The comparison between the square of the functional value and the sum of squares of the L^2 norms of function and its Laplacian - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T06:14:23Z http://mathoverflow.net/feeds/question/32752 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/32752/the-comparison-between-the-square-of-the-functional-value-and-the-sum-of-squares The comparison between the square of the functional value and the sum of squares of the L^2 norms of function and its Laplacian Analysis Now 2010-07-21T06:42:08Z 2010-07-22T12:22:57Z <p>I was reading a paper where I came across the following argument :</p> <p>For any x in M and for a geodesic ball B(x; epsilon) 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 :</p> <p>the square of f(x) is &lt;= C times ( the square of L^2 norm of f over B(x;epsilon) + square of L^2 norm of L(f) over B(x:epsilon)), </p> <p>where L(f)= Laplacian of f, where C is independent of the Riemannian metric on M.</p> <p>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.</p> <p>N.B.: ihe 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 R^n is infinity etc.</p> <p>Any help ? Thanks in adavance !</p> http://mathoverflow.net/questions/32752/the-comparison-between-the-square-of-the-functional-value-and-the-sum-of-squares/32811#32811 Answer by Willie Wong for The comparison between the square of the functional value and the sum of squares of the L^2 norms of function and its Laplacian Willie Wong 2010-07-21T16:16:57Z 2010-07-21T16:16:57Z <p>You are working on a Riemann surface. That bit of information is rather important, as <a href="http://en.wikipedia.org/wiki/Sobolev_inequality" rel="nofollow">Sobolev inequalites</a> depends rather much on the dimension of the space. The basic Sobolev inequality is <code>$$\| f \|_{L^q(\Omega)} \leq C (\| \partial f \|_{L^p(\Omega)} + \| f\|_{L^p(\Omega)})$$</code> 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). </p> <p>In any case, in two dimensions by iterating the derivatives, you can actually show that for smooth $f$ <code>$$|f| \leq C( \|f\|_{L^2} + \|\partial^2 f\|_{L^2})$$</code> 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.) </p> <p>Now, a consequence of <a href="http://en.wikipedia.org/wiki/G%C3%A5rding%27s_inequality" rel="nofollow">Garding's inequality</a> states that for an uniformly elliptic differential operator $L$ of order $k$, one has that <code>$$\| \partial^k f\|_{L^2} \leq C (\| Lf\|_{L^2} + \| f\|_{L^2})$$</code> 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. </p> <p>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 &amp; Trudinger <em>Elliptic Partial Differential Equations of Second Order</em> or Adams <em>Sobolev Spaces</em></p>