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 \(\mathbb{R}^n\).
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 \(\mathrm{esssup}\)).
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 \[\int_{\mathbb{R}^n} \left\lvert f(x)g(x)\right\rvert dx \leq \left\lVert f\right\rVert_{L^\infty}|K|^{\frac{1}{q}} \left( \int_{K}\left\lvert g(x)\right\rvert^pdx\right)^{\frac{1}{p}}.\]
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 \[\sup_{x\in \mathbb{R}^n}\left\lvert f\ast g (x)\right\rvert=\left\lVert f\ast g\right\rVert_{L^\infty\left( \mathbb{R}^n\right)}\leq \left\lVert f\right\rVert_{L^p\left( \mathbb{R}^n\right)} \left\lVert g\right\rVert_{L^q\left( \mathbb{R}^n\right)}.\]
Exercise 2 (Poincaré-like inequalities).
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 \(p\).
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 \(r=\infty\).
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 \[\overline{f} = \frac{1}{b-a}\int_a^b f(s)ds.\]
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 \[\begin{equation} \nonumber %\label{eq:poincare_dirichlet} \left\lVert f\right\rVert_{L^p\left( [a,b]\right)} \leq \left( b-a\right)\left\lVert f^\prime\right\rVert_{L^p\left( [a,b]\right)}. \end{equation}\]
Exercise 3 (The Poincaré constant).
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 \[\lambda =
\frac{\displaystyle \int_\Omega |\nabla u(x)|^2 \, d x}{\displaystyle
\int_\Omega \left\lvert u(x)\right\rvert^2 \, d x} \geq
0.\]
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.
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 \[\nonumber \int_{E}\left\lvert f(x)g(x)\right\rvert dx \leq \left( \int_{E}\left\lvert f(x)\right\rvert^2dx\right)^{\frac{1}{2}} \left( \int_{E}\left\lvert g(x)\right\rvert^2dx\right)^{\frac{1}{2}}\]↩︎