real-analysis-final-exam-with-solution-2008

real-analysis-final-exam-with-solution-2008 - Real Analysis...

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Real Analysis Final Exam, 2008-1-8 Name: ID #: Show detailed argument to each problem. 1. (10 points) Let E R n be a measurable set and q ( x ) 0 be a nonnegative measurable function on E . Show that for any constant α > 0 we have the inequality Z E qdx · Z E q α dx | E | · Z E q 1+ α dx. solution: By Hölder inequality Z E qdx μZ E q 1+ α dx 1 1+ α μZ E dx α 1+ α , Z E q α dx μZ E ( q α ) 1+ α α dx α 1+ α μZ E dx 1 1+ α the proof is done. ¤ 2. (10 points) Assume | E | < and u is a measurable function on E, which is everywhere positive. For 0 < p < , set Φ p ( u ) := μ 1 | E | Z E | u ( x ) | p dx 1 /p = μ 1 | E | Z E u p ( x ) dx 1 /p . (0.1) Assume that (1) . u p ( x ) 1 uniformly on E as p 0 + . (2) . d dp ¡R E u p ( x ) dx ¢ = R E d dp ( u p ( x )) dx. (3) . | log u ( x ) | C for all x E, for some constant C > 0 . Evaluate the limit lim p 0 + Φ p ( u ) in terms of an integral involving u. (Compare with the limit lim p →∞ Φ p ( u ) = k u k . ) solution: As p 0 + , Φ p ( u ) has the form 1 . Hence we can use Lopital rule . Note that log Φ p ( u ) = log ³ 1 | E | R E u p ( x ) dx ´ p μ 0 0 form and so lim p 0 + log Φ p ( u ) = lim p 0 + d dp log ³ 1 | E | R E u p ( x ) dx ´ 1 = lim p 0 + 1 | E | d dp R E u p ( x ) dx 1 | E | R E u p ( x ) dx = lim p 0 + 1 | E | R E d dp ( u p ( x )) dx 1 | E | R E u p ( x ) dx = lim p 0 + 1 | E | R E u p ( x ) log u ( x ) dx 1 | E | R E u p ( x ) dx = 1 | E | Z E log u ( x ) dx. Hence we have the formula lim p 0 + Φ p ( u ) = exp μ 1 | E | Z E log u ( x ) dx .
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