Partial Differential Equations III & V
Problem Class 5 Solution

Exercise 1 (Hölder inequality). The so-called Hölder inequality which states that for for (measurable) set \(E\subseteq \mathbb{R}^n\) and any \(f\in L^p\left( E\right)\) and \(g\in L^q\left( E\right)\) where \(p\) and \(q\) Hölder conjugate, i.e. \(p,q\in [1,\infty]\) and \(1/p+1/q=1\) we have that1 \[\begin{equation} \nonumber %\label{eq:holder} \int_{E}\left\lvert f(x)g(x)\right\rvert dx \leq \left\lVert f\right\rVert_{L^p\left( E\right)}\left\lVert g\right\rVert_{L^q\left( E\right)}. \end{equation}\] where \(L^p\left( E\right)\) is defined like \(L^p\left( \mathbb{R}^n\right)\) where the set on which we integrate is \(E\) instead of \(\mathbb{R}^n\).

  1. Prove Hölder inequality when \(E=\mathbb{R}^n\), \(p=\infty\) and \(q=1\). You may assume that \(f\) is bounded on \(\mathbb{R}^n\) (and as such use \(\sup\) instead of \(\mathrm{esssup}\)).

  2. Show that if \(f\in C_c\left( \mathbb{R}^n\right)\) and \(g\in L^p\left( \mathbb{R}^n\right)\) then there exists a compact set \(K\subset \mathbb{R}^n\) such that \[\int_{\mathbb{R}^n} \left\lvert f(x)g(x)\right\rvert dx \leq \left\lVert f\right\rVert_{L^\infty}|K|^{\frac{1}{q}} \left( \int_{K}\left\lvert g(x)\right\rvert^pdx\right)^{\frac{1}{p}}.\]

  3. Use Hölder inequality to show that if \(f\in L^p\left( \mathbb{R}^n\right)\) and \(g\in L^q\left( \mathbb{R}^n\right)\) with \(p\) and \(q\) Hölder conjugates we have that \[\sup_{x\in \mathbb{R}^n}\left\lvert f\ast g (x)\right\rvert=\left\lVert f\ast g\right\rVert_{L^\infty\left( \mathbb{R}^n\right)}\leq \left\lVert f\right\rVert_{L^p\left( \mathbb{R}^n\right)} \left\lVert g\right\rVert_{L^q\left( \mathbb{R}^n\right)}.\]

Solution.

  1. We have that \[\int_{\mathbb{R}^n}\left\lvert f(x)g(x)\right\rvert dx \leq \int_{\mathbb{R}^n}\left( \sup_{x\in\mathbb{R}^n}\left\lvert f(x)\right\rvert\right)\left\lvert g(x)\right\rvert dx = \left\lVert f\right\rVert_{L^\infty\left( \mathbb{R}^n\right)}\int_{\mathbb{R}^n}\left\lvert g(x)\right\rvert dx\] \[=\left\lVert f\right\rVert_{L^\infty\left( \mathbb{R}^n\right)} \left\lVert g\right\rVert_{L^1\left( \mathbb{R}^n\right)},\] which is the desired inequality.

  2. Since \(f\in C_c\left( \mathbb{R}^n\right)\) we know that there exists a compact set \(K\subset \mathbb{R}^n\) such that \(f\vert_{K^c}=0\) and that \(\left\lVert f\right\rVert_{\infty}<\infty\). We conclude that \[\int_{\mathbb{R}^n} \left\lvert f(x)g(x)\right\rvert dx = \int_{K}\left\lvert f(x)g(x)\right\rvert dx \leq \left\lVert f\right\rVert_{L^\infty}\int_{K}\left\lvert g(x)\right\rvert dx = \left\lVert f\right\rVert_{L^\infty}\int_{K}1\;\left\lvert g(x)\right\rvert dx\] \[\leq \left\lVert f\right\rVert_{L^\infty} \left\lVert 1\right\rVert_{L^q\left( K\right)}\left\lVert g\right\rVert_{L^p\left( K\right)}=\left\lVert f\right\rVert_{L^\infty}|K|^{\frac{1}{q}} \left( \int_{K}\left\lvert g(x)\right\rvert^pdx\right)^{\frac{1}{p}}.\]

