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FUNCTIONAL ANALYSIS LECTURE NOTES CHAPTER 3. BANACH SPACES CHRISTOPHER HEIL 1. Elementary Properties and Examples Notation 1.1. Throughout, F will denote either the real line R or the complex plane C. All vector spaces are assumed to be over the field F. Definition 1.2. Let X be a vector space over the field F. Then a semi-norm on X is a function k · k: X ! R such that (a) kxk ≥ 0 for all x 2 X, (b) kαxk = jαj kxk for all x 2 X and α 2 F, (c) Triangle Inequality: kx + yk ≤ kxk + kyk for all x, y 2 X. A norm on X is a semi-norm which also satisfies: (d) kxk = 0 =) x = 0. A vector space X together with a norm k · k is called a normed linear space, a normed vector space, or simply a normed space. Definition 1.3. Let I be a finite or countable index set (for example, I = f1; : : : ; Ng if finite, or I = N or Z if infinite). Let w : I ! [0; 1). Given a sequence of scalars x = (xi)i2I , set 1=p jx jp w(i)p ; 0 < p < 1; 8 i kxkp;w = > Xi2I <> sup jxij w(i); p = 1; i2I > where these quantities could be infinite.:> Then we set p `w(I) = x = (xi)i2I : kxkp < 1 : n o p p p We call `w(I) a weighted ` space, and often denote it just by `w (especially if I = N). If p p w(i) = 1 for all i, then we simply call this space ` (I) or ` and write k · kp instead of k · kp;w. Date: April 11, 2006. These notes closely follow and expand on the text by John B. Conway, \A Course in Functional Analysis," Second Edition, Springer, 1990. 1 2 CHRISTOPHER HEIL p Exercise 1.4. Show that if 1 ≤ p ≤ 1 then k · kp;w defines a semi-norm on `w, and it is a norm if w(i) > 0 for all i. p n In particular, if I = f1; : : : ; ng then `w = F , and each choice of p and w gives a semi-norm or norm on Fn. Hints: The Triangle Inequality on `p is often called Minkowski's Inequality. It is easy to prove if p = 1 or p = 1. There are several ways to prove it for other p. One ways is to begin with p p p−1 kx + ykp = jxi + yij = jxi + yij jxi + yij Xi2I Xi2I p−1 p−1 ≤ jxi + yij jxij + jxi + yij jyij: Xi2I Xi2I Then apply H¨older's Inequality to each sum using the exponent p0 on the first factor and p 0 p−1 for the second (recall that p = p=(p − 1)). Then divide both sides by kx + ykp . Definition 1.5. Let (X; Ω; µ) be a measure space. Given a measurable f : X ! [−∞; 1] (if F = R) or f : X ! C (if F = C), set 1=p p 8 jf(x)j dµ(x) ; 0 < p < 1; kfk = ZX p > <ess sup jf(x)j; p = 1; x2X > where these quantities could be:>infinite. Define p L (X) = f : X ! [−∞; 1] or C : kfkp < 1 : n o Other notations for Lp(X) are Lp(µ), Lp(X; µ), Lp(dµ), Lp(X; dµ), etc. When we write Lp(Rn), it will be assumed that µ is Lebesgue measure on Rn, unless specifically stated otherwise. In this case we will write dx instead of dµ(x). p p The space `w(I) is a special case of L (X), where X = I and µ is a weighted counting measure on I. p Exercise 1.6. Show that if 1 ≤ p ≤ 1 then k · kp is a semi-norm on L (X), and it is a norm if we identify functions that are equal almost everywhere. The Triangle Inequality on Lp is often called Minkowski's Inequality, and its proof is similar to the proof of Minkowski's Inequality for `p. Exercise 1.7. Show that every subspace of a normed space is itself a normed space (using the same norm). Definition 1.8 (Distance). Let k · k be a norm on X. Then the distance from x to y in X is d(x; y) = kx − yk. CHAPTER 3. BANACH SPACES 3 Exercise 1.9. Show that d(·; ·) defines a metric on X (see Appendix A). Since X has a metric and hence has an associated topology, all the standard topological notions (open/closed sets, convergence, etc.) apply to X. For convenience, we give some explicit definitions and facts relating to these topics for the setting of normed spaces. Definition 1.10 (Convergence). Let X be a normed linear space (such as an inner product space), and let ffngn2N be a sequence of elements of X. (a) We say that ffngn2N converges to f 2 X, and write fn ! f, if lim