MIT18_102s09_lec05

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MIT OpenCourseWare http://ocw.mit.edu 18.102 Introduction to Functional Analysis Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .
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22 LECTURE NOTES FOR 18.102, SPRING 2009 Lecture 5. Thursday, 19 Feb I may not get quite this far, since I do not want to rush unduly! Let me denote by L 1 ( R ) the space of Lebesgue integrable functions on the line as defined last time. Proposition 4. L 1 ( R ) is a linear space. Proof. The space of all functions on R is linear, so we just need to check that L 1 ( R ) is closed under multiplication by constants and addition. The former is easy enough multiplication by 0 gives the zero function which is integrable. If g ∈ L 1 ( R ) then by definition there is an absolutely summable series of step function with elements f n such that (5.1) | f n | < , f ( x ) = f n ( x ) x s.t. | f n ( x ) | < . n n n Then if c = 0 , cf n ‘works’ for cg. The sum of two functions g and g ∈ L 1 ( R ) is a little trickier. The ‘obvious’ thing to do is to take the sum of the series of step functions. This will lead to trouble! Instead, suppose that f n and f n are series of step functions showing that g, g ∈ L 1 ( R ) . Then consider (5.2) h n ( x ) = f k n = 2 k 1 f k n = 2 k. This is absolutely summable since (5.3) | h n | = | f k | + | f k | < . n k k More significantly, the series | h n ( x ) | converges if and only if both the series n | f k ( x ) | and the series | f k ( x ) converge. Then, because absolutely convergent k k series can be rearranged, it follows that (5.4) | h n ( x ) | < = h n ( x ) = f k ( x ) + f k ( x ) = g ( x ) + g ( x ) . n n k k Thus, g + g ∈ L 1 ( R ) . So, the message here is to be a bit careful about the selection of the ‘approxi- mating’ absolutely summable series. Here is another example. Definition 4 . A set E R is of measure zero if there exists an absolutely summable series of step functions f n such that (5.5) | f n ( x ) | = ∞ ∀ x E.
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