MA677 Assignment #3 Morgan Schreffler Due 09/19/12 Exercise 1 Using Hölder s inequality, prove Minkowski s inequality for f, g L p (R d ), p 1:

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1 Exercise 1 Using Hölder s inequality, prove Minkowski s inequality for f, g L p (R d ), p 1: f + g p f p + g p. Proof. If f, g L p (R d ), then since f(x) + g(x) max {f(x), g(x)}, we have f(x) + g(x) p p max { f(x) p, g(x) p } p ( f(x) p + g(x) p ). It follows that f + g p p f p + p g p <, so f + g L p (R d ). Now, note that f + g p f + g p 1 f + g f + g p 1 ( f + g ). Thus, ( f + g p ) p (f + g) p 1 ( f + g ) 1 (f + g) p 1 f + (f + g) p 1 g (f + g) p 1 p ( f p + g p 1 p ) by Hölder s inequality ( ) p 1 ( f + g p 1 ) p p p 1 ( f p + g p ) ( ( ) ) 1 p 1 f + g p p ( f p + g p ) ( f + g p ) p 1 ( f p + g p ). Finally, since we showed f + g p is finite, dividing both sides of the above inequality by ( f + g p ) p 1 gives us the desired result. Page 1 of 5

2 Page 195 Exercise 11 Let P be the orthogonal projection associated with a closed subspace S in a Hilbert space H, that is, P (f) f if f S and P (f) 0 if f S. (a) Show that P P and P P. (b) Conversely, if P is any bounded operator satisfying P P and P P, prove that P is the orthogonal projection for some closed subspace of H. (c) Using P, prove that if S is a closed subspace of a separable Hilbert space, then S is also a separable Hilbert space. Proof. (a) If f S, then P (f) P (P (f)) P (f), and if f S, then P (f) P (P (f)) P (0). But P is linear, so P (0) 0 P (f). Hence, P P. Now, let f S. Then (P f, f) (f, f) (f, P f), but (P f, f) (f, P f). Since P is unique, we have P f P f. Similarly, f S gives (P f, f) 0 (f, P f), so P P. (b) First, we will show that S P (H) is a subspace of H. Let f, g S. Then f P (h 1 ) and g P (h ) for some h 1, h H. Consequently, so f + λg S. f + λg P (h 1 ) + λp (h ) P (h 1 + λh ), Moreover, S is closed. Indeed, suppose {f n } n1 is a sequence in S such that f n f. Then f n P g n for some g n H and each n N. Since P is bounded, it is continuous, so P f n P f. This gives us P (P g n ) P g n P g n f n P f. But H is a metric space, so H is Hausdorff. In particular, this tells us limits are unique, so f P (f). As a direct result, S is closed. Also, since each f S is the limit of a convergent sequence in S, we have that P (f) f for all f S. Further, if P f f, then f S, so S {f H P f f}. Finally, we will show that S ker(p ). To begin, note that if {f n } n1 ker(p ) converges to f H, then P f lim P f n 0, so f ker(f), making ker(p ) closed. Next, we know that ker(p ) S. To see this, we note that S S {0} and S {f H P f f}, so if f ker(p ), then either f 0 or P f f, i.e., f S S. Now, suppose that 0 f S. Then for all g H, (P f, g) (f, P g) (f, P g). If g S, then P g g, and since f S, (f, P g) (f, g) 0. Similarly, if g / S, then P g is, so (f, P g) 0. Hence, (P f, g) 0 for all g H, so P f 0, putting f ker(p ). Thus, P is the orthogonal projection for the closed subspace S P (H). Page of 5

