I’m gonna be sick

rowan kennedy

Need to know

Real Analysis

Lebesgue Integration

Definition 1. A characteristic function or indicator function of a set EE is defined as χE(x):={1xE0xE\chi_E(x):= \begin{cases} 1 & x\in E \\ 0 & x\notin E \end{cases}

Definition 2. A simple function is a function which takes finitely many values. Thus for any simple function there is a set of values a1,a2,ana_1, a_2\dots , a_n and a collection of sets E1,E2,EnE_1,E_2,\dots E_n such that s(x)=i=1naiχEi(x)s(x)= \sum^n_{i=1}a_i\chi_{E_i}(x)

Definition 3 (Lebesgue Integration). Suppose ss a simple function such that s(x)=i=1naiχEi(x)s(x)= \sum^n_{i=1}a_i\chi_{E_i}(x) and μ\mu is a measure on XX. We define sdμ=i=1naiμ(Ei).\int s \ d\mu = \sum^n_{i=1} a_i\mu(E_i).

For a non-negative measureable function f:Xf:X\to \mathbb R, we define fdμ=sup{sdμ:sf,s is simple}\int f \ d\mu = \sup\left\{\int s \ d\mu : s\le f, s \text{ is simple}\right\}

Proposition 1 (integration over a set and useful consequences). We define Afdμ:=fχAdμ\int_Af \ d\mu :=\int f\chi_A \ d\mu.
Note that any characteristic function is a simple function and so for any AA with μ(A)<\mu (A)< \infty we find μ(A)=χAdμ=A1dμ.\mu(A) = \int \chi_A \ d \mu = \int_A 1 \ d\mu.

Theorem 2. If ff is Riemann integrable, then ff is Lebesgue integrable and the integrals agree.

Standard Limit Theorems

Theorem 3 (Lebesgue’s Monotone Convergence Theorem). Suppose (X,𝒜,μ)(X,\mathcal A, \mu) is a measure space and {fn}n=0\{f_n\}_{n=0}^\infty is a family of real-valued measurable functions with 0f1(x)f2(x)0\le f_1(x)\le f_{2}(x)\le \cdots for all xx and limnfn(x)=f\lim_{n\to \infty}f_n(x)=f for all xx. Then limnfndμ=fdμ.\lim_{n\to \infty}\int f_n \ d\mu =\int f \ d \mu.

Theorem 4 (Fatou’s Lemma). Suppose (X,𝒜,μ)(X,\mathcal A, \mu) is a measure space and {fn}n=0\{f_n\}_{n=0}^\infty is a family of real-valued nonnegative measurable functions. Then liminfnfndμliminfnfndμ.\int\liminf_{n\to \infty} f_n \ d\mu \le \int\liminf_{n\to \infty} f_n \ d \mu.

Theorem 5. (Dominated Convergence Theorem)Suppose (X,𝒜,μ)(X,\mathcal A, \mu) is a measure space and {fn}\{f_n\} is a family of measurable real-valued functions where fn(x)f(x)f_n(x) \to f(x) pointwise. If there exists a non-negative integrable function gg such that |fn(x)|g(x)|f_n(x)|\le g(x) for all xx for each nn. Then fndμfdμ.\int f_n \ d\mu\to \int f\ d\mu .

Differentiation

Definition 4 (Absolutely Continuous). Let (X,𝒜)(X,\mathcal A) be a measurable space and let μ,ν\mu, \nu be measures on 𝒜\mathcal A. If μ(E)=0ν(E)=0\mu(E)=0\implies\nu(E)=0 for all E𝒜E\in \mathcal A, then we say ν\nu is absolutely continuous with respect to ν\nu, denoted νμ\nu \ll \mu.

Theorem 6 (Radon-Nikodym). Suppose μ,ν\mu, \nu are measures on a σ\sigma-algebra 𝒜\mathcal A such that μ\mu is a σ\sigma-finite positive measure and ν\nu a finite positive measure with νμ\nu \ll \mu. Then there exists a μ\mu-integrable function ff measurable with respect to 𝒜\mathcal A such that ν(A)=Afdμ\nu(A)=\int_Af \ d \mu unique up to almost everywhere equivalence.

Other

Theorem 7 (Hölder’s inequality). For functions f,gf,g and 1<p,q<1< p,q< \infty such that 1p+1q=1\frac{1}{p}+\frac{1}{q}=1 then |fg|dμfpgq\int|fg|\ d\mu \le \|f\|_p\|g\|_q

Theorem 8 ().

Lemma 9 (Chebyshev’s Inequality). If 1p<1\le p< \infty, then μ({x:|f(x)|>1})|f|pdμap\mu(\{x: |f(x)|>1\})\le \frac{\int |f|^p \ d \mu }{a^p}

Theorem 10 (Egorov’s Theorem). Suppose μ\mu is a finite measure, ε>0\varepsilon> 0, and fnff_n\to f almost everywhere. There exists a measureable set AA such that μ(A)<ε\mu(A)<\varepsilon and fnff_n\to f uniformly on AcA^c

Complex

Theorem 11 (Liouville’s Theorem). If f:f:\mathbb C\to \mathbb C is an entire function and there exists a KK such that f(z)Kf(z)\le K for all zz\in \mathbb C then ff is constant

Theorem 12 (Morera’s Theorem). A function f:Uf:U\to \mathbb C is holomorphic on UU if and only if Tf(z)dz=0\int_Tf(z)\ dz=0 for any triangle TT contained by UU.

