Applications of Hardy's inequality - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T08:38:05Z http://mathoverflow.net/feeds/question/48292 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/48292/applications-of-hardys-inequality Applications of Hardy's inequality Mark Kim 2010-12-04T19:05:49Z 2011-01-08T14:40:03Z <p>Every so often I would encounter Hardy's inequality:</p> <blockquote> <p><strong>Theorem 1 (Hardy's inequality)</strong>. If $p>1$, $a_n \geq 0$, and $A_n=a_1+a_2+\cdots+a_n$, then $$\sum_{n=1}^\infty \left(\frac{A_n}{n}\right)^p &lt; \left(\frac{p}{p-1}\right)^p \sum_{n=1}^\infty a_n^p,$$ unless $(a_n)_{n=1}^\infty$ is identically zero. The constant is the best possible.</p> </blockquote> <p>and its integral version:</p> <blockquote> <p><strong>Theorem 2 (Hardy's integral inequality)</strong>. If $p>1$, $f(x) \geq 0$, and $F(x) = \int_0^x f(t) \ dt$, then $$\int_0^\infty \left(\frac{F}{x}\right)^p \ dx &lt; \left(\frac{p}{p-1}\right)^p \int_0^\infty f^p(x) \ dx,$$ unless $f \equiv 0$. The constant is the best possible.</p> </blockquote> <p>with a comment or two emphasizing how <em>important</em> and <em>fundamental</em> they are. Nevertheless, I have yet to see a good application of the above inequalities. So...</p> <blockquote> <p>Could you give an application of Theorem 1 or Theorem 2 that you think is particularly useful or instructive?</p> </blockquote> <p>Thanks in advance!</p> http://mathoverflow.net/questions/48292/applications-of-hardys-inequality/48297#48297 Answer by Terry Tao for Applications of Hardy's inequality Terry Tao 2010-12-04T19:45:36Z 2010-12-04T22:29:22Z <p>In analysis, individual inequalities or estimates are usually not so useful <em>per se</em> (though there are some notable exceptions, such as the Sobolev embedding inequality, or the Cauchy-Schwarz inequality), but are instead representative examples of a larger useful class of estimates. (cf. Gowers' "<a href="http://www.dpmms.cam.ac.uk/~wtg10/2cultures.pdf" rel="nofollow">Two cultures of mathematics</a>".) </p> <p>In this particular case, the classical Hardy inequality exemplifies two useful principles; firstly, that an inverse power weight such as $1/|x|^\alpha$ is "dominated" in some $L^p$ sense by the corresponding derivative $|\nabla|^\alpha$ (or, to put it somewhat facetiously, $\frac{1}{x} = O(\frac{d}{dx} )$; compare with the uncertainty principle $dx \cdot d\xi \gtrsim 1$); and secondly, that a maximal average of a function is often dominated in an $L^p$ sense by the function itself. The first principle is captured by a number of higher-dimensional generalisations of Hardy's inequality (which typically take a shape such as</p> <p>$$\| \frac{f}{|x|^\alpha} \| _ {L^p({\bf R}^n)} \leq C_{p,\alpha,n} \| |\nabla|^\alpha f \|_{L^p({\bf R}^n)}$$</p> <p>under suitable assumptions on $p,n,\alpha,f$) which are fundamental to the analysis of any PDE that involves singular potentials or weights such as $\frac{1}{|x|^\alpha}$. The second principle is captured by a different family of generalisations of Hardy's inequality, namely the maximal inequalities for which the <a href="http://en.wikipedia.org/wiki/Hardy%25E2%2580%2593Littlewood_maximal_function" rel="nofollow">Hardy-Littlewood maximal inequality</a> is the model example. This inequality is the foundation of a large part of real-variable harmonic analysis, and in particular in the analysis of singular integral operators such as the Hilbert transform or pseudo-differential operators.