Indeed, the family is still orthogonal if we consider a complex valued inner product ( or an inner product on complex vector space)

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1 Fourier series of complex valued functions Suppose now f is a piecewise continuous complex valued function on [, π], that is f(x) = u(x)+iv(x) such that both u and v are real valued piecewise continuous functions on [, π]. Of course one could compute Fourier series for both u and v before adding it up. However, it is more customary to use the following expansion: f(x) = c k e ikx with c k = 1 k Z 2π f(x)e ikx dx. Note that it is not true that the family {e ikx : k Z} is orthogonal (in our old inner product). But it has similar property: e imx e inx dx = when m + n. Indeed, the family is still orthogonal if we consider a complex valued inner product ( or an inner product on complex vector space) (f, g) = Note that we now have: f(x)g(x)dx if f and g are complex valued functions. It follows from this observation that by formal calculation, we have the formula for c k. It is customary to write ˆf(k) instead of c k and it is usually being called the Fourier coefficients of the function f. 45

2 46 In case f is only real valued, then note that c k = c k and hence its Fourier series expansion is again real valued and just the same as before. Indeed, we are still interested at real valued functions most of the time. Some advantages for using this notations: Convolution of functions and their Fourier coefficients: let f and g are both periodic piecewise continuous functions on [, π]. Then we define its convolution as Then f g(n) = ˆf(n)ĝ(n). f g(x) = 1 f(x y)g(y)dy. 2π

3 47 Examples of convolutions: f D n (x) = 1 2π f(y) n k= n e ik(x y) dx = n k= n ˆf(k)e ikx. Some other properties of convolutions: (1) f g = g f. (2) (f g) h = f (g h). (3) (αf 1 + f 2 ) g = α(f 1 g) + f 2 g. (4) f g is a continuous function. Fact to be used: If f is a periodic piecewise continuous function on [, π] (of period 2π), then there exists a sequence {f k } of continuous periodic functions (of period 2π) such that lim k f k (x) f(x) dx = and sup f k (x) sup f(x). x [,π] x [,π] Remark A proof for the case where the functions are Riemann integrable periodic functions can be found in the main reference book: Fourier Analysis, Section 2. Note that its proof are similar to ours except that it uses a stronger fact. A useful lemma. If f is a periodic function of period L and f is piecewise continuous on [, L], then Note that it is clear that f(x + y)dy = f(x y)dy = f(y)dy = f(y)dy for any x R, /2 L/2 f(y)dy for any x R, ( ). ( ). Proof of Lemma. Let z = x y. Then f(x y)dy = ( ) is now clear by letting t = z + L x. x L x f(z)dz = x x L f(z)dz.

4 48 Proof of Properties of convolution. Fact (1) follows from change of variable, indeed, x x+π f(x y)g(y)dy = f(z)g(x z)dz = f(z)g(x z)dz = f(z)g(x z)dz. x+π x Fact (3) is obvious. Fact (2) clearly follows from Fubini s theorem if we are given that f, g, h are all continuous. Here is the Fubini s theorem for a given continuous function F on [a, b] [c, d]: b d a c F (x, y)dydx = d b c a F (x, y)dxdy. Indeed, f(y)g(x y)dyh(z x)dx = f(y) g(x y)h(z x)dxdy = f(y) g(t)h(z y t)dtdy. Fact (4) is also clear when f is continuous. Indeed, let g(y) M. Then f g(x 1 ) f g(x 2 ) = 1 2π 1 y) f(x 2 y)]g(y)dy [f(x M 2π f(x 1 y) f(x 2 y) dy = M f(x 1 x 2 y) f( y) dy. 2π However, since f is uniformly continuous, given any ε >, there exists δ > such that f(z 2 ) f(z 1 ) < ε whenver z 2 z 1 < δ. Hence f g(x 1 ) f g(x 2 ) < Mε, whenver x 2 x 1 < δ. It is now clear that f g is continuous. Now suppose f is only piecewise continuous, let {f k } be a sequence of continuous periodic functions (of period 2π) such that lim k f k (x) f(x) dx = and sup f k (x) sup f(x). x [,π] x [,π]

5 Next, again, let g(y) M. Then f k g(x) f g(x) 1 2π f k (x y) f(x y) g(y) dy M 2π 49 f k (x y) f(x y) dy = M f k (y) f(y) dy. 2π It is now clear that f k g f g uniformly on [, π]. Since {f k g} is a sequence of continuous functions, f g is also continuous. Let us now complete the proof of part (2). Again, for any piecewise continuous function f, let us choose {f k } as above. Then if h and g are both continuous, we have f k (g h) = (f k g) h Moreover, we know from the previous argument that f k (g h) converges to f (g h) uniformly. Also, f k g f g uniformly and hence (f k g) h = (f g) h. Hence (2) holds if g and h are both continuous. Next, choose a sequence of continuous functions {g k } as above and use the fact from the previous case. Finally choose a sequence of continuous functions {h k } as above and we can complete the proof of (2) by similar argument.

6 5 We now show that f g(n) = ˆf(n)ĝ(n). First, if both f and g are continuous, we can apply Fubini s theorem to get Next note that Hence if f g(n) = = 1 π (2π) 2 1 π (2π) 2 = ˆf(n)ĝ(n). f(x y)g(y)dye inx dx f(x y)e in(x y) dxg(y)e iny dy ˆf k (n) ˆf(n) 1 f k (x) f(x) e inx dx. (2π) f k (x) f(x) dx, then ˆf k (n) ˆf(n). Thus, if f and g are both piecewise continuous functions, we only need to choose sequences of continuous functions {f k }, {g k } such that lim k f k (x) f(x) dx = lim k g k (x) g(x) dx =. Then ˆf(n)ĝ(n) = lim ˆfk (n)ĝ k (n) = lim f k g k (n). k k However, by similar approach, we can show that lim f k g k (n) = f g(n) EXERCISE. k