Theorem 13. For a function f:Uf:U\to \mathbb C and a closed, piecewise C1C^1 curve γ\gamma 12πiγf(z)dz=\frac{1}{2\pi i}\int_\gamma f(z) \ dz =

Theorem 14 (Riemann Mapping Theorem). Every simply connected open subset of \mathbb C is conformally equivalent to the open unit disk.

Theorem 15 (Riemann’s Removable Singularity Theorem). Suppose a function f:Uf:U\to \mathbb C has an isolated singularity at z0z_0. There exists a ε>0\varepsilon> 0 and a KK such that |f(z)|<K|f(z)|<K for all zD(z0,ε)z\in D(z_0,\varepsilon) if and only if z0z_0 is a removable singularity

Proposition 16 (Cauchy’s Estimates). Suppose fH(U)f\in H(U). If there exists a ε>0\varepsilon>0 such that |f(z)|<M|f(z)|<M in a disk D(z0,ε)D(z_0, \varepsilon), then |f(n)(z0)|Mn!εn|f^{(n)}(z_0)|\le \frac{Mn!}{\varepsilon^n} for n=0,1,2,n=0,1,2, \dots

Problems and Solutions

Real analysis

Question 1. Define f:22f:\mathbb R^2\to \mathbb R^2 by f(x,y):=(x4y3,x2+y)f(x,y):=(x^4-y^3, x^2+y) and H:={(x,y)2:x>0}H:=\{(x,y)\in \mathbb R^2 : x>0 \}. Show that for every (a,b)H(a,b)\in H, there is an open neighborhood UU of (a,b)(a,b) with UHU\subset H, on which ff is injective, and there is a differentiable g:f(U)Ug:f(U)\to U such that g(f(x,y))=(x,y)g(f(x,y))=(x,y) for all (x,y)U(x,y)\in U

Answer 1. Set u(x,y)=x4y3u(x,y)=x^4-y^3 and v(x,y)=x2+yv(x,y)=x^2+y so that f=(u,v)f=(u,v). (Scratch: Taking all of the first order partial derivatives ux=4x3u_x=4x^3, uy=3y2u_y=-3y^2, vx=2xv_x=2x and vy=1v_y=1) . For all (x,y)2(x,y)\in \mathbb R^2, the determinant of the Jacobian of ff at (x,y)(x,y) is 4x3+6y2x=2x(2x2+6y2)4x^3+6y^2x=2x(2x^2+6y^2). Given any (a,b)H(a,b)\in H, since a>0a>0 by construction and b20b^2\ge 0 for any bb, 2a(2a2+6b2)02a(2a^2+6b^2)\ne 0. Therefore, the Inverse Function Theorem states that there is such a neighborhood UU of (a,b)(a,b) and such a function g:f(U)Ug:f(U)\to U. 0◻

Question 2. For fL1(,,m)f\in L^1(\mathbb R, \mathcal L , m) and tt\in \mathbb R define f̂(t):=f(x)exp(2πixt)dm(x).\widehat{f}(t):=\int _\mathbb R f(x)\exp(-2\pi ixt) \ dm(x). Show that f̂(x)\widehat{f}(x) is a continuous function of tt.

Question 3. Let (X,,μ)(X, \mathcal M, \mu ) be a measure space and fL1(X,,μ)f\in L^1(X, \mathcal M, \mu). Show that, for all ε>0\varepsilon > 0, there is a δ>0\delta>0 so that, for all EE \in \mathcal M, μ(E)<δE|f|dμ<ε.\mu(E) < \delta \implies \int_E |f | d\mu < \varepsilon.

Answer 2. Since fL1(X,,μ)f\in L^1(X,\mathcal M, \mu), it must be the case |f||f| a non-negative measurable functions. Thus, there is a sequence of non-negative simple functions bounded above by |f||f| and converging to |f||f| pointwise. Given ε>0\varepsilon>0 we can choose some simple function ss such that |f|dμsdμ+ε\int |f|\ d\mu \le \int s \ d\mu + \varepsilon.

Question 4. Let (X,,μ)(X, \mathcal M, \mu ) be a measure space. A family \mathcal F of measurable functions is said to be uniformly integrable if, for every ε>0\varepsilon > 0, there is a δ>0\delta>0 so that, for all EE \in \mathcal M and f,f\in \mathcal F, μ(E)<δE|f|dμ<ε.\mu(E) < \delta \implies \int_E |f | d\mu < \varepsilon. Show that any finite L1(X,,μ)\mathcal F\subset L^1(X,\mathcal M, \mu ) is uniformly integrable. (You may ignore =\mathcal F=\emptyset)

4. Let D={z|z|<1}D = \{z \in \mathbb C \mid |z| < 1\} be the unit disc. Either construct a holomorphic function f:DDf: D \to D with f(1/2)=3/4f(1/2) = 3/4 and f(1/2)=2/3f'(1/2) = 2/3 or prove that one does not exist.
5. Set 𝔻={z:|z|<1}\mathbb D = \{z \in \mathbb C : |z| < 1\} and Ω=𝔻\((1,1/2][1/2,1))\Omega = \mathbb D \setminus \big((-1, -1/2] \cup [1/2, 1)\big). Give an explicit conformal equivalence f:Ωf: \Omega \to \mathbb H.