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Pointwise Convergence Ofofof Complex Fourier Series Pointwise convergence ofofof complex Fourier series Let f(x) be a periodic function with period 2 l defined on the interval [- l,l]. The complex Fourier coefficients of f( x) are l -i n π s/l cn = 1 ∫ f(s) e d s 2 l -l This leads to a Fourier series representation for f(x) ∞ i n π x/l f(x) = ∑ cn e n = -∞ We have two important questions to pose here. 1. For a given x, does the infinite series converge? 2. If it converges, does it necessarily converge to f(x)? We can begin to address both of these issues by introducing the partial Fourier series N i n π x/l fN(x) = ∑ cn e n = -N In terms of this function, our two questions become 1. For a given x, does lim fN(x) exist? N→∞ 2. If it does, is lim fN(x) = f(x)? N→∞ The Dirichlet kernel To begin to address the questions we posed about fN(x) we will start by rewriting fN(x). Initially, fN(x) is defined by N i n π x/l fN(x) = ∑ cn e n = -N If we substitute the expression for the Fourier coefficients l -i n π s/l cn = 1 ∫ f(s) e d s 2 l -l into the expression for fN(x) we obtain N l 1 -i n π s/l i n π x/l fN(x) = ∑ ∫ f(s) e d s e n = -N(2 l -l ) l N = ∫ 1 ∑ (e-i n π s/l ei n π x/l) f(s) d s -l (2 l n = -N ) 1 l N = ∫ 1 ∑ ei n π (x-s)/l f(s) d s -l (2 l n = -N ) The expression in parentheses leads us to make the following definition. The Dirichlet kernel is the function defined as N 1 i n π x/l KN(x) = ∑ e 2 l n = -N In terms of the Dirichlet kernel, we can write the expression for fN(x) as l fN(x) = ∫ KN(x-s)f(s) d s -l Some properties ofofof thethethe Dirichlet kernel By rewriting the expression for the Dirichlet kernel, we can recogize that the Dirichlet kernel is actually a geometric series. N N n 1 i n π x/l 1 i π x/l KN(x) = ∑ e = ∑ (e ) 2 l n = -N 2 l n = -N Because this is a geometric series, it can be summed explicitly. N N+1 -N n i π x/l i π x/l 1 i π x/l 1 (e ) - (e ) KN(x) = ∑ (e ) = 2 l n = -N 2 l ei π x/l - 1 N+1/2 -( N+1/2) ( i π x/l) ( i π x/l) = 1 e - e 2 l 1/2 -1/2 (ei π x/l) - (ei π x/l) N+1/2 -( N+1/2) ( i π x/l) ( i π x/l) = 1 e - e 2 l ei π x/(2 l) - e-i π x/(2 l) sin (2 N+1) π x ( ) = 2 l 2 l sin π x (2 l ) Some explicit integrations show that 0 l 1 ∫ KN(x) d x = ∫ KN(x) d x = -l 0 2 Convolutions The integral we saw earlier 2 l fN(x) = ∫ KN(x-s)f(s) d s -l is an example of what is known as a convolution integral . Specifically, if g(x) and h(x) are two periodic functions with period 2 l defined on [- l,l] the convolution of g and h is defined by l (g*h)( x) = ∫ g(x-s) h(s) d s -l Here is an important property of convolution integrals. From the definition we have that l (g*h)( x) = ∫ g(x-s) h(s) d s -l If we introduce a change of variables z = x - s in the integral, the integral becomes x-l x+l ∫ g(z)h(x-z)(-1) d z = ∫ g(z)h(x-z) d z x+l x-l Since both g and h are assumed to be periodic with the same period, if we shift the range of integration by a factor of x, the integral has the same value. x+l l ∫ g(z)h(x-z) d z = ∫ g(z)h(x-z) d z x-l -l Replacing the variable z with s in the final integral gives l l (g*h)( x) = ∫ g(x-s) h(s) d s = ∫ g(s)h(x-s) d s = (h*g)(x) -l -l This is an important symmetry property of the convolution of periodic functions. For our present purposes, because both the Dirichlet kernel KN(x) and our function f(x) are periodic, we have that l l fN(x) = ∫ KN(x-s) f(s) d s = (KN*f)( x) = (f*KN)(x) = ∫ KN(s) f(x-s) d s -l -l This latter form is a more convenient form to work with. The pointwise convergence theorem A function f(x) is said to be piecewise smooth on an interval [- l,l] if the function has at most a finite number of isolated discontinuities in that interval, and at each point where the function is discontinuous it has a finite limit on either side of the discontinuity. That is, lim f(s) = f(x-) s→x- lim f(s) = f(x+) s→x+ both exist and are finite. 