this inequality must be globally true in the distribution sense, even if $|A|$ vanishes at various points."

I suppose the problem here is, that since $|A|$ involves a square root, any derivative term would yield infinity if $|A| = 0$. Since with the identity $|A| \, \Delta |A| = \frac 1 2 \Delta |A|^2 - |\nabla|A||^2$ there is no more square root in the first term, I suppose the problematic term is the latter. Now, it should be somehow possible to get rid of the derivatives by multiplying with a test function and integrating by parts. However, I can't see how this works.

EDIT 1

What I tried is to get rid of derivatives of $|A|$ (without the square). Let $\varphi \in C^\infty_c(M)$, then

$$\int |\nabla |A||^2 \, \varphi = \underbrace{\int \nabla \cdot (|A|\, \nabla |A| \, \varphi)}_{=0} - \int |A| \, \nabla |A| \cdot \nabla \varphi - \int |A| \, \Delta |A| \, \varphi$$

Now the first term is total divergence, so it will vanish. But how about the second? How do I get rid of the derivative in front of $|A|$?

EDIT 2

Eric Silva mentioned in the comments below, that $|A|$ would satisfy an elliptic PDE, as shown in Simons' paper. However, in Simons' paper (and in Holck-Minicozzi's book, eq. (2.16)) I can only find an elliptic PDE for $|A|^2$, whence I still have no clue about the regularity of $|A|$. Or am I wrong here?

EDIT 3

I think what could work is a convergence argument. Observe that upon multiplication with $|A|^2$ all the (possibly) singular terms behave smoothly. Now I want to show that for $\varphi \in C^\infty_c(M)$ with $\varphi_\varepsilon = \frac 1 \varepsilon \, \min\{\varepsilon,|A|^2\} \to \varphi \,\, (\varepsilon \to 0)$ we have $$\int_M |\nabla |A||^2 \, \varphi = \lim_{\varepsilon \to 0} \int_M |\nabla |A||^2 \, \varphi_\varepsilon < \infty$$

Does someone have an idea or a tip for me?

Thanks a lot in advance!

