MATH 6337 Second Midterm April 1, 2014

Transcription

1 You can use your book and notes. No laptop or wireless devices allowed. Write clearly and try to make your arguments as linear and simple as possible. The complete solution of one exercise will be considered more that two half solutions. In your solution you can use only statements that were proven in class. Name: Question: Total Points: Score:

2 1. (10 points) Let f(x) be a bounded measurable function from n to such that lim sup x > f(x) x α = C < where x = x x 2 n. For which values of α can you say that f L 1 ( n )? Is your condition on α necessary? Solution: Since f is bounded we have that f is in L 1 (B ) where B = {x x }. Form the property of f we know that there exists such that that is sup f(x) x α C + 1 x > f(x) C + 1 x α for x > From Corollary 2.52 in the book we know that if α > n then f is in L 1 (B c ). Thus α > n implies that f L 1 ( n ). Take now f such that f(x) = 1 if i x i + 2 i, for every integer i. Clearly we have lim sup f(x) = 1 while lim x > sup x > f(x) x α = for every α > 0. Notwithstanding this, f L 1 ( n ). Page 1 of 8

3 2. Given f L 1 () define F (x) = x f(y)dy. (a) (10 points) Show that F (x) is a continuous function. Is F (x) differentiable? Solution: Let y n be such that lim n y n = y. Call f n (x) = χ {x yn}(x)f(x). Clearly lim f n(x) = f(x)χ {x y} (x) a.e. n while f n f. Thus lim F (y n) = lim f n (x)dx = n n so that F is continuous. Clearly if f is discontinuous in x, F (x) does not exists. f(x)χ {x y} (x)dx = F (y) Page 2 of 8

4 (b) (15 points) Show that for every g C0() 1 we have: F (x)g (x)dx = f(x)g(x)dx. emember that C 1 0() is the set of all function in C 1 () that are 0 outside a compact set. Moreover C 1 () is the set of all function that are continuous and admit a continuous derivative. Solution: From the definition we know that there are K, M such that g(x) = g (x) = 0 for x > K and g (x) < M for x K. Let h(x, t) = f(t)g (x)χ {t<x} (x, t) We have h(x, t) dxdt 2 f(t) g (x) dxdt 2KM 2 Thus h L 1 ( 2 ). From Tonelli Theorem we get F (x)g (x)dx = h(x, t)dxdt = g (x)f(t)dt = 2 t f(t) dt < f(t)g(t)dt. Page 3 of 8

5 3. Let f(x) be in L 1 () and let ˆf(k) = e ikx f(x)dx. (a) (15 points) Assume that xf(x) is in L 1 (). Show that the derivative ˆf (k) of f(k) exists and is continuous. Moreover ˆf (k) = i e ikx xf(x)dx. Solution: Let h n be such that h n 0 as n. We have ˆf(k + h n ) ˆf(k) = (e ihnx 1)e ikx f(x)dx We first observe that, (e ihnx 1)e ikx 2 so that, by the Dominated Convergence Theorem, we have that ˆf(k) is continuous. Moreover e ix 1 x so that, for every n, e ihnx 1 e ikx f(x) h n x f(x) Since xf(x) is in L 1 () the Dominated Convergence Theorem implyes that: d ˆf(k) = dx eikx f(x)dx = i e ikx xf(x)dx. Finally reasoniong as above we get that f (k) is continuous. Page 4 of 8

6 (b) (15 points) Assume that e λ x f(x) is in L 1 (). Show that, for h small enough, we have h ˆf(k n + h) = n! ˆf (n) (k). How big can h be? Solution: We have ˆf(k + h) = k=0 e ihx e ikx f(x)dx = h n (ix) n e ikx f(x)dx n! n=0 Observe that h n (ix) n e ikx f(x) n! f(x) h n x n n! n=0 n=0 = e h x f(x) If h λ we can apply again Dominated Convergence and obtain ˆf(k + h) = n=0 h n (ix) n e ikx f(x)dx n! Clearly we have that x n f(x) is in L 1 () so that iterating point a) we get (ix) n e ikx f(x)dx = ˆf (n) (k). Page 5 of 8

7 4. Let f n, n = 1, 2,..., and f be measurable functions from a measure space (X, M, µ) to and let h n (x) = max{f n (x), 0} and h(x) = max{f(x), 0}. (a) (10 points) Assume that f n f µ-a.e. Show h n h µ-a.e. Solution: Let x be such that f n (x) f(x). If f(x) > 0 then f n (x) is definitively positive and thus h n (x) = f n (x) f(x) = h(x). If f(x) < 0 then f n (x) < 0 definitively so that h n (x) = h(x) = 0. Finally if f(x) = 0 then f n (x) 0 and clearly h n (x) 0. Page 6 of 8

8 (b) (15 points) Assume that f n f in L 1 (X, µ). Show h n h in L 1 (X, µ). (Hint: let A n = {x f n (x) > 0} and A = {x f(x) > 0}. Write X = (A A n ) (A c A c n) (A c A n ) (A A c n). Convergence in the first two set is trivial. For the other two use the inequalities that define A and A n.) Solution: Observe that on A A n we have h n h = f n f, on A c A c n we have h n h = 0, on A c A n we have h n h = h n f n f = f n f and finally on A A c n we have h n h = h f f n = f n f. Thus for every x X we have h n h f n f. Thus if f n f in L 1 (X, µ) then h n h dµ X X f n f dµ 0 Page 7 of 8

9 (c) (15 points) Assume that f n f in measure. Show h n h in measure. (Hint: see the hint to point b).) Solution: It follows immediately from {x h n h > ɛ} {x f n f > ɛ}. Page 8 of 8