# The comparison between the square of the functional value and the sum of squares of the $L^2$ norms of function and its Laplacian

I was reading a paper where I came across the following argument :

For any $x$ in $M$ and for a geodesic ball $B(x; \varepsilon)$ in a compact Riemannian manifold $M$ with injectivity radius bigger than or equal to epsilon, and for any smooth eigenfunction $f$ of Laplacian on $M$, we have :

the square of $f(x)$ is $\leq C$ times ( the square of $L^2$ norm of $f$ over $B(x;\varepsilon)$ $+$ square of $L^2$ norm of $L(f)$ over $B(x:\varepsilon)$),

where $L(f)=$ Laplacian of $f$, where $C$ is independent of the Riemannian metric on $M$.

I was unable to see, with my limited Analysis knowledge, why this is true, but they mentioned that it follows from Sobolev's and Garding's inequality, for which they referred to S. Agmon's "Lectures on Elliptic boundary value problems"... still it is unclear to me.

N.B.: the injectivity radius of a manifold is the smallest of all numbers r such that I can have a geodesic ball of radius r around each point of M. e.g. injectivity radius of the sphere of radius $1$ with standard metric is $\pi$, injectivity radius of $\Bbb R^n$ is infinity etc.

Any help ? Thanks in advance.

• I would guess that the "eigenfunction" hypothesis is probably redundant,probably it should hold for any smooth function f on compact M, at least that is apparent from the way the steps are written in the paper. I cannot attach the paper here, because it is not available from mathscinet etc. – Analysis Now Jul 21 '10 at 6:45
• You should probably make the title more descriptive if you want answers. – Harry Gindi Jul 21 '10 at 6:47
• The dimension of M is not larger than 3, is it? – Pietro Majer Jul 21 '10 at 7:12
• Too add to Pietro's comment: roughly speaking Garding allows you to control the Sobolev norm $\|f\|_{H^2}$ by $\|f\|_{L^2} + \|\triangle_g f\|_{L^2}$. And in dimensions $\leq 3$ Sobolev controls $L^\infty$ by $H^2$. If there's any reason why they specify it only for eigenfucntions, it would most likely be for the fact that they want $C$ to be independent of the metric. – Willie Wong Jul 21 '10 at 10:07
• @SPal: is the paper published (in English)? If so you can just give us the title and the author. If you otherwise have permission to share the paper, you can privately e-mail me and I'll take a look at it. Or you can try to include more context in your question (which will definitely help us to decipher it). In particular, as the set given by the geodesic ball is dependent on the metric chosen, I am really curious about how $C$ can be independent of the metric. – Willie Wong Jul 21 '10 at 10:16

You are working on a Riemann surface. That bit of information is rather important, as Sobolev inequalites depends rather much on the dimension of the space. The basic Sobolev inequality is $$\| f \|_{L^q(\Omega)} \leq C (\| \partial f \|_{L^p(\Omega)} + \| f\|_{L^p(\Omega)})$$ where the condition $\frac1p \geq\frac1q \geq \frac1p - \frac1n$ is satisfied (and $\Omega$ needs to be suitably regular). and $C$ depends on the set $\Omega$ and the coefficients $p,q$. If you want the sup norm on the left hand side, you can morally speaking replace $q$ by $\infty$ (so $1/q = 0$ and ask that the second inequality be strict).

In any case, in two dimensions by iterating the derivatives, you can actually show that for smooth $f$ $$|f| \leq C( \|f\|_{L^2} + \|\partial^2 f\|_{L^2})$$ using that $0 > 1/2 - 2/2$. (The 2 in the numerator is the number of derivatives. In the denominator in the first term is the Lebesgue exponent, and in the second term is the dimension.)

Now, a consequence of Garding's inequality states that for an uniformly elliptic differential operator $L$ of order $k$, one has that $$\| \partial^k f\|_{L^2} \leq C (\| Lf\|_{L^2} + \| f\|_{L^2})$$ so using that the Laplacian is uniformly elliptic of order 2, you can plug Garding's inequality into Sobolev inequality and square the whole expression to get what the authors claim.

As to the actual dependence of the constant $C$ on various parameters: off the top of my head I can't remember the details. So I suggest you look it up either in Agmon's book as the authors suggest, or in Gilbarg & Trudinger Elliptic Partial Differential Equations of Second Order or Adams Sobolev Spaces

• Oh, and by the way, the fact that $f$ is an eigenfunction is not used in this step. It is used in the next step where the Laplacian $\triangle_g f$ is replaced by $\lambda^2 f$, the eigenvalue. – Willie Wong Jul 21 '10 at 16:19
• Thanks very much Mr. Willie Wong, I really appreciate your answer ! – Analysis Now Jul 21 '10 at 16:26
• It seems to me that more needs to be said about the constant $C$. As far as I can tell, the constant $C$ depends on the metric $g$ and its Christoffel symbols with respect to local co-ordinates, and I don't think just knowing that the domain is within the injectivity radius is enough to get uniform ellipticity or uniform bounds on the Christoffel symbol. However, the paper in question appears to assume that the metric is hyperbolic (constant Gauss curvature equal to -1). That would be enough. – Deane Yang Jul 21 '10 at 16:49
• Deane: is it not enough that $\epsilon$ is sufficiently small? Suppose we have a compact Riemann surface, the Gauss curvature is everywhere bounded, so on geodesic balls of size $\epsilon$ we can bound the Christoffel symbols (in geodesic normal coordinates, say) by something like $\epsilon R$, where $R$ is the upper bound of the Gauss curvature, and so the variations in the metric is bounded by roughly $\epsilon^2 R$ and so in the patch we have uniform ellipticity. I hope I am not missing something obvious. Of course, I agree with you that just "within injectivity radius" is not enough. – Willie Wong Jul 21 '10 at 17:12
• Willie, your comment is essentially on the mark, but I didn't see any mention of curvature in either the question or your answer. If you have upper and lower bounds on the Gauss curvature, then you can definitely find co-ordinates in which you get full control over the constant $C$. Surprisingly, geodesic normal co-ordinates don't seem to work, but Jost and Karcher showed that "almost linear co-ordinates" do. But everything is a lot easier, if you just assume the metric is hyperbolic. – Deane Yang Jul 21 '10 at 18:14