cesare borgia
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  • $|A|$ is at least locally Lipschitz, so it's differentiable almost everywhere, and so its zero set has measure zero so the problematic points don't seem to matter in the end. –  Nov 16 '17 at 08:51
  • how do i know that $|A|$ is at least locally Lipschitz? – cesare borgia Nov 16 '17 at 09:40
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    If $M$ is an immersed minimal submanifold of $N$, then its second fundamental form $A$ satisfies a linear elliptic PDE of second order (this is the main result of the paper of Simons which Schoen-Simon-Yau reference in the very beginning of the paper) and so is a fortierori $C^{1}$. –  Nov 16 '17 at 10:11
  • isn't it a problem, that $|A|$ is not globally lipschitz? in this case, how do you show that the zero set is of measure zero? – cesare borgia Nov 22 '17 at 20:06
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    You can write $M$ as a countable union of neighborhoods in which $|A|$ is Lipschitz so it's not a problem. –  Nov 22 '17 at 20:09
  • another question: in prop. 2.2.1 of simons paper, he states that $A$ is a $C^\infty$ section in $HM$, therefore $|A|$ is $C^\infty$ too, or am I wrong? so no need to invoke the elliptic PDE for this result, right? – cesare borgia Nov 23 '17 at 20:14
  • There really isn't any need to invoke the elliptic pde result, it's overkill, hence my a fortierori comment, I was simply trying to emphasize that it's actually much better than you need. but mearly being $C^{\infty}$ doesn't guarantee $|A|$ is. I mean the identity map from $\mathbb{R}^{n}$ to itself is $C^{\infty}$ but $f(x) = |x|$ is not. But $A$ being smooth does guarantee enough to apply Rademachers theorem so you're fine. –  Nov 23 '17 at 21:37
  • i'm sorry for the silly questions, i'm also a novice in measure theory, but i'm trying to learn; could you maybe give some more detail to your last comment? how do i see, that $|A|$ is differentiable a.e.? – cesare borgia Nov 25 '17 at 11:33
  • Does $|\nabla |A||^2$ even make sense as a distribution? I know the product of distributions is in general not well defined, and it certainly feels to me like you can't integrate-by-parts away all the derivatives of $|A|.$ – Anthony Carapetis Nov 27 '17 at 11:45
  • i get your point, but we should be able to make sense out of it; since $A$ is smooth, $|A|$ is smooth on $M \smallsetminus \{ |A| = 0\}$; maybe with some convergence argument – cesare borgia Nov 27 '17 at 12:31
  • I was confusing myself, it's not really problematic: since $|A|$ is Lipschitz, its distributional derivative is just the a.e. derivative from Rademacher (since $C^{0,1} = W^{1,\infty}$), which is in $L^\infty$, so it makes sense to square it. I agree that some kind of approximation argument should work: the idea should be that the terms that are singular at $|A|=0$ have enough powers of $|A|$ multiplying them to not be an issue. – Anthony Carapetis Nov 28 '17 at 08:48
  • @AnthonyCarapetis i know the result you mentioned ($C^{0,1} = W^{1,\infty}$), however i don't see how and why $|A|$ should be Lipschitz? In Colding-Minicozzi there is only an elliptic PDE for $|A|^2$. What am I missing? – cesare borgia Nov 28 '17 at 09:16
  • @cesareborgia: $x\mapsto A(x)$ is smooth (and thus locally Lipschitz) and the matrix norm $A \mapsto |A|$ is a Lipschitz function of the matrix components; so the composition $x \mapsto |A(x)|$ is locally Lipschitz. You don't need the minimality here - it's true for any $C^3$ surface. – Anthony Carapetis Nov 28 '17 at 09:42
  • @AnthonyCarapetis: I suppose you mean the one-norm, ie. $|A| = \sum_{i,j} |h_{ij}|$. However, please observe that in the question I stated we consider the two-norm, ie. $|A|^2 = \sum_{i,j} h_{ij}^2$. I agree that the one-norm of $A$ is (locally) Lipschitz, but the two-norm would only be Hölder continuous. Or am I missing your point? – cesare borgia Nov 28 '17 at 10:54
  • The 1-norm and 2-norm are equivalent, so it doesn't matter which you use. They're both Lipschitz. (In fact any norm is Lipschitz with constant 1, so long as you are using the same norm to measure the inputs: this is the reverse triangle inequality $| |A| - |B| | \le |A - B|.$) – Anthony Carapetis Nov 28 '17 at 11:03
  • @AnthonyCarapetis thank you so much! (for your patience) Would you post your last comment as an answer? This exactly solves my question and I would like to give the bounty to you. – cesare borgia Nov 28 '17 at 23:17

1 Answers1

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Since $A$ is smooth, it is locally Lipschitz, so from the reverse triangle inequality we see that $|A|$ is too. Since the distributional derivative of a Lipschitz function is in $L^\infty$, we conclude that $\nabla |A|\in L^\infty_\mathrm{loc}$ and thus also $|\nabla|A||^2 \in L^\infty_\mathrm{loc}.$

After using the given identity to eliminate $\Delta|A|,$ the inequality we have on $|A| \ne 0$ becomes $$Z = \frac12\Delta|A|^2 + |A|^4 + C_1|\nabla|A||^2 + C_2|A|+C_3|A|^2+C_4^2 \ge 0$$ for some constants $C_i$. Note that this now contains no singular terms: $Z$ is a genuine $L^\infty_\mathrm{loc}$ function defined on the whole surface. This means that to show this inequality holds globally, we can show separately that $Z \ge 0 $ on $|A| \ne 0$ - which we already know - and $Z\ge 0$ on $|A|=0$, which is easy: when $|A|=0$ we are at a local minimum of $|A|^2,$ so $\Delta|A|^2 \ge 0$, and thus $$Z = \frac12 \Delta|A|^2 + C_1 |\nabla|A||^2 + C_4^2$$ is manifestly non-negative.

Anthony Carapetis
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