3 (c) Note that every closed subspace S H has an orthogonal projection P S associated with it. Also, we have that P S (S) S. Now, let A H be a countable dense subset, and let U S P S (S) be open in S. Since P S is continuous, (P S ) 1 (U) is open in H. Since A is dense, there is an a A which is also in (P S ) 1 (U), i.e., P S (a) U. Since U was arbitrary, P S (A) is countable and dense in S, making S separable. Page 197 Exercise 18 Let H be a Hilbert space and L(H) the vector space of all B.L.T.s on H. Given T L(H), we define the operator norm T inf{b : T v B v, for all v H}. (a) Show that T 1 + T T 1 + T whenever T 1, T L(H). (b) Prove that d(t 1, T ) T 1 T defines a metric on L(H). (c) Show that L(H) is complete in the metric d. Proof. (a) For any f H, (T 1 + T )(f) T 1 f + T f T 1 f + T f, so T 1 + T T 1 + T. (b) Using part (a) we can show that the triangle inequality holds. d(t 1, T 3 ) T 1 T 3 T 1 T + T T 3 T 1 T + T T 3 d(t 1, T ) + d(t, T 3 ). Further, it is clear that d(t 1, T ) d(t, T 1 ), since for every f H, (T 1 T )(f) T 1 f T f T f T 1 f (T T 1 )(f). Thus, we need only show that d(t 1, T ) 0 iff T 1 T. Page 3 of 5

4 If T 1 T, then for all f H, T 1 f T f 0 0. Thus, for all B [0, ), (T 1 T )(f) B f, so d(t 1, T ) T 1 T 0. Conversely, if d(t 1, T ) 0, then for all f H, (T 1 T )(f) 0. But the norm is always nonnegative, so (T 1 T )(f) 0. This implies that (T 1 T )(f) 0, i.e., T 1 f T f 0 or T 1 f T f. Since f was arbitrary, we have T 1 T. (c) Let {T n } n1 be a Cauchy sequence in L(H), ε > 0, and f H. Then there is an N N such that n, m N implies T n T m < ε. By definition, this means that 3 (T n T m )(f) < ε. Hence, {T 3 nf} n1 is a Cauchy sequence of vectors in H, which converges since H is complete. Let T be defined by T f lim T n f. We will show that T L(H). First, note that T is linear, since for any f, g H and λ C, T (f + λg) lim T n (f + λg) lim [T n f + λt n g] lim T n f + λ lim T n g T f + λt g. To show T is bounded, let {f i } i1 be a convergent sequence in H with f i f. Now, if we let N be chosen as above, then we see for each m N and g H that T m g T g lim T m g T n g ε 3. Further, since each T m is bounded (and thus continuous), for any m N we can choose N such that j N gives T m f j T m f lim k T m f j T m f k ε 3. Hence, if we let N max N, N, then j, m N gives T f j T f T f j T m f j + T m f j T m f + T m f T f ε 3 + ε 3 + ε 3 ε. This proves that T is continuous (and hence bounded), so T L(H). Page 4 of 5

5 Page 04 Exercise 8 Suppose {T k } is a collection of bounded operators on a Hilbert space H, with T k 1 for all k. Suppose also that T k Tj Tk T j 0 for all j k. Let S N N T k. Show that S N (f) converges as N for every f H. If T (f) denotes the limit, prove that T 1. Proof. First, note that the ranges of the T k (and similarly the Tk ) are mutually orthogonal, for if f, g H and j k, then (T j f, T k g) (f, Tj T k g) (f, 0) 0. Thus we have that for each k, Tk P ktk, where P k is the orthogonal projection of some closed subspace S k containing the range of T k, namely, S k (Range(Tk ) ). Moreover, T k T k P k (since Pk P k). Now, consider the norm of S N (f). S N (f) T k f T k f T k P k f T k P k f P k f. All of the S k are orthogonal to each other, so we can use Proposition 4. inductively to represent f by f ˆf N + N f k, where each f k S k, (f j, f k ) 0 for all j k, and ( ˆf N, f k ) 0 for all N k N. Further, f ˆf N + N f k. Hence, P k f f k ˆf N + f k f. Therefore, if T (f) lim N S N (f) exists, then T (f) f for all f H, i.e., T 1. To show that S N (f) converges, note that S N (f) converges to some value L 1. Thus, for each ε > 0 there is an N N such that n N implies L S n (f) < ε. As a result, if m, n N, we have S n (f) S m (f) S n (f) S m (f) < ε, since no two points inside a circle of radius ε can be more than ε units apart. Consequently, {S N (f)} N1 is a Cauchy sequence in H, and thus convergent. Page 5 of 5