</p> <p>There are two nice features of Hardy's original inequality that are also worth pointing out. Firstly, it is an $L^p$ inequality with an explicit optimal constant, which is something of a rarity in analysis (there are maybe only a dozen or so other such sharp inequalities known for the fundamental operators in analysis). The other is that the inequality is never actually satisfied with equality (except in the trivial case when the function vanishes); one can construct sequences of near-extremisers that get arbitrarily close to attaining equality, but they do not converge to a limit that actually attains that equality. (The function $f = x^{-1/p}$ formally attains equality for Theorem 2, but there is a logarithmic divergence on both sides.) This is perhaps one of the simplest examples of such a situation, and one which is well worth studying if one is interested in using variational methods to find optimal constants for other inequalities, as one needs to have a good intuition as to when one expects optimisers to actually exist or not.</p> http://mathoverflow.net/questions/48292/applications-of-hardys-inequality/48300#48300 Answer by Denis Serre for Applications of Hardy's inequality Denis Serre 2010-12-04T20:16:23Z 2010-12-04T20:16:23Z <p>This is a testimony rather than a answer. In the early eighties, I began looking at the inviscid limit in systems of conservation laws. These are PDEs (Partial differential equations) with time and space variables. The viscous system is second order in space, whereas the inviscid one is only first order. A typical example is Navier-Stokes <em>vs</em> Euler, for a compressible fluid. In presence of a boundary, the viscous system must be supplemented with a full boundary condition, for instance the Dirichlet condition $u=u_b$ (with $u(t,x)\in\mathbb R^n$ the unknown vector). On the contrary, the inviscid system needs a number of boundary conditions equal to the number $p$ of <em>entering modes</em> and we have in general $p&lt; n$. Therefore the limit, as the coefficients of the viscous tensor tend to zero, is singular: there appears a <em>boundary layer</em>.</p> <p>The first result (assuming that the boundary is non-characteristic) in this direction was establish in the PhD thesis of my student M. Gisclon. The Hardy inequality (with $p=2$) was crucial in the proof. Since then, this work has been extended in many directions by various authors.</p> <p>I apologize for being a bit topical, if not technical. </p> http://mathoverflow.net/questions/48292/applications-of-hardys-inequality/48307#48307 Answer by Willie Wong for Applications of Hardy's inequality Willie Wong 2010-12-04T20:50:57Z 2010-12-04T20:50:57Z <p>Just to add a bit more on what Terry and Denis said (it shouldn't be surprising the people who jumped at this question all work in PDEs): a simple example of the power of the Hardy Inequality can be illustrated by the following "qualitative" description of the inequality:</p> <blockquote> <p>The Hardy Inequality allows you to control low derivative norms by "weighted" high derivative norms subject to boundary conditions. </p> </blockquote> <p>This is particularly useful in the study of hyperbolic partial differential equations. Consider the linear wave equation:</p> <p>$$\Box u = 0$$</p> <p>Ignoring for now Fourier analytic methods, by just integration by parts in physical space, you get the "energy estimate" </p> <p>$$\frac{d}{ds}E(s) = 0, \quad E(s) = \int_{t = s} (\partial_tu(t,x))^2 + (\nabla u(t,x))^2 dx$$</p> <p>which provides you with global-in-time control of the $L^2$ norm of one derivative of a solution $u$. </p> <p>In general, you can get more complicated energy estimates by integrating the equation against different "weighted derivatives" of the solution $u$. This is often called the "ABC-method" of Morawetz or the "vector-field method" (depending on whom you talk to). The method can often give you a conserved, almost conserved, or monotonically decreasing scalar quantity that dominates a weighted $L^2$ norm of some <strong>positive</strong> number of derivatives of the function $u$. The key here is that we have a somewhat systematic way of constructing these energy estimates for hyperbolic PDEs. </p> <p>But sometimes you may need to estimate $L^2$ of the function itself, without