Ricardo A. Sáenz

Escuela de Análisis matemático Análisis de Fourier y operadores de multiplicación Si $latex f\in L^p(\mathbb R^d)$ y $latex g\in L^1(\mathbb R^d)$, entonces $latex f*g\in L^p(\mathbb R^d)$ y $latex ||f*g||_{L^p} \le ||f||_{L^p} ||g||_{L^1}.$ Si $latex f\in L^1(\mathbb R^d)\cap L^2(\mathbb R^d)$, entonces $latex ||\hat f||_{L^2} = ||f||_{L^2}.$ ( Sugerencia: Considera $latex h=f*g$, con $latex g(x) = \overline{f(-x)}$, y nota que $latex h(0) = \int \hat h$.) El operador $latex f\mapsto \hat f$ es sobreyectivo en $latex L^2(\mathbb R^d)$. ( Sugerencia: Si no, existiría $latex g\in L^2$ tal que $latex \int \hat f g = 0$ para toda $latex f\in L^2$. Utiliza el pasito del sombrero.) Sea $latex h(x) = \begin{cases} 0 & x<0\\1 & x\ge 0\end{cases}$ la función de Heaviside. Como distribuciones, $latex h' = \delta,$ la función delta de Dirac. Sea $latex \mathbf 1$ la función constante igual a 1. Como distribuciones, $latex \hat{\mathbf 1} = \delt...

Escuela de Análisis matemático Análisis de Fourier y operadores de multiplicación Si $latex T:V\to V$ es simétrico y $latex \lambda_i\not=\lambda_j$ son eigenvalores reales distintos de $latex T$, con eigenvectores $latex u_i, u_j$ correspondientes, entonces $latex u_i\perp u_j$. Sea $latex T:\mathscr H\to\mathscr H$ un operador acotado diagonalizado con sucesión multiplicadora $latex \lambda_k$. T es unitario si, y solo si, $latex |\lambda_k|=1$ para todo k . T es una proyección ortogonal si, y solo si, todo $latex \lambda_k = 0\text{ o } 1$. T es un operador compacto si, y solo si, $latex \lambda_k \to 0$. Si $latex \sum a_n$ es una serie convergente, entonces es Cesàro-sumable. Si $latex \sum a_n$ es Cesàro-sumable, entonces es Abel-sumable.

Park City Mathematics Institute Undergraduate Summer School 2018 Introduction to Harmonic Analysis Prove by induction, for the interval case, that $latex \min\{ \mathscr E_m(u): u|_{\{0,1\}}=v\} = \mathscr E_0(v) = (v(0)-v(1))^2,$ with the minimizer satisfying $latex \displaystyle u\Big(\frac{2k+1}{2^m}\Big) = \frac{1}{2}\Big(u\Big(\frac{k}{2^{m-1}}\Big) + u\Big(\frac{k+1}{2^{m-1}}\Big)\Big).$ The minimum of $latex f(x,y,z) = (a-x)^2 + (x-y)^2 + (y-a)^2 + (x-b)^2 + (b-z)^2 + (z-x)^2 + (y-z)^2 + (z-c)^2 + (c-y)^2$ is attained at $latex \displaystyle x^* = \frac{2a+2b+c}{5},\; y^* = \frac{2a+b+2c}{5},\; z^* = \frac{a+2b+2c}{5},$ with $latex f(x^*, y^*, z^*) = \dfrac{3}{5}\big((a-b)^2 + (b-c)^2 + (c-a)^2\big).$ Prove that one can obtain the values $latex u(F_2(q_1)) = x, u(F_3(q_1)) = y$ of a harmonic function in terms of the values $latex a,b,c$ at the points $latex p_2, q_1, p_3$, respectively  (as in the figure below). Use the previous problem to show that, if $latex u$ is a ha...

Park City Mathematics Institute Undergraduate Summer School 2018 Introduction to Harmonic Analysis Show that $latex ||\cdot||_{H^1}$ is a norm, by showing it is induced by an inner product. Consider, for a connected domain $latex \Omega$, the energy form $latex \mathscr E(u,v) = \int_\Omega \nabla u\cdot \nabla v$. $latex \mathscr E(u,v)$ is an inner product on $latex H^1$ modulo constants. $latex \mathscr E(u,v)$ is an inner product on $latex H_0^1$. Show the equivalences of the Dirichlet principle.

