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# Untitled - Chapter 16 Fourier series Sudarshan Balakrishnan...

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Chapter 16: Fourier series Sudarshan Balakrishnan June 17, 2011 Periodic Functions Exercise 16.1.1 Show that every real number x can be written in exactly one way in the form x = k + y , where k is an integer and y [0 , 1) . Proof In order to prove this statement, we will ﬁrst introduce certain concepts that will be useful to keep in mind. In the above problem, k is called the integer part of the real number x , denoted by [ x ] and y is called the fractional part of x denoted by { x } .Now, let us consider the set: k = sup { l Z : l x } , (1) from ( ?? ) it is immediate that we have to prove 0 x - k < 1 . (2) In ( ?? ), we can see that the ﬁrst side of inequality is clear since x - k 0 as x is the upper bound of the the set k . Now, to prove the other inequality, let us try and obtain a contradiction by assuming x - k > 1.This implies x > k + 1. Clearly, this is a contradiction because this would imply k + 1 would be in the set k and so we have shown that x can be as a sum of its integer and fractional part. We still have to show uniqueness of this representation. Let x = k + y , now suppose x = k 0 + y 0 where k 0 Z and y 0 [0 , 1) then we have that: k - k 0 + y - y 0 = 0 (3) 1

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This implies that y = y 0 and k = k 0 because the diﬀerence of the frac- tional parts has to equal zero for the expression to equal zero, uniqueness follows. Exercise 16.1.2 (a) If f C(R/Z; C) , then f is bounded( i.e., there exists a real number M > 0 such that | f ( x ) | ≤ M for all x R . Proof Let us consider the closed interval [0 , 1], extending the result to the general case follows immediately. For ± = 1, we can choose a δ s.t | x - y | ≤ δ implies | f ( x ) - f ( y ) | ≤ 1 . Now, we have to choose a suitable lower bound for δ , this can be done by selecting an integer z s.t δ 1 z . Next, we substitute x k = k z for k ∈ { 1 ,...z - 1 } . For a suitable x [0 , 1], there exists k ∈ { 1 ,...,z - 1 } such that | x - x k | ≤ 1 z δ . All that remains to show is that we can select a suitable bound for f ( x ), this can be done as follows: | f ( x ) | ≤ | f ( x ) - f ( x k ) | + | f ( x k ) | ≤ 1 + max 1 k z - 1 | f ( x k ) | (4) As we can see, the upper bound is clearly independent of x . The claim follows. Exercise 16.1.2 (b)If f,g C(R/Z;C) , then the functions f + g , f - g , and fg are also in C(R/Z;C) . Also, if c is any complex number, then the function cf is also in C(R/Z;C) . Proof Let f,g C(R/Z;C) be continuous and x R/Z . Now, consider y R/Z . Then f ( x ) + g ( x ) - f ( y ) - g ( y ) = ( f ( x ) - f ( y )) + ( g ( x ) - g ( y )) (5) Let ± > 0 be arbitrary, we can choose δ 1 > 0 such that for all | x - y | < δ 1 | ( f ( x ) - f ( y )) | < ± 2 , (6) similarly, we choose a δ 2 > 0 such that for all | x - y | < δ 2 2
| ( g ( x ) - g ( y )) | < ± 2 . (7) Let δ = min( δ 1 2 ) and | x - y | < δ . Then we have that | ( f ( x ) - f ( y )) | + | ( g ( x ) - g ( y )) | < ± 2 + ± 2 = ± (8) Therefore f + g is continuous and

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## This note was uploaded on 11/19/2011 for the course MATH 101 taught by Professor Wormer during the Spring '08 term at UCSD.

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Untitled - Chapter 16 Fourier series Sudarshan Balakrishnan...

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