derivatives, in the course of the argument. The construction of energy described above does not generally extend to work on the case of zero derivatives. This is where Hardy inequalities becomes useful. For hyperbolic equations, if the prescribed initial data has compact support, the "finite speed of propagation" property (often available for these systems) implies that for any future time, the solution will also have compact spatial support: therefore the boundary requirement of Hardy's inequality is satisfied. With this we can convert a weighted energy estimate on derivatives of the function $u$ to a weighted (with different weight) $L^2$ control for the function itself. </p> <p>To give a limited sample of how this is used in the context of wave equations</p> <ul> <li>A Hardy-type inequality is used in <a href="http://arxiv.org/abs/math/0605023" rel="nofollow">Rodnianski and Sterbenz's proof of stable blow-up for the co-rotational wave-map in 1+2 dimensions</a> for a similar reason to what I described above: they need to extend a certain energy control available for higher order derivatives to lower order derivatives, in order to control the lower-order terms in the evolution. </li> <li>Hardy-type inequalities is also used quite often in the work of Dafermos and Rodnianski studying the decay of linear waves on black hole backgrounds. See, e.g. <a href="http://arxiv.org/abs/1010.5132" rel="nofollow">this recent paper</a>. </li> </ul> http://mathoverflow.net/questions/48292/applications-of-hardys-inequality/48318#48318 Answer by Helge for Applications of Hardy's inequality Helge 2010-12-04T22:35:49Z 2010-12-04T22:35:49Z <p>AS for physical meaning of Hardy's inequality for p=2. Consider the Schroedinger operator $$H = -\frac{d^2}{dx^2} - \frac{c}{x^2}$$ on $(0,\infty)$ with Dirichlet boundary condition at $0$. A natural question is: When is this operator positive on $L^2(0,\infty)$ and the answer is just Hardy's inequality, since $$\langle \psi, H \psi \rangle \geq 0$$ is equivalent to $$\int_0^{\infty} |\psi'(x)|^2 dx \geq + c \int_0^{\infty} \left|\frac{\psi(x)}{x}\right|^2 dx$$ an inspection of Hardy's inequality now shows that $c = \frac{1}{4}$ is critical.</p> <p>One can actually use this reasoning to prove Hardy's inequality, since the ODE $-u''(x) = c u(x)/x^2$ is explicitly solvable.</p> http://mathoverflow.net/questions/48292/applications-of-hardys-inequality/48331#48331 Answer by Dale Roberts for Applications of Hardy's inequality Dale Roberts 2010-12-05T00:40:12Z 2010-12-05T00:40:12Z <p>To hopefully complement all the other great posts, here are a few examples where I have seen them used, either directly or lurking around in the background machinery (in their classic or generalised form):</p> <ul> <li>Studying the elliptic equation $\mathscr{A} u = f$ where $\mathscr{A}$ is the second-order elliptic operator $$(\mathscr{A} u)(x) := - \sum_{i=1}^n \frac{\partial}{\partial x_i} \left(a_i(x) \frac{\partial u}{\partial x_i}\right)(x) + a_0(x) u(x)$$ on some smooth bounded domain $U \subset \mathbb{R}^n$, where either $a_i(x) \to 0$ (i.e. <em>degenerates</em>) or $a_i(x) \to \infty$ (i.e. <em>has a singularity</em>) as $x \to x_0 \in \partial U$. Then we can use a weight to control the blow-up or degeneracy.</li> <li>Suppose on the other hand that your elliptic operator $\mathscr{A}$ is "nice" but your data is bad, i.e. consider <code>\left\{\begin{align} - \Delta u &amp;= f, &amp;&amp; x \in U,\\u &amp;= g,&amp;&amp; x \in \partial U.\end{align}\right.</code> If $g \in W^{-1/2,2}(\partial U)$ and $f \in W^{-1,2}(U)$ then we can obtain a weak solution $u \in W^{1,2}_0(U)$. However, if $g$ has a singularity at $x_0 \in \partial U$ then $g \notin L^2(\partial U)$ and we cannot use the "standard approach". Introducing a weight that disappears nears $x_0 \in \partial U$ allows us to proceed.