Park City Mathematics Institute Undergraduate Summer School 2018 Introduction to Harmonic Analysis Extend the result that the Hilbert transform is of weak-type $latex L^1$ to any operator of the form $latex Tf = K*f$ where $latex K$ satisfies: $latex \hat K\in L^\infty(\R)$ (say, taking the Fourier transform in the $latex L^2$ sense); there exists a constant $latex A>0$ such that $latex |K'(x)| \le A/x^2$ for any $latex x\in\R, \; x\not=0$. Show that we can replace (2) above by the condition 2'. There exists a constant $latex A>0$ such that $latex \displaystyle \int_{|x|\ge 2|t|} |K(x-t) - K(x)| dx \le A$ for any $latex t\in\R$. (Chebyshev Inequality) If $latex f\in L^p$, for some $latex 1 < p <\infty$, then $latex |\{x : |f(x)|>\alpha\}| < \dfrac{1}{\alpha^p} ||f||_{L^p}^p$ Prove that we can substitute the condition $latex \hat K\in L^\infty$ with the boundedness of $latex T$ on any $latex L^q(\R), \; q>1$, that is, 1'. there exists a...

Park City Mathematics Institute Undergraduate Summer School 2018 Introduction to Harmonic Analysis Let $latex \{\Phi_t\}_{t>0}$ be a collection that satisfies, for some constants $latex c_0, c_1, c_2$, and all $latex x\in\R^d$ and $latex t>0$, $latex \int \Phi_t = c_0$, $latex |\Phi_t(x)| \le c_1 t^{-d}$, $latex |\Phi_t(x)| \le c_2 t/|x|^{d+1}$. Then there exists $latex A>0$ such that $latex |\Phi_t*f(x)|\le A Mf(x)$, for all $latex x\in\R^d$, and thus $latex \Phi_t*f(x)\to f(x)$ as $latex t\to 0$ for almost every $latex x\in\R^d.$ (Needs complex analysis) $latex \displaystyle \hat Q _y(\xi) = \lim_{N\to\infty}\frac{1}{\pi} \int_{-N}^N \frac{x e^{-2\pi ix\xi}}{x^2+y^2} dx = -i\text{sgn}\xi e^{-2\pi y |\xi|}$. ( Hint: Use the residue theorem from complex analysis.) (Needs complex analysis) Give another proof that the Hilbert transform is bounded on $latex L^2$ by considering the Cauchy integral $latex \displaystyle Cf(z) = \frac{1}{i\pi} \int_{-\infty}^\infty \frac...

Park City Mathematics Institute Undergraduate Summer School 2018 Introduction to Harmonic Analysis Write the details of the proof of Vitalli's Lemma Use the Hardy-Littlewood maximal theorem to prove the Lebesgue differentiation theorem: if $latex f\in L^1(\R^d)$, then $latex \displaystyle \lim_{r\to 0} \frac{1}{|B_r(x)|} \int_{B_r(x)} f(y) dy = f(x)$ for almost every $latex x\in\R^d.$ State conditions on $latex \Phi$ such that, for $latex f\in L^1(\R^d)$, $latex \displaystyle \lim_{t\to 0} \Phi_t*f(x) = f(x)$ for almost every $latex x\in\R^d.$ Let $latex f(x) = \begin{cases} \dfrac{1}{|x|(\log|x|)^2} & 0<|x|<1/2\\0&\text{otherwise.}\end{cases}$ Then $latex f\in L^1(\R)$, but $latex Mf$ is not locally integrable.

Park City Mathematics Institute Undergraduate Summer School 2018 Introduction to Harmonic Analysis Verify the integral $latex \displaystyle \int_{\R^d}\frac{dx}{(|x|^2+1)^\frac{d+1}{2}} = \frac{\pi^\frac{d+1}{2}}{\Gamma(\frac{d+1}{2})}.$ Use Minkowski inequality to prove that, if $latex K\in L^1(\R^d)$ and $latex f\in L^p(\R^d)$, then $latex ||K*f||_{L^p}\le ||K||_{L^1}||f||_{L^p}.$ Prove that, if $latex f\in C_c(\R^d)$ and $latex y\in\R^d$, then $latex ||f(\cdot - ty) - f||_{L^p} \to 0$ as $latex t\to 0$. Let $latex \Phi\in L^1(\R^d)$ with $latex \int \Phi = 1$, and $latex \Phi_t(x) = t^{-d}\Phi(x/t).$ Then $latex \int \Phi_t = 1$ for all $latex t>0.$ There exists some $latex M>0$ such that $latex \int |\Phi_t| \le M$ for all $latex t > 0.$ For each $latex \delta>0$, $latex \displaystyle \int_{|x|\ge\delta} |\Phi_t(x)|dx \to 0$ as $latex t\to 0.$ If $latex f\in C_c(\R^d)$, then $latex \Phi_t*f(x) \to f(x)$ uniformly. State conditions on $latex \Phi$ (as i...

