royden6 - Royden, Real Analysis 3rd ed. Chapter 6 Problem...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Royden, Real Analysis 3rd ed. Chapter 6 Problem 6-1 For any ε > 0 , we have from the de f nition of k·k that | f ( x ) | k f k + ε and | g ( x ) | k g k + ε . for x a.e. Hence | ( f + g )( x ) | k f k + k g k +2 ε a.e. and thus k f + g k k f k + k g k +2 ε . Now let ε 0 . Problem 6-2 Let E ε = { x : | f ( x ) | > k f k ε } . By de f nition m ( E ε ) > 0 for any ε > 0 . Hence k f k p = μZ 1 0 | f ( x ) | p 1 /p μZ E ε | f ( x ) | p 1 /p ( k f k ε ) · ( m ( E ε )) 1 /p . Because lim x →∞ a 1 /x =1 for any a> 0 , we have lim k f k p k f k ε . (1) Fortheotherd irect ion ,wehave k f k p = μZ 1 0 | f ( x ) | p 1 /p μZ 1 0 ( k f k + ε ) p 1 /p = k f k + ε (2) Now let ε 0 in (1) and (2) . Problem 6-7 De f ne f ( x )= X k =1 ξ k χ [ k,k +1) ( x ) and g ( x )= X k =1 η k χ [ k,k +1) ( x ) . Then k f k p = k < ξ k > k p , k g k p = k < η k > k p and k f · g k 1 = P k =1 | ξ k · η k | . Now apply both inequalities to f and g. Problem 6-8 Let us cite the following Young’s Inequality: Let f be a real-valued, continuous and strictly increasing function on [0 ,c ] . If f (0) = 0 ,a [0 ,c ] and b [0 ,f ( c )] , then Z a 0 f ( x ) dx + Z b 0 f 1 ( x ) dx a ·
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Equality holds in (1) if and only if b = f ( a ) . You may draw the f gure and get a geometric proof. a. Let f ( x )= x p 1 in (1) . Then f 1 ( x )= x 1 / ( p 1) = x q 1 and Z a 0 f ( x ) dx = a p p , Z b 0 f 1 ( x ) dx = b q q . Here we need 1 /p +1 /q =1 which implies 1+1 / ( p 1) = q. Hence a p p + b q q ab (2) and equality holds if and only if b = a p 1 . b. Letting a = | f ( x ) | / k f k p and b = | g ( x ) | / k g k q in (2) and then integrating over x, we have Z 1 0 | f ( x ) · g ( x ) | k f k p ·k g k q 1 p · R | f | p k f k p p + 1 q · R | g | q k g k q q = 1 p + 1 q =1 . That is, k f · g k 1 k f k p ·k g k q and equality holds if and only if | g ( x ) | = c ·| f ( x ) | p 1 a.e., which is equivalent to | g ( x ) | q = c 0 ·| f ( x ) | p a.e. (3) Intheprev iousargumentwehaveassumedimp l ic ity lythat k f k p ·k g k q > 0 . The conclusion still holds even if k f k p ·k g k q =0 . Theon lychangeisthat (3) is replaced by α | f | p = β | g | q a.e. for some constants α and β with α 2 + β 2 > 0 . c. If 0 <p< 1 , then q< 0 and p 1 + q 1 =1 implies 1 p =1+ 1 q or 1= 1 P + 1 Q where P =1 /p and Q = q/p lie in (1 , ) . Let A =( a · b ) p and B = b p . By (2) A P P + B Q Q A · B with ”=” held if and only if B = A P 1 . As imp leca lcu lat
Background image of page 2
which is the same as a p p + b q q ab for a, b > 0 . Note that a geometric proof similar to that in
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 10

royden6 - Royden, Real Analysis 3rd ed. Chapter 6 Problem...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online