kf − fnk = 0; n!1 i.e., 8 " > 0; 9 N > 0 such that n > N =) kf − fnk < ": (b) We say that ffngn2N is Cauchy if 8 " > 0; 9 N > 0 such that m; n > N =) kfm − fnk < ": Exercise 1.11. Let X be a normed linear space. Prove the following. (a) Reverse Triangle Inequality: kfk − kgk ≤ kf − gk. (b) Continuity of the norm: fn ! f =) k fnk ! kfk. (c) Continuity of vector addition: fn ! f and gn ! g =) fn + gn ! f + g. (d) Continuity of scalar multiplication: fn ! f and αn ! α =) αnfn ! αf. (e) All convergent sequences are bounded, and the limit of a convergent sequence is unique. (f) Every Cauchy sequence is bounded. (g) Every convergent sequence is Cauchy. Exercise 1.12. Let ffngn2N be a sequence of vectors in a normed space X. Show that if −n kfn − fn+1k < 2 for every n, then ffngn2N is Cauchy. p p Exercise 1.13. Let `w(I) be the weighted ` space defined in Exercise 1.3, where we assume p w(i) > 0 for all i 2 I. Let fxngn2N be a sequence of vectors in `w(I), and let x be a vector in p `w(I). Write the components of xn and x as xn = (xn(1); xn(2); : : : ) and x = (x(1); x(2); : : : ). (a) Prove that if xn ! x (i.e., kx − xnkp;w ! 0), then xn converges componentwise to x, i.e., for each fixed k we have limn!1 xn(k) = x(k). (b) Prove that if I is finite then the converse is also true, i.e., componentwise convergence implies convergence with respect to the norm k · kp;w. (c) Prove that if I is infinite then componentwise convergence does not imply convergence w in the norm of `p (I). 4 CHRISTOPHER HEIL It is not true in an arbitrary normed space that every Cauchy sequence must converge. Normed spaces which do have the property that all Cauchy sequences converge are given a special name. Definition 1.14 (Banach Space). A normed space X is called a Banach space if it is complete, i.e., if every Cauchy sequence is convergent. That is, ffngn2N is Cauchy in X =) 9 f 2 X such that fn ! f: p p Exercise 1.15. Show that the weighted ` space `w(I) defined in Exercise 1.3 is a Banach space if w(i) > 0 for all i 2 I. p Hints: Consider the case I = N. Suppose that fxngn2N is a Cauchy sequence in `w. Each xn is a sequence of scalars. Write out the components of xn as xn = (xn(1); xn(2); : : : ): Prove that for a fixed component k we have jxm(k) − xn(k)j ≤ C kxm − xnkp;w; where C is a fixed constant (determined by the weight and by k but independent of m and n). Conclude that with k fixed, fxn(k)gn2N is a Cauchy sequence of scalars, and hence con- verges. Define x(k) = limn!1 xn(k) and set x = (x(1); x(2); : : : ). Then we have constructed a candidate limit for the sequence fxngn2N. However, so far we only have that each indi- vidual component of xn converges to the corresponding component of x, i.e., xn converges componentwise to x. This is not enough: to complete the proof you must show that xn ! x p in the norm of `w. Use the fact that fxngn2N is Cauchy together with the componentwise p convergence to show that kx − xnk`w ! 0 as n ! 1. Compare this proof to the proof of Theorem 2.18 given below. Exercise 1.16. Show that the space Lp(X) defined in Example 1.5 is a Banach space if we identify functions that are equal almost everywhere. This is called the Riesz{Fisher Theorem. Hint: The argument is similar in spirit but more subtle than the one used to prove that p `w(I) is a Banach space. First find a candidate limit and then show that the sequence converges in norm to this limit. The next two exercises will be useful to us later. Exercise 1.17. Let X be a normed linear space. If ffngn2N is a Cauchy sequence in X and there exists a subsequence ffnk gk2N that converges to f 2 X, then fn ! f. Solution " N Choose any " > 0. Since ffngn2 is Cauchy, there is an N such that kfm − fnk < 2 for m, " n > N. Also, there is a k such that nk > N and kf − fnk k < 2 . Hence for n > N we have " " kf − f k ≤ kf − f k + kf − f k < + = ": n nk nk n 2 2 CHAPTER 3.
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