3 We are now in a position to state Pointwise convergence theorem forforfor complex Fourier series If f(x) is a piecewise smooth periodic function defined on the interval [- l,l] then lim fN(x) = f(x) N→∞ whereever f(x) is continuous. At points where f(x) has a jump discontinuity, 1 lim fN(x) = (f(x-) + f(x+) ) N→∞ 2 Proof We will show a somewhat stronger pair of results. 0 1 lim ∫ KN(s)f(x-s) d s = f(x+) N→∞ -l 2 l 1 lim ∫ KN(s)f(x-s) d s = f(x-) N→∞ 0 2 both proofs are similar, so we will only show the proof of the second equality. To start with, we will use a fact about the Dirichlet kernel I mentioned above. l 1 ∫ KN(s) d s = 0 2 Using this gives us l 1 f(x-) = ∫ f(x-) KN(s) d s 2 0 Thus, to show that l 1 lim ∫ KN(s)f(x-s) d s = f(x-) N→∞ 0 2 we can instead prove the equivalent l lim ∫ KN(s)(f(x-s)- f(x-) ) d s = 0 N→∞ 0 Earlier I showed that sin (2 N+1) π s ( 2 l ) KN(s) = 2 l sin π s (2 l) Substituting this into the integral gives 4 sin (2 N+1) π s l ( ) lim ∫ 2 l (f(x-s)- f(x-) ) d s = 0 N→∞ 0 2 l sin π s (2 l) or l lim ∫ f(x-s)- f(x-) sin (2 N+1) π s d s = 0 N→∞ 0 2 l sin π s ( 2 l ) (2 l) Next, we introduce f(x-s)- f(x-) F(x)(s) = 2 l sin π s (2 l) To proceed beyond this point we are now going to need a pair of lemmas. Lemma 111 lim - df (x-s) +( ) lim F (s) = lim f(x-s)- f(x-) = s→0 dx + (x) + s→0 s→0 2 l sin π s lim π cos π s + (2 l) s→0 (2 l (2 l)) Even if x is a point of discontinuity, if we assume that f is piecewise smooth, then lim - df (x-s) + s→0 ( dx ) exists and is finite. Thus, lim F (s) = - df (x-) + (x) s→0 dx Lemma 222 (Bessel's Inequality) If {φN(s)} is a sequence of orthogonal functions defined on [0, l] then for all N and all functions F(s) we have ∞ 2 ∑ (F(s), φN(s)) ≤ (F(s), F(s)) N = 0 (φN(s), φN(s)) Here (,) is any inner product for our function space. In practice, this is usually the standard complex inner product l (F(s), φN(s)) = ∫ F(s) φN(s) d s 0 5 We now use these two lemmas to continue with the proof of our main result. We need to prove that l (2 N+1) π s lim ∫ F(x)(s) sin d s = 0 N→∞ 0 ( 2 l ) To prove this, we apply Bessel's inequality with F(s) = F(x)(s) and φN(s) = sin ((2 N+1) π s/2 l). The first thing to note here is that the sequence of functions (2 N+1) π s φN(s) = sin ( 2 l ) is in fact a sequence of orthogonal functions defined on the interval [0, l]. Now consider the inner product l 2 (F(s), F(s)) = (F(x)(s), F(x)(s)) = ∫ (F(x)(s)) d s 0 The only thing that could keep this integral from being finite is a singularity at s = 0. By lemma 1 above, lim F (s) = - df (x-) + (x) s→0 dx so there is no such singularity. Thus, the right hand side in the inequality ∞ (F(x)(s), φN(s)) ∑ ≤ (F(x)(s), F(x)(s)) N = 0 (φN(s), φN(s)) must be finite, and hence the sum on the left must converge. For that sum to converge, a necessary condition is that 2 (F (s), φ (s)) lim (x) N = 0 N→∞ (φN(s), φN(s)) Since 2 l (2 N+1) π s (φN(s), φN(s)) = ∫ sin d s 0 ( ( 2 l )) 1 (sin (2 N π) + 2 N π + π) l = 2 2 N π + π = l 2 saying that 2 (F (s), φ (s)) lim (x) N = 0 N→∞ (φN(s), φN(s)) 6 means that we must have lim (F(x)(s), φN(s)) = 0 N→∞ This translates into the condition that l (2 N+1) π s lim ∫ F(x)(s) sin d s = 0 N→∞ 0 ( 2 l ) and the result is proved.
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