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Unformatted text preview: Mathematics 245B Terence Tao Midterm, Feb 4, 2009 Instructions: Try to do all three problems; they are all of equal value. There is plenty of working space, and a blank page at the end. You may freely use the axiom of choice in your work, and cite any theorem or result from the class notes or textbook. Good luck! Name: Student ID: Signature: Problem 1. Problem 2. Problem 3. Total: Problem 1. Let 1 ≤ p < ∞ , and let f ∈ L p ( R ), where we give R the usual Lebesgue measure and σ-algebra. For every t ∈ R , let f t ∈ L p ( R ) be the translated function f t ( x ) := f ( x- t ). Show that f t converges in L p norm to f in the limit t → 0 (i.e. lim t → k f t- f k L p ( R ) = 0). ( Hint: First verify this when f is a step function 1 [ a,b ] , then when f is an indicator function 1 E of some set of finite measure E , then as a simple function, then as an arbitrary L p function. Alternatively, show that the space of f ∈ L p ( R ) which obeys the desired properties is a closed linear subspace of L p that contains the step functions.) Let V be the space of all f ∈ L p ( R ) such that lim t → k f t- f k L p ( R ) = 0, thus V is a subset of L p ( R ). It is closed under addition: if f,g ∈ V , then f + g ∈ V , because k ( f + g ) t- ( f + g ) k L p ( R ) ≤ k f t- f k L p ( R ) + k g t- g k L p ( R ) thanks to the triangle inequality (and the fact that translation f 7→ f t is a linear operation). For similar reasons, it is closed under multiplication: if c ∈ C and f ∈ V , then cf ∈ V , because k ( cf ) t- cf k L p ( R ) = | c |k f t- f k L p ( R ) . Thus V is a vector space. Also observe that V is closed in L p ( R ): if f ( n ) ∈ V converges in L p ( R ) to f , then for any ε > we have k f ( n )- f k L p ( R ) ≤ ε for some sufficiently large n . Since translation does not affect Lebesgue measure (and thus does not affect the L p norm), we also have k f ( n ) t- f t k L p ( R ) ≤ ε for all t ∈ R (it is important here that the RHS is uniform in t ). Fixing n , we then see that as f ( n ) ∈ V , we also have k f ( n ) t- f ( n ) k L p ( R ) ≤ ε for all sufficiently small t . Putting this all together using the triangle inequality, we see that...
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