  3. We have that \[\left\lvert f\ast g (x)\right\rvert = \left\lvert\int_{\mathbb{R}^n}f(x-y)g(y)dy\right\rvert\leq \int_{\mathbb{R}^n}\left\lvert f(x-y)g(y)dy\right\rvert \leq \left\lVert f\left( x-\cdot\right)\right\rVert_{L^p\left( \mathbb{R}^n\right)}\left\lVert g\right\rVert_{L^q\left( \mathbb{R}^n\right)}.\] Since \[\left\lVert f\left( x-\cdot\right)\right\rVert_{L^p\left( \mathbb{R}^n\right)}=\left\{\begin{array}{ll} \left( \int_{\mathbb{R}^n}\left\lvert f(x-y)\right\rvert^pdy\right)^{\frac{1}{p}}, & p\in [1,\infty),\\ \mathrm{esssup}_{y\in\mathbb{R}^n}\left\lvert f(x-y)\right\rvert, & p=\infty, \end{array} \right.\] \[= \left\{\begin{array}{ll} \left( \int_{\mathbb{R}^n}\left\lvert f(z)\right\rvert^pdz\right)^{\frac{1}{p}}, & p\in [1,\infty),\\ \mathrm{esssup}_{z\in\mathbb{R}^n}\left\lvert f(z)\right\rvert, & p=\infty, \end{array} \right.= \left\lVert f\right\rVert_{L^p\left( \mathbb{R}^n\right)}\] we conclude the desired result.

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Exercise 2 (Poincaré-like inequalities).

  1. Using the formula \[\begin{equation} \nonumber %\label{eq:FTC} f(x)-f(y)=\int_{y}^x f^\prime (s)ds, \end{equation}\] which holds for any \(C^1\) function on an interval that contains \(x\) and \(y\), together with Hölder’s inequality show that for any Hölder conjugates \(p\) and \(q\), any \(f\in C^1\left( [a,b]\right)\), and any \(c\in [a,b]\) we have that \[\begin{equation} \nonumber %\label{eq:poincare_general_infty} \sup_{x\in [a,b]}\left\lvert f(x)-f(c)\right\rvert \leq \left( b-a\right)^{\frac{1}{q}}\left\lVert f^\prime\right\rVert_{L^p\left( [a,b]\right)}, \end{equation}\] where \(q\) is the Hölder conjugate of \(p\).

  2. Conclude that for any \(p\in [1,\infty]\) and \(r\in [1,\infty)\), any \(f\in C^1\left( [a,b]\right)\), and any \(c\in [a,b]\) we have that \[\begin{equation} \nonumber %\label{eq:poincare_general_I} \left( \int_{a}^b\left\lvert f(x)-f(c)\right\rvert^r dx\right)^{\frac{1}{r}} \leq \left( b-a\right)^{\frac{1}{q}+\frac{1}{r}}\left\lVert f^\prime\right\rVert_{L^p\left( [a,b]\right)}. \end{equation}\] or equivalently \[\begin{equation} \nonumber %\label{eq:poincare_general_II} \left\lVert f-f(c)\right\rVert_{L^r\left( [a,b]\right)}\leq \left( b-a\right)^{\frac{1}{q}+\frac{1}{r}}\left\lVert f^\prime\right\rVert_{L^p\left( [a,b]\right)}. \end{equation}\] Note that due to our previous sub-question the above remains true when \(r=\infty\).

