Math 6210 Lec 01

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<ul><li><p>7/30/2019 Math 6210 Lec 01</p><p> 1/3</p><p>Sean LiMath 6210 Notes Fall 2013</p><p>Grad Measure Theory and Lebesgue Integration</p><p>Lecture 1 8/28/13</p><p>Course Info</p><p>Raul Gomez, Malott 593, rg558@cornell.edu</p><p>TA: Sergio Da Silva, Malott 120, smd322@cornell.edu</p><p>Homework: Given Wednesday, due Wednesday.</p><p>Prelim: October 23rd.</p><p>Grading: The higher of 20% HW, 40% Prelim, 40% Final; and 10% HW, 30% Prelim, 60%Final.</p><p>Text: The Elements of Integration and Lebesgue Measure by Bartle</p><p>Motivation. Riemann integral not good enough.</p><p>Let [a, b] be an interval. A partition of [a, b] is a set p = {x0, . . . , xn} such that a = x0 0, there exists &gt; 0 s.t. if p = {x0, . . . , xn},ck [xk1, xk] is a tagged partition with p &lt; , then</p><p>n</p><p>k=1</p><p>f(ck)xk I &lt; </p><p>In this case we defineba</p><p>f(x)dx = I.</p><p>Definition. Let V be a vector space. A norm on V is a map : V R 0, such that</p></li><li><p>7/30/2019 Math 6210 Lec 01</p><p> 2/3</p><p>(a) x + y x + y, x, y V,</p><p>(b) cx = |c|x if c is a scalar,</p><p>(c) x = 0 iffx = 0.</p><p>Definition. Let (V, ) be a normed vector space. We say that a sequence {xn} convergesto an element x if for all &gt; 0, there exists N N such that if n &gt; N, then x xn &lt; .</p><p>Definition. Let (V, ) be as before. Then {xn} is Cauchy if &gt; 0, N N s.t. ifn,m &gt; N, then xn xm &lt; .</p><p>Definition. A normed vector space (V, ) is complete if every Cauchy sequence has alimit.</p><p>Definition. A Banach space is a normed vector space that is complete.</p><p>Example. Let C([a, b]) = {f : [a, b] C |f is continuous.}. We define a norm on C([a, b])by</p><p>f = supx[a,b]</p><p>|f(x)|.</p><p>This is a Banach space.</p><p>Proof: Let {fn} C([a, b]) be Cauchy. In particular, {fn(x)} is Cauchy. Set f(x) =limn fn(x). Want to show the convergence is uniform. Let &gt; 0. We have f fn f fm + fm fn, and by Cauchy there exists N s.t. ifn,m &gt; N, then fm fn &gt; 0 s.t. |f(x) fm(x)| &lt; 2 . This shows that fn funiformly, and since all the fns are continuous, we know that f is continuous.</p><p>Definition. A Hilbert space is a Banach space (H, ) together with a sesquilinear map</p><p>, : H H C</p><p>such thatv =</p><p>v, v.</p><p>Definition.Let (H, , ) be a Hilbert space. A Hilbert basis is a set {vi}i</p><p>I such that anyv H can be uniquely expressed as v =</p><p>iI aivi where ai = vi, v.</p><p>Theorem. Every Hilbert space has a Hilbert basis.</p><p>Definition. Let C(R /Z) = {f : R C | f is cont. and f(x + n) = f(x)n Z}.</p><p>Page 2</p></li><li><p>7/30/2019 Math 6210 Lec 01</p><p> 3/3</p><p>Definition. L2R(R /Z)0 = {f : R C | f is Z-periodic and</p><p>10</p><p>|f(x)|2dx &lt; }. (The Rmeans Riemann-integrable.)</p><p>On L2R(R /Z) we define a norm f, g =</p><p>1</p><p>0f(x)g(x)dx. Set f =</p><p>1</p><p>0|f(x)|2dx.</p><p>Ex. Let</p><p>f(x) =</p><p>0 if x [0, 1], x = 1/2,</p><p>1 if x = 1/2.</p><p>Then10</p><p>|f(x)|2dx = 0.</p><p>Definition. Given f, g L2R(R /Z)0, we say f g if</p><p>10</p><p>|f(x) g(x)|2dx = 0.</p><p>Set L2R(R /Z) = L2R(R /Z)</p><p>0/ .</p><p>Theorem. L2</p><p>R(R</p><p>/Z</p><p>) is not a Hilbert space.</p><p>Theorem. Let L2(R /Z) = L2(R /Z)0/ where the requirement is Lebesgue integrable,not Riemann integrable. Then L2(R /Z) is a Hilbert space with Hilbert basis {fn}nZ wherefn(x) = e</p><p>2inx.</p><p>Page 3</p></li></ul>