f(s) e -i n π s/l d s

Transcription

1 Pointwise convergence of complex Fourier series Let f(x) be a periodic function with period l defined on the interval [,l]. The complex Fourier coefficients of f( x) are This leads to a Fourier series representation for f(x) We have two important questions to pose here. c n = 1 l f(s) e -i n π s/l d s l f(x) = c n e i n π x/l n = - 1. For a given x, does the infinite series converge?. 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 In terms of this function, our two questions become 1. For a given x, does f (x) exist?. If it does, is f (x) = f(x)? The Dirichlet kernel f (x) = c n e i n π x/l n = - To begin to address the questions we posed about f (x) we will start by rewriting f (x). Initially, f (x) is defined by f (x) = c n e i n π x/l n = - If we substitute the expression for the Fourier coefficients into the expression for f (x) we obtain c n = 1 l f(s) e -i n π s/l d s l f (x) = 1 l f(s) e ( -i n π s/l d s e i n π x/l n = - l ) = l 1 (e -i n π s/l e i n π x/l ) f(s) d s ( l n = - ) 1

2 = l 1 e ( i n π (x-s ) /l f(s) d s l n = - ) The expression in parentheses leads us to make the following definition. The Dirichlet kernel is the function defined as K (x) = 1 l e i n π x/l n = - In terms of the Dirichlet kernel, we can write the expression for f (x) as Some properties of the Dirichlet kernel f (x) = l K (x-s)f(s) d s By rewriting the expression for the Dirichlet kernel, we can recogize that the Dirichlet kernel is actually a geometric series. K (x) = 1 e i n π x/l = 1 (e i π x/l ) n l n = - l n = - Because this is a geometric series, it can be summed explicitly. K (x) = 1 (e i π x/l ) n = 1 (e i π x/l ) +1 - (e i π x/l ) - l n = - l e i π x/l - 1 = 1 (e i π x/l ) +1/ - (e i π x/l ) -(+1/) l (e i π x/l ) 1/ - (e i π x/l ) -1/ = 1 l (e i π x/l ) +1/ - (e i π x/l ) -(+1/) e i π x/(l) - e-i π x/(l) sin (+1) π x ( = l ) l sin π x ( l ) Some explicit integrations show that Convolutions The integral we saw earlier 0 K (x) d x = l K (x) d x = 1 0

3 f (x) = l K (x-s)f(s) d s is an example of what is known as a convolution integral. Specifically, if g(x) and h(x) are two periodic functions with period l defined on [,l] the convolution of g and h is defined by (g*h)(x) = l g(x-s) h(s) d s Here is an important property of convolution integrals. From the definition we have that (g*h)(x) = l g(x-s) h(s) d s If we introduce a change of variables z = x - s in the integral, the integral becomes x x+l g(z)h(x-z)(-1) d z = x+l g(z)h(x-z) d z 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 x Replacing the variable z with s in the final integral gives x g(z)h(x-z) d z = l g(z)h(x-z) d z (g*h)(x) = l g(x-s) h(s) d s = l g(s)h(x-s) d s = (h*g ) (x) This is an important symmetry property of the convolution of periodic functions. For our present purposes, because both the Dirichlet kernel K (x) and our function f(x) are periodic, we have that f (x) = l K (x-s) f(s) d s = (K *f)(x) = (f*k ) (x) = l This latter form is a more convenient form to work with. The pointwise convergence theorem K (s) f(x-s) d s A function f(x) is said to be piecewise smooth on an interval [,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 it on either side of the discontinuity. That is, both exist and are finite. f(s) = f(x-) - s x f(s) = f(x+) s x + 3

4 We are now in a position to state Pointwise convergence theorem for complex Fourier series If f(x) is a piecewise smooth periodic function defined on the interval [,l] then f (x) = f(x) whereever f(x) is continuous. At points where f(x) has a jump discontinuity, Proof We will show a somewhat stronger pair of results. f (x) = 1 ( f(x-) + f(x+)) 0 K (s)f(x-s) d s = 1 f(x+) l K (s)f(x-s) d s = 1 f(x-) 0 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. Using this gives us Thus, to show that we can instead prove the equivalent Earlier I showed that l 0 K (s) d s = 1 1 f(x-) = l f(x-) K (s) d s 0 l K (s)f(x-s) d s = 1 f(x-) 0 l K (s)(f(x-s) - f(x-)) d s = 0 0 sin (+1) π s ( K (s) = l ) l sin π s ( l) Substituting this into the integral gives 4

5 l 0 sin (+1) π s ( l ) ( f(x-s) - f(x-)) d s = 0 l sin π s ( l ) or l 0 f(x-s) - f(x-) sin( l sin π s ( l ) (+1) π l s) d s = 0 ext, we introduce F (x) (s) = f(x-s) - f(x-) l sin π s ( l ) To proceed beyond this point we are now going to need a pair of lemmas. Lemma 1 s 0 +F (x)(s) = f(x-s) - f(x-) = s 0 + ( - df dx (x-s) ) s 0 + l sin π s π cos( π s ( l) s 0 + ( l l)) Even if x is a point of discontinuity, if we assume that f is piecewise smooth, then exists and is finite. Thus, Lemma (Bessel's Inequality) s 0 + ( - df dx (x-s) ) s 0 +F (x)(s) = - df dx (x-) If {φ (s)} is a sequence of orthogonal functions defined on [0,l] then for all and all functions F(s) we have Here = 0 (F(s), φ (s)) (F(s),F(s)) (φ (s), φ (s)) is any inner product for our function space. In practice, this is usually the standard complex inner product (, ) (F(s), φ (s)) = l 0 F(s) φ (s) d s 5

6 We now use these two lemmas to continue with the proof of our main result. We need to prove that l 0 F (x) (s) sin (+1) π ( l s) d s = 0 To prove this, we apply Bessel's inequality with F(s) = F (x) (s) and φ (s) = sin((+1) π s/ l ). The first thing to note here is that the sequence of functions φ (s) = sin (+1) π s ( l ) is in fact a sequence of orthogonal functions defined on the interval [0,l]. ow consider the inner product (F(s),F(s)) = (F (x) (s),f (x) (s)) = l 0 (F (x)(s) ) d s The only thing that could keep this integral from being finite is a singularity at s = 0. By lemma 1 above, s 0 +F (x)(s) = - df dx (x-) so there is no such singularity. Thus, the right hand side in the inequality = 0 (F (x) (s), φ (s)) (F(x) (s), F (x) (s)) (φ (s), φ (s)) must be finite, and hence the sum on the left must converge. For that sum to converge, a necessary condition is that Since (F (x) (s), φ (s)) = 0 (φ (s), φ (s)) (φ (s), φ (s)) = l 0 ( sin (+1) π s ( l )) d s 1 ( sin( π) + π + π) l = π + π = l saying that (F (x) (s), φ (s)) = 0 (φ (s), φ (s)) 6

7 means that we must have (F (x)(s), φ (s)) = 0 This translates into the condition that and the result is proved. l 0 F (x) (s) sin (+1) π ( l s) d s = 0 7