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Unformatted text preview: Modern Analysis, Homework 9 Possible solutions Review problems for chapter 9. Recommended problems are marked with a *. Through out, α is an increasing function on the interval [ a,b ], where a < b . 1. * [10] Rudin, Ch 6, #1: Suppose that a ≤ x ≤ b , α is continuous at x , f ( x ) = 1, and f ( x ) = 0 if x 6 = x . Prove that f ∈ R ( α, [ a,b ]) and that R b a f dα = 0. Solution: Assume x 6 = a or b (a similar argument to the following applies if x is one of the endpoints). Let ε > 0. By the continuity of α at x , there exists 0 < δ < min(  x a  ,  x b  ) such that  x x  ≤ δ implies  α ( x ) α ( x )  < ε . Consider the following partition of [ a,b ]: P = { a,x δ,x + δ,b } . For this partition, we have L ( P,f,α ) = 0 and U ( P,f,α ) < ε , since for all intervals except the one containing x , f is identically zero. On the interval [ x δ,x + δ ], M ( f ) = 1, m ( f ) = 0, and Δ α < ε . This shows that f ∈ R ( α, [ a,b ]) and that R b a f dα = 0. 2. * [14] Rudin, Ch 6, #2: Suppose that f ≥ 0, f is continuous on [ a,b ] and that R b a f ( x ) dx = 0. Prove that f ( x ) = 0 for all x ∈ [ a,b ]. Solution: Suppose that f ( x ) > 0 for some x ∈ [ a,b ]. As in the previous problem, assume x 6 = a or b , as those cases can be handled similarly to the following argument. By the continuity of f at x , there exists min(  x a  ,  x b  ) > δ > 0 such that  x x  < δ implies  f ( x ) f ( x )  < f ( x ) / 2, in particular, f ( x ) > f ( x ) / 2 > 0. Define a function g : [ a,b ] → R as follows: g ( x ) = f ( x ) / 2 if  x x  < δ otherwise Then for all x ∈ [ a,b ], we have f ( x ) ≥ g ( x ). Therefore, Z b a f ( x ) dx ≥ Z b a g ( x ) dx = δf ( x ) > a contradiction! 3. [15] Rudin, Ch 6, #11: For u ∈ R ( α, [ a,b ]) define k u k 2 = Z b a  u  2 dα 1 / 2 . Given f,g,h ∈ R ( α, [ a,b ]), prove that k f h k 2 ≤ k f g k 2 + k g h k 2 . Solution: Given f,g ∈ R ( α, [ a,b ]), define h f,g i = Z b a fg dx. Notice that h· , ·i satisfies the following properties: 1 1. h f,g i = h g,f i for any pair f,g ∈ R ( α, [ a,b ]) 2. h cf + g,h i = c h f,h i + h g,h i for any triple f,g,h ∈ R ( α, [ a,b ]) and scalar c ∈ R 3. h f,f i = k f k 2 2 ≥ 0 for any f ∈ R ( α, [ a,b ]) Now we can imitate our proof of the CauchySchwarz inequality: fix f,g ∈ R ( α, [ a,b ]) and consider variable t ∈ R . Then k f + tg k 2 2 = h f + tg,f + tg i = k f k 2 2 + 2 t h f,g i + t 2 k g k 2 2 ≥ . Since the quadratic function (of t ) k f k 2 2 + 2 t h f,g i + t 2 k g k 2 2 is everywhere nonnegative, it follows that its discriminant is nonpositive: 4 h f,g i 2 4 k f k 2 2 k g k 2 2 ≤ So h· , ·i satisfies a sort of CauchySchwarz inequality: h f,g i ≤ k f k 2 k g k 2 for all f,g ∈ R ( α, [ a,b ]). The triangle inequality follows just as it does in the case of the usual CauchySchwarz inequality on...
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This note was uploaded on 10/18/2011 for the course MATH S4061X taught by Professor Peters during the Spring '11 term at Columbia.
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