</li> <li>Suppose $U$ is an unbounded domain. Sometimes when solving boundary value problems, we may want to introduce a weight to specify <em>conditions at infinity</em>. For example, $$\int_{|x|>1} |u(x)|^2(1+|x|)^\epsilon dx &lt; \infty, \quad \epsilon \in \mathbb{R}.$$</li> <li>Suppose now that $\partial U$ is no longer a smooth boundary but is quite "nasty", i.e. $\partial U$ might have sharp corners, cusps, etc. Again, when considering a PDE in this domain we can use a weight to control the bad behaviour near the nasty parts of the boundary. This is useful for numerical schemes, such as finite elements.</li> <li>Another BVP situation, we might have points on the boundary where the boundary conditions change, for example, from Dirichlet boundary conditions to Neumann boundary conditions. Weights can be used here too.</li> <li>When considering the stochastic partial differential equation (<em>SPDE</em>) $$\frac{\partial u}{\partial t} + \Delta u = \dot W(t)$$ where $\dot W$ is a space-time Gaussian white noise, we can generally only obtain solutions in $S'(\mathbb{R}^n)$. As it is nice to obtain function-valued solutions, this can sometimes be achieved if we look for solutions in the weighted space $L^2(\mathbb{R}^n, e^{-\alpha |x|}dx)$.</li> <li>Weights also appear in the theory of semilinear equations. For example, for the problem <code>\left\{\begin{align}\frac{\partial u}{\partial t} -\Delta u &amp;= f(u),&amp;&amp; x \in U, t &gt; 0,\\ u &amp;= 0,&amp;&amp; x \in \partial U, t &gt;0,\\ u(t,0) &amp;= u_0(x),&amp;&amp; x \in U\end{align}\right.</code> the concept of a <em>very weak solution</em> may be defined (as a first step to studying blow-up, etc). The definition involves the use of the weighted space $L^1(U,\delta)$ where $\delta(x):=\text{dist}(x,\partial U)$.</li> <li>As already mentioned, they appear in Littlewood-Paley type estimates. For example, take the <em>square function</em> $$S^2 f(x_0) = \int_{\Gamma(x_0)} |t \sqrt{\Delta} e^{-t \sqrt{\Delta}} f(x)|^2 \frac{dx dt}{t^n}$$ where <code>$\Gamma(x_0):= \{(x,t):|x-x_0| \le t\}$</code> with vertex at $x_0 \in \mathbb{R}^{n-1}$ and the <em>nontangential maximal function</em> <code>$u^*(x_0) := \sup\{e^{-t \sqrt{\Delta}}f (x): (x,t) \in \Gamma(x_0)\}$</code>. Then we can show that <code>$\|u^*\|_{L^p(\mathbb{R}^{n-1})}$</code> and <code>$\|S(f)\|_{L^p(\mathbb{R}^{n-1})}$</code> are equivalent, and <code>$\|u^*\|_{L^p(\mathbb{R}^{n-1})}$</code> can be controlled by <code>$\|f\|_{L^p(\mathbb{R}^{n-1})}$</code>. This has implications (again!) for boundary value problems but is also an interesting concept to study on its own.</li> </ul> <p>Finally, in its discrete form:</p> <ul> <li>I've heard it has applications in the game of Cricket ;-P</li> </ul> <p><strong>Note</strong>: I've made this a community wiki. Please feel free to correct any misconceptions, typos, etc.</p> http://mathoverflow.net/questions/48292/applications-of-hardys-inequality/51486#51486 Answer by S. Sra for Applications of Hardy's inequality S. Sra 2011-01-08T14:40:03Z 2011-01-08T14:40:03Z <p>I would like to add that perhaps in the following books / theses you find some interesting applications of Hardy's inequality (or Hardy-type inequalities), e.g., to financial analysis:</p> <ol> <li><p>B. Opic, A. Kufner. <em>Hardy-type inequalities</em>. Pitman Research Notes in Mathematics 219, 1990.</p></li> <li><p>A. Wedestig. <em>Weighted Inequalities of Hardy-Type and their Limiting Inequalities,</em> Ph.D. thesis, Department of Mathematics, Lulea University of Technology, 2003.</p></li> <li><p>A. Kufner, L. E. Persson. <em>Weighted inequalities of Hardy type.</em> World Scientific. 2003 </p></li> <li><p><a href="http://books.google.com/books?id=4Rs7jyIwq2AC&amp;lpg=PP1&amp;dq=mitrinovic&amp;pg=PA175#v=onepage&amp;q=hardy%2520inequality&amp;f=false" rel="nofollow">This book here</a> (several applications are mentioned, e.g., to differential geometry)</p></li> </ol>