New version page

UCSD MATH 142B - Lecture

Upgrade to remove ads
Upgrade to remove ads
Unformatted text preview:

Introduction to Normed Vector SpacesAudrey TerrasMarch 29, 20091 Why worry about in…nite dimensional normed vector spaces?We want to unde rstand the integral from Lang’s perspective rather than that of your calculus book. Secondly we want tounderstand convergence of series of functions - something that proved problematic for Cauchy in the 1800s. These things areimportant for many applications in physics, engineering, statistics. We will be able to study vibrating things such as violinstrings, drums, buildings, bridges, spheres, planets, stock values. Quantum physics, for example, involves Hilbert space,which is a type of normed vector space with a scalar product where all Cauchy sequences of vectors converge.The theory of such normed vector spaces was created at the same time as quantum mechanics - the 1920s and 1930s.So with this chapter of Lang you are moving ahead hundreds of years from Newton and Leibnitz, perhaps 70 years fromRiemann.Fourier series involve orthogonal sets of vectors in an in…nite dimensional normed vector space:C[a; b] = ff : [a; b] ! R jf continuousg:The L2norm of a continuous function f in C[a; b] [email protected](x)j2dx1A1=2:This is an analog of the usual idea of length of a vector f = (f (1); :::; (f(n)) 2 Rn:[email protected]=1jf(j)j21A1=2:There are other natural norms for f 2 C[a; b] such as:kfk1=bZajf(x)jdx:kfk1= maxaxbjf(x)j:On …nite dimensional vector spaces such as Rnit does not matter what norm you use when you are trying to …gure outwhether a sequence of vectors has a limit. However, in in…nite dimensional normed vector spaces convergence can disappearif a di¤erent norm is used. Not all norms are equivalent in in…nite dimensions. See Homework 9, problem 6.Note that C[a; b] is in…nite dimensional since the set f1; x; x2; x3; :::; xn; :::g is an in…nite set of linearly independentvectors. Prove this as follows. Suppose that we have a linear dependence relationnXj=0cjxj= 0; for all x in [a; b]. Thisimplies all the constants cj= 0: Why? Extra Credit. Prove this.In…nite dimensional vector spaces are thus more interesting than …nite dimensional ones. Each (inequivalent) norm leadsto a di¤erent notion of convergence of sequences of vectors.12 What is a Normed Vector Space?In what follows we de…ne normed vector space by 5 axioms. We will not put arrows on our vectors. We will try to keepvectors and scalars apart by using Greek letters for scalars. Our scalars will be real. Maybe by the end of next quarter wemay allow complex s calars. It simpli…es Fourier series.De…nition 1 A vector space V is a set of vectors v 2 V which is closed under addition and closed under multiplicationby scalars  2 R. This means u + v 2 V and v 2 V and the following 5 axioms must hold for all u; v; w 2 V and all;  2 R:VS1. u + (v + w) = (u + v) + wVS2. 9 0 2 V s.t. 0 + v = vVS3. 8 v 2 V 9 v02 V s.t. v + v0= 0:VS4. v + u = u + vVS5. 1v = v; (v) = ()v; ( + )v = v + v; (u + v) = u + vYou may say we cheated by putting 4 axioms into VS5.De…nition 2 A vector space V is a normed vector space if there is a norm function mapping V to the non-negative realnumbers, written kvk; for v 2 V; and satisfying the following 3 axioms:N1: kvk  0 8v 2 V and kvk = 0 if and only if v = 0:N2: kvk = jjkvk; 8v 2 V and8 2 R: Here jj = absolute value of :N3: ku + vk  kuk + kvk; 8u; v 2 V: Triangle Inequality:De…nition 3 The distance between 2 vectors u; v in a normed vector space V is de…ned byd(u; v) = ku  vk:Example 1. 3-Space.R3=8<:[email protected]x1; x2; x32 R9=;:De…ne addition and multiplication by scalars as usual:[email protected][email protected][email protected]+ y1x2+ y2x3+ y31A:[email protected][email protected]x1x2x31A; 8  2 R:The usual norm iskxk2=qx21+ x22+ x23ifx [email protected]:Other norms are possible:kxk2= jx1j + jx2j + jx3j or kxk1= max fjx1j; jx3j; jx3jg:The proof that these de…nitions make R3a normed vector space is tedious. No doubt we will make it a homeworkproblem.2Example 2. The space of continuous functions on an interval.C[a; b] = ff : [a; b] ! R jf continuousg:For f; g 2 C[a; b]; de…ne (f + g)(x) = f(x) + g(x) for all x 2 [a; b] and de…ne for  2 R (f )(x) = f (x) for allx 2 [a; b]: We leave it as an Exercise (see Homework 9) to check the axioms for a vector space. The most non-trivial one isthe one that says f + g and f are both continuous functions on [a; b].Again there are many possible norms. We will look at 3:[email protected](x)j2dx1A1=2:kfk1=bZajf(x)jdx:kfk1= maxaxbjf(x)j:Most of the axioms for norms are easy to check. Let’s do it for the kf k1norm.N1. kvk  0 8 v 2 V and kvk = 0 if and only if v = 0.N2. kvk = jjkvk; 8 v 2 V and 8 2 R .N3. Triangle Inequality. ku + vk  kuk + kvk; 8 u; v 2 V:Proof of N1.Since jf(x)j  0 for all x we know that the integral is  0; because the integral preserves inequalities (Lang, Thm. 2.1,p. 104 or earlier in these n otes). Suppose thatbZajf(x)jdx = 0: Since f is continuous by Theorem 2.4 of Lang, p. 104 orthese notes near Figure 1, this implies f(x) = 0 for all x 2 [a; b]:Proof of N2.Also for any  2 R and f 2 C[a; b]; we have kf k1=bZajf(x)jdx =bZajjjf (x)jdx = jjbZajf(x)jdx = jjkfk1: Thisproves N2 for norms. Here we used the multiplicative property of absolute value as well as the linearity of the integral (i.e.,scalars come out of the integral lecture notes p. 80).Proof of N3. Using the de…nition of the 1-norm, and the triangle inequality for real numbers as wellas the fact that the integral preserves ; we see thatkf + gk1=bZajf(x) + g(x)jdx bZa(jf(x)j+ jg(x)j) dx:To …nish the proof, use the linearity of the integral to see thatbZa(jf(x)j+ jg(x)j) dx =bZajf(x)jdx +bZajg(x)jdx = kfk1+ kg k1:Putting it all together gives kf + gk1 kf k1+ kg k1:33 Scalar Products.You have seen the dot (or scalar or inner) product in R3: It [email protected][email protected]= x1y1+ x2y2+ x3y3:It turns out there is a similar thing for C[a; b]: First let’s de…ne the scalar produ ct on a vector space and see how to geta norm if in addition the scalar product is positive de…nite.De…nition 4 A (positive de…nite) scalar product < v; w > for vectors v; w in a vector space V is a real number < v; w >such that the following axioms hold:SP1: < v; w >=< w; v >; 8 v; w 2 V (symmetry)SP2: < u; v + w >=< u; v > + < u; w >; 8 u; v; w 2 VSP3: < v; w >=  < v; w >; 8 v; w 2 V and 8RSP4: < v; v >  0; 8 v 2 V


View Full Document
Download Lecture
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Lecture and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Lecture 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?