Owen's Math Portal

The following is a short presentation on Hardy Spaces for MATH507 - Functional Analysis with Professor Maxime Van de Moortel. The presentation includes a brief introduction to H^p spaces and a proof that H^p is a Banach space for all p from 1 to infinity.

My mathematical interests lie in geometric analysis and differential geometry. Specifically, my research is concerned with the analysis of singularities of energy minimizing maps and geometric flows. In geometric analysis, the question of the uniqueness of tangent maps is notoriously opaque, and is the subject of much modern research. My thesis consists of two parts, the first of which focuses on the analysis of harmonic maps following the pioneering work of Leon Simon[1]. Here, I explicate the theory of singularities and tangent maps of energy minimizers on Riemannian manifolds.

The following exercises are an assortment of homework problems from MATH 502 - Theory of Functions of a Real Variable II with Professor Dennis Kriventsov at Rutgers University. Topics include Radon Measures, weak convergence, Haar measures, Fourier analysis, PDEs, distributions, and probability theory.

Today, we will evaluate the integral $$ \int_{-\infty}^{\infty} \frac{\cos(x)}{x^2+1} dx $$ using Cauchy’s Residue Theorem and contour integration. This is one of the most beautiful results in all of calculus, and the solution is rather elegant. To begin, let us define the following function in \(\mathbb{C}\) : $$ f(z)=\frac{e^{iz}}{z^2+1} $$ and let us use the canonical contour, that is, the top half of a semicircle in the complex plane. Let us define this contour as \(C\) , with radius \(R\) . Note that \(f(z)\) had two singularities, one at \(z=-i\) and the other at \(z=i\) . Only \(z=i\) is inside of our contour \(C\) , as seen below:

Here we will solve the Dirichlet Integral using two different techniques. The Dirichlet Integral is defined as $$ \int_{0}^{\infty} \frac{\sin(x)}{x} dx $$

Today, we will be evaluating the integral $$ \int_{-\infty}^{\infty} e^{ax^2+bx+c} dx \hspace{1cm} a \in \mathbb{R^{-}}, b \in \mathbb{R}, c \in \mathbb{R} $$ The restriction of \(a\) to \(\mathbb{R^{-}}\) is required for the convergence of this integral. My thought process for this integral was initially to minipulate the integrand in such a way that we can employ the result of the Gaussian integral, namely, \(\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}\) . To do this, we complete the square on \(f(x)=ax^2+bx+c\) : $$ y=ax^2+bx+c \implies y-c=ax^2+bx=a(x^2+\frac{b}{a}x) \implies \frac{y-c}{a}=x^2+\frac{b}{a}x $$ Further, $$ \frac{y-c}{a}+\frac{b^2}{4a^2}=x^2+\frac{b}{a}x+\frac{b^2}{4a^2} \implies \frac{y-c}{a}=(x+\frac{b}{2a})^2-\frac{b^2}{4a^2} $$ Lastly, $$ y=a(x+\frac{b}{2a})^2-\frac{b^2}{4a}+c $$ Thus we have that $$ I=\int_{-\infty}^{\infty} e^{ax^2+bx+c} dx=\int_{-\infty}^{\infty} e^{a(x+\frac{b}{2a})^2-\frac{b^2}{4a}+c} dx=e^{-\frac{b^2}{4a}+c} \int_{-\infty}^{\infty} e^{a(x+\frac{b}{2a})^2} dx $$ Given that this integrand spans all of \(\mathbb{R}\) we can employ the substitution \(t=x+\frac{b}{2a}\) and our bounds will remain the same. Thus, $$ I=e^{-\frac{b^2}{4a}+c} \int_{-\infty}^{\infty} e^{at^2} dt $$ and since \(a\) is negative, we can write $$ I=e^{-\frac{b^2}{4a}+c} \int_{-\infty}^{\infty} e^{-(\sqrt{-a}t)^2} dt $$ Now letting \(v=\sqrt{-a}t\) we have that \(dv=\sqrt{-a} dt \implies \frac{dv}{\sqrt{-a}}=dt\) and thus $$ I=e^{-\frac{b^2}{4a}+c} \int_{-\infty}^{\infty} e^{-(v)^2} \frac{dv}{\sqrt{-a}} \implies I=\frac{e^{-\frac{b^2}{4a}+c}}{\sqrt{-a}} \sqrt{\pi} = e^{-\frac{b^2}{4a}+c} \sqrt{-\frac{\pi}{a}} $$ and we’re done.

To prove that the space of all invetible matricies is dense in the space of all matricies, it will suffice to show that the space of all invertible matricies is dense the space of matricies with distinct eigenvalues, as the space of matricies with distinct eigenvalues is dense in the space of all matricies. Denote \(E(n, \mathbb{R}) \) as the space of matricies with n distinct eigenvalues.

In this article, we will examine the nature and utility of the first fundamental form, a quadratic form on a surface. Given a curve parametrized by \(u=u(t)\) and \(v=v(t)\) which lies on a surface \(X\) parametrized by \(X=X(u,v)\) we have that

$$ ds = |\frac{dX}{dt}| dt = |X_u \frac{du}{dt} + X_v \frac{dv}{dt}| dt = \sqrt{(X_u \dot{u} + X_v \dot{v}) \cdot (X_u \dot{u} + X_v \dot{v})} $$

Now defining the coefficients, called the First Fundamental Form coefficients E, F and G, we have

The Gaussian Integral is given by $$ \int_{-\infty}^{\infty} e^{-x^2} dx $$ This is a beautiful integral with many important applications to probability and statistics, namely normal distributions. Here, we will evaluate this integral using techniques from Calculus 3.

For a set of orthonormal vectors \((v_1,v_2,...,v_m)\) in an m-dimensional vector space , and set of constants \(r_1,r_2,...r_m \in \mathbb{R}\) , if we consider the equation \(r_1v_1+r_2v_2+...+r_mv_m=0\)

We will evalute the integral \(\int_0^1 \frac{\ln(x)}{1+x} dx \) using Taylor Series expansion and switching the order of integration and summation.

Firstly, we notice that $$ \frac{1}{1+x}=\sum_{n=0}^{\infty} (-1)^n x^n $$

by taylor series expansion. This series coverges for \(x \in (0,1)\)

so we can substitute this into the integral:

$$ \int_0^1 \frac{\ln(x)}{1+x} dx = \int_0^1 \ln(x) \sum_{n=0}^{\infty} (-1)^n x^n dx $$

Further, by changing the order of integration and summation, we have $$ \int_0^1 \ln(x) \sum_{n=0}^{\infty} (-1)^n x^n dx = \sum_{n=0}^{\infty} (-1)^n \int_0^1 x^n \ln(x) dx $$

The Basel Problem, named after Leonard Euler’s hometown of Basel, Switzerland, is the solution to the infinite series \(\sum_{n=1}^{\infty} \frac{\pi^2}{6}\) Euler was the first to publish his solution one year after having solved this problem in 1734. Since Euler’s solution, many alternative proofs have been offered, using complex analysis, symmetric polynomials, and so on. These alternative proofs are more rigorous than Euler’s liberal use of infinite polyomial tails, but Euler’s solution is perhaps the most elegant. This is the outline of the proof.