Park City Mathematics Institute Undergraduate Summer School 2018 Introduction to Harmonic Analysis A polynomial $latex p\in\mathscr P_k$ if and only if $latex \sum x_j \partial_j p = kp.$ $latex \dim \mathscr P_k = \binom{k+d-1}{k}$ Let $latex p\in\mathscr P_k$, and consider its orthogonal projection $latex |x|^2q$ onto the space $latex |x|^2\mathscr P_{k-2}$ with respect to the inner product seen in class. Then $latex p - |x|^2q$ is harmonic. ( Hint : Prove $latex \langle r,\Delta(p - |x|^2q\rangle = 0$ for every $latex r\in\mathscr P_{k-2}$.) If $latex p\in\mathscr H_k$, then $latex \dfrac{p}{|x|^{2k+d-2}}$ is harmonic on $latex \R^d\setminus\{0\}$. The spaces are invariant under rotations ,  i.e. if $latex f\in H_k$ then $latex f\circ R\in H_k$ for any rotation $latex R$.

Park City Mathematics Institute Undergraduate Summer School 2018 Introduction to Harmonic Analysis The function $latex x\mapsto P(x,\xi)$ is harmonic for each $latex \xi$, where $latex P(x,\xi)$ is the Poisson kernel for the ball. (Symmetry Lemma) For any $latex x\in\mathbb B$ and $latex \xi\in\mathbb S$, $latex \displaystyle \Big| |x|\xi - \frac{x}{|x|}\Big| = |x-\xi|$. (Harnack's inequality) If $latex u$ is a positive harmonic function on $latex \bar{\mathbb B}$, then $latex \displaystyle \frac{1-|x|}{(1+|x|)^{d-1}} u(0) \le u(x) \le \frac{1 + |x|}{(1-|x|)^{d-1}}u(0).$ (Hopf Lemma) If $latex u$ is a nonconstant harmonic function on $latex \bar{\mathbb B}$ and attains its maximum at $latex \zeta\in\mathbb S$, there exists $latex c>0$ such that $latex u(\zeta) - u(r\zeta) \ge c(1-r)$ for any $latex 0 < r < 1.$ If $latex u$ is harmonic on $latex \bar{\mathbb B}$ and its normal derivative is 0 everywhere on $latex \mathbb S$, then $latex u$ is constant.

Park City Mathematics Institute Undergraduate Summer School 2018 Introduction to Harmonic Analysis If $latex u$ is harmonic in the connected domain $latex \Omega$ and is not constant, then $latex u(\Omega)$ is open in $latex \R$. Suppose $latex \Omega$ is bounded and that its boundary $latex \partial\Omega$ is connected. If $latex u$ is harmonic in $latex \Omega$, then $latex u(\Omega)\subset u(\partial\Omega)$. A radial harmonic function on $latex \mathbb B$ is constant. A positive harmonic function on $latex \R^d$ is constant.

Park City Mathematics Institute Undergraduate Summer School 2018 Introduction to Harmonic Analysis Let $latex P:\R^d\to\R^d$ be orthogonal. $latex \Delta(u\circ P) = (\Delta u)\circ P$ If $latex u$ is harmonic, then $latex u\circ P$ is also harmonic. If $latex T$ is a translation and $latex u$ is harmonic, then $latex u\circ T$ is harmonic. Use polar coordinates to verify $latex \displaystyle \int_{\R^2}e^{-\pi |x|^2} dx = 1.$ Use (3.1) to prove $latex \displaystyle \int_{\R^d} e^{-\pi |x|^2} dx = 1$, for any $latex d\ge 1$. Use spherical coordinates and the previous result to prove $latex \omega_d = \dfrac{2\pi^{d/2}}{\Gamma(d/2)}$, where $latex \displaystyle \Gamma(s) = \int_0^\infty t^{s-1} e^{-t} dt$ is the Gamma function. Use the divergence theorem $latex \displaystyle \int_\Omega \nabla\cdot \vec F dx = \int_{\partial\Omega} \vec F \cdot \hat{\mathbf n} \; d\sigma$ to prove Green's identity. ( Hint: Consider $latex \vec F = u \nabla v - v\nabla u$.) The f...

Park City Mathematics Institute Undergraduate Summer School 2018 Introduction to Harmonic Analysis Let $latex R$ be a rotation in the plane. Consider the change of variables $latex (\xi, \eta) = R(x,y)$. Then $latex \displaystyle \frac{\partial^2 u}{\partial\xi^2} + \frac{\partial^2 u}{\partial\eta^2} = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}$. If $latex u$ is harmonic, then $latex u\circ R$ is also harmonic. Let $latex (r,\theta)$ be the polar coordinates of the plane. Then $latex \displaystyle \Delta u = \frac{\partial^2 u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{\partial r} + \frac{1}{r^2} \frac{\partial^2 u}{\partial\theta^2}$ Let $latex u$ be a harmonic function on $latex \R^2$. Then there exists a harmonic function $latex v$ that is conjugate to $latex u$, so $latex f = u + iv$ is holomorphic. ( Hint: Consider a line integral of the 1-form $latex \displaystyle - \frac{\partial u}{\partial y} dx + \frac{\partial u}{\partial x} dy$.) ...