  3. Use sub-question [item:basic_ineq] to conclude that for any \(f\in C^1\left( [a,b]\right)\) and any \(p\in[1,\infty]\) we have \[\begin{equation} \nonumber %\label{eq:poincare_average} \left\lVert f-\overline{f}\right\rVert_{L^p\left( [a,b]\right)} \leq \left( b-a\right)\left\lVert f^\prime\right\rVert_{L^p\left( [a,b]\right)} \end{equation}\] where \[\overline{f} = \frac{1}{b-a}\int_a^b f(s)ds.\]

  4. Use sub-question [item:basic_ineq] to conclude that for any \(f\in C^1\left( [a,b]\right)\) such that \(f(c)=0\) for some \(c\in [a,b]\) and any \(p\in[1,\infty]\) we have \[\begin{equation} \nonumber %\label{eq:poincare_dirichlet} \left\lVert f\right\rVert_{L^p\left( [a,b]\right)} \leq \left( b-a\right)\left\lVert f^\prime\right\rVert_{L^p\left( [a,b]\right)}. \end{equation}\]

Solution.

  1. We have \[\left\lvert f(x)-f(c)\right\rvert = \left\lvert\int_{c}^x f^\prime(s)ds\right\rvert \leq \int_{c}^x \left\lvert f^\prime (s)\right\rvert ds = \int_{c}^x 1\cdot \left\lvert f^\prime (s)\right\rvert ds \leq \int_{a}^b 1\cdot \left\lvert f^\prime (s)\right\rvert ds.\] Using Hölder inequality we find that \[\left\lvert f(x)-f(c)\right\rvert \leq \left\lVert 1\right\rVert_{L^q\left( [a,b]\right)}\left\lVert f^\prime\right\rVert_{L^p\left( [a,b]\right)} = \left( b-a\right)^{\frac{1}{q}}\left\lVert f^\prime\right\rVert_{L^p\left( [a,b]\right)} .\]

  2. Since \[\int_{a}^b \left\lvert f(x)-f(c)\right\rvert^r dx \leq \left( \sup_{x\in [a,b]}\left\lvert f(x)-f(c)\right\rvert\right)^r\int_a^b 1 dx = \left( b-a\right)\left( \sup_{x\in [a,b]}\left\lvert f(x)-f(c)\right\rvert\right)^r\] we conclude from the previous sub-question that \[\left( \int_{a}^b \left\lvert f(x)-f(c)\right\rvert^r\right)^{\frac{1}{r}} dx \leq \left( b-a\right)^{\frac{1}{r}}\sup_{x\in [a,b]}\left\lvert f(x)-f(c)\right\rvert\] \[\leq \left( b-a\right)^{\frac{1}{q}+\frac{1}{r}}\left\lVert f^\prime\right\rVert_{L^p\left( [a,b]\right)}\] which is the desired result.

  3. The mean value theorem for integrals says that for any two continuous functions on \([a,b]\), \(f\) and \(g\), we have that there exists \(c\in [a,b]\) such that \[\int_{a}^b f(x)g(x)dx = f(c)\int_{a}^b g(x)dx.\] Choosing \(g\equiv 1\) we find that there exists \(c_\ast\in [a,b]\) such that \[f(c_\ast) = \frac{\int_{a}^b f(x)dx }{\int_{a}^b 1dx }=\frac{1}{b-a}\int_a^b f(x)dx=\overline{f}.\] plugging that into sub-question [item:basic_ineq] with \(r=p\) gives the desired result.

  4. Similarly, choosing \(c=a\) and \(r=p\) in sub-question [item:basic_ineq] gives us the desired result.

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Exercise 3 (The Poincaré constant).

  1. Let \(\Omega\) be an open bounded domain in \(\mathbb{R}^n\) with smooth boundary. Consider the PDE \[\label{EVP} \begin{align} - \Delta u(x) & = \lambda u(x) \quad \textrm{in } \Omega, \\ u(x) & = 0 \quad \; \, \; \textrm{on } \partial \Omega, \end{align}\] where \(u\in C^2\left( \overline{\Omega}\right)\) and \(\lambda \in \mathbb{R}\). By multiplying the above by \(u\) and integrating by parts show that if \(u\ne 0\) we have that \[\begin{equation} \label{Rayleigh} \lambda = \frac{\displaystyle \int_\Omega |\nabla u(x)|^2 \, d x}{\displaystyle \int_\Omega u^2(x) \, d x} \geq 0. \end{equation}\] This is known as the Rayleigh quotient formula for the eigenvalue \(\lambda\) in terms of the eigenfunction \(u\).
    Remark: This procedure, where we multiplied by a function, integrated, and recovered a functional that helps us understand our equation better is known as the energy method (the functional we found is our “energy”). This is an important method which will repeat in this module.
    Remark: We could allow \(u\) to have values in \(\mathbb{C}\). In that case we would multiply our PDE with \(\overline{u}\) and conclude that \[\lambda = \frac{\displaystyle \int_\Omega |\nabla u(x)|^2 \, d x}{\displaystyle \int_\Omega \left\lvert u(x)\right\rvert^2 \, d x} \geq 0.\]

  2. Show that any \(\lambda\) that satisfies our PDE must satisfy \[\lambda \geq \frac{1}{C_P^2}\] where \(C_P\) is the Poincaré inequality that is associated to \(\Omega\) and the Dirichlet boundary condition, i.e. the best constant for which \[\int_{\Omega}\left\lvert u(x)\right\rvert^2 dx \leq C_P^2 \int_{\Omega}\left\lvert\nabla u(x)\right\rvert^2dx.\] Remark: The above implies that \[\min_{\textrm{eigenvalues}}\lambda \geq \frac{1}{C_P^2}.\] One can in fact show that there is equality in the above. This is connected to minimising the energy \[E[u] = \frac{\displaystyle \int_\Omega |\nabla u(x)|^2 \, d x}{\displaystyle \int_\Omega |v(x)|^2 \, d x}\] over an appropriate space of functions.

Solution.

  1. Multiplying our equation by \(u\) and integrating we find that \[-\int_{\Omega}u(x)\Delta u(x)dx = \lambda \int_{\Omega}u^2(x)dx.\] Since \[\mathrm{div} \left( u(x)\nabla u(x)\right) = \left\lvert\nabla u(x)\right\rvert^2 + u(x)\Delta u(x)\] we conclude that \[-\int_{\Omega}u(x)\Delta u(x)dx = \int_{\Omega}\left\lvert\nabla u(x)\right\rvert^2dx - \int_{\Omega}\mathrm{div}\left( u(x)\nabla u(x)\right)dx\] \[=\int_{\Omega}\left\lvert\nabla u(x)\right\rvert^2dx - \int_{\partial \Omega}u(y)\nabla u(y)\cdot n(y)dS(y)=\int_{\Omega}\left\lvert\nabla u(x)\right\rvert^2dx\] where we have used the fact that \(u\vert_{\partial \Omega}=0\). We conclude that \[\int_{\Omega}\left\lvert\nabla u(x)\right\rvert^2dx = \lambda \int_{\Omega}u^2(x)dx\] which is the desired result.

  2. We know that any \(u\) which is \(C^1\) and is zero on the boundary satisfies \[\int_{\Omega}\left\lvert u(x)\right\rvert^2 dx \leq C_P^2 \int_{\Omega}\left\lvert\nabla u(x)\right\rvert^2dx\] and as such \[\frac{\int_{\Omega}\left\lvert\nabla u(x)\right\rvert^2dx} {\int_{\Omega}\left\lvert u(x)\right\rvert^2}\geq \frac{1}{C_P^2}.\] This implies that if \(u\) solves our PDE and \(u\ne 0\) then \[\lambda = \frac{\int_{\Omega}\left\lvert\nabla u(x)\right\rvert^2dx} {\int_{\Omega}\left\lvert u(x)\right\rvert^2} \geq \frac{1}{C_P^2}.\]

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  1. One particular case that is worth to mention is when \(p=q=2\). This case is the famous Cauchy-Schwarz inequality and it reads as \[\nonumber \int_{E}\left\lvert f(x)g(x)\right\rvert dx \leq \left( \int_{E}\left\lvert f(x)\right\rvert^2dx\right)^{\frac{1}{2}} \left( \int_{E}\left\lvert g(x)\right\rvert^2dx\right)^{\frac{1}{2}}\]↩︎