Problem Set 4. GeneralizedProblem functions.set 3
MathMath 212a 212a OctoberSept. 30, 2, 2014,2008 due Due Oct. Oct. 15 1
The purpose of this exercise set, which is mainly computational, is to give you some experience with the ideas associated to generalized functions. Recall that in our study of the Fourier transform we introduced the space consisting of all functions on R which have derivatives of all order which vanishS rapidly at infinity, and that on this space we have a countable family of norms
m n f m,n := x D f , where m and n are non-negative integers. Convergence in means convergence with respect to all of these norms. So a linear functionS on this space is continuous if (fk) 0 whenever fk m,n 0 for every m and n.Welet denote this dual| space,| the space of continuous linear functions on . S For example, consider the “Heaviside function” S
1 for x>0 H(x):= . 0 for x 0 This function does not belong to , but it defines a continuous linear function on by S S H, f = f(x)H(x)dx = f(x)dx. 0 Of course, any element of certainly defines a linear function on by the same procedure, the linear functionS associated to g is just the HilbertS space scalar product S f (f,g)= f(x)g(x)dx. This also works for any g L2(R), but H L2(R). From the Riesz representa- tion theorem we thus know that the linear function given by H is not continuous with respect to the L2 norm, but it is continuous relative to the topology given above by the countable family of norms . m,n Contents
1Di erentiation of generalized functions. 2
1 1 2 Weak convergence. 4 2.1 Di⌅erentiating under the limit...... 5
3 Extending the domain of definition. 6
4 Other test spaces. 6
5 Cauchy principal value. 8
6 Anti-derivatives of elements of 0.8 D 7 The tensor product of two generalized functions. 9
8 The wave equation in one dimension. 12
9 Fractional integration. 13
1Di erentiation of generalized functions.
The operation of di⌅erentiation d d : ,f f 0 = f dx S S dx is a continuous linear operator. Hence it has a well defined transpose
d ⇤ d ⇤ d : 0 0 ,f := , f . dx S S dx dx ✓ ◆ ⌧✓ ◆ ⌧ ✓ ◆ For example, if we consider the linear function associated to an element g of , transpose is given by S
f (f 0,g)= f 0(x)g(x)dx = f(x)g (x)dx = (f,g0) 0 Z Z d by integration by parts. Thus the transpose of dx when applied to the linear function associated to g is the linear function associated to g .This 0 suggests making the following S definition: We define S d : 0 0 dx S S ✓ ◆ by d d ⇤ := . dx dx ✓ ◆ ✓ ◆ For example,
d 1 H, f = H, f0 = f 0(x)dx = f(0). dx ⇣ ⌘ ⌧✓ ◆ Z0
2 So if we define the Dirac delta “function” = 0 by 0 S ,f := f(0) ⇣ ⌘ We have d H = . (1) dx We have learned how to di⌅erentiate discontinuous functions! When Heaviside wrote down formulas like this at the beginning of the twentieth century he was derided by mathematicians. When Dirac did the same, mathematicians were puzzled, but a bit more respectful because of Dirac’s fantastic achievements in physics. The advent of theorems like the Riesz representation theorem gave rise to the idea that we might want to think of a function as a functional, i.e. as a linear function on a suitable space of functions. The choices of the appropriate spaces of “test functions” such as (there are others which are convenient for di⌅erent purposes) with their topologiesS and the whole subject of “generalized functions” or “distributions” was developed by Laurent Schwartz around 1950.
1. What is the k-th derivative of the Dirac delta function?
We may generalize the Heaviside function by considering the function
x⇤ for x>0 x⇤ := . (2) + 0 for x 0 ⇢ ⌃ As a function of x, this is defined for all complex values of ⇧ by the rule
x⇤ = e⇤ log x.
For ⇧ = 0 we get H(x). For Re⇧ > 1 the integral
1 ⇤ f(x)x+dx Z0 ⇤ converges for all f and so x+ defines a continuous linear function on . (I am following the standard S convention here and not using a complex conjugation;S equally well, just consider ⇧ real.) So for these values of ⇧, we can consider ⇤ x+ as defined by (2) as an element of 0. We never have any trouble with the convergence of the above integral at infinity.S But we have convergence problems at 0 when Re ⇧ 1. We will therefore have to modify the definition (2) for these values of ⇧⌃ . Notice that for Re ⇧ > 0 we have
d ⇤ ⇤ 1 (x )=⇧x (3) dx + + both as functions and as elements of 0. S
3 2. Compute the derivative d (x⇤ ) for 1 < Re ⇧ < 0. In fact, show that dx + ⇤ 1 ⇤ 1 (x )0,f = ⇧x [f(x) f(0)]dx. ⇣ + ⌘ Z0 ⇤ ⇤ [Hint: Write 1 x f 0(x)dx as lim 0 1 x f 0(x)dx and do an integration by 0 ! parts with f (x)dx = du, u = f(x)+C, and v = x⇤. Choose C appropriately 0 R R so as to be able to evaluate the limit.]
Notice that the function f(x) f(0) will not belong to the space if f(0) = 0. ⇤ 1 S What makes the above integral converge at infinity is that x vanishes rapidly enough at infinity if Re ⇧ < 0. This suggests the following strategy: For Re ⇧ > 1 we have 1 1 x⇤dx = . ⇧ +1 Z0 So we can write 1 1 f(0) x⇤ ,f = x⇤ [f(x) f(0)]dx + x⇤f(x)dx + . (4) ⇣ + ⌘ + ⇧ +1 Z0 Z1 Notice that the right hand side of this equation makes sense for all ⇧ such that Re ⇧ > 2 with the exception of the single point ⇧ = 1. (In the language of ⇤ complex variable theory we would say that x+ is a meromorphic function of ⇧ with a pole at 1 with residue .) We will therefore take the right hand side of (4) as a new definition of x⇤ valid for all ⇧ = 1 such that Re ⇧ > 2. + 3. Show that in the range 2 < Re ⇧ < 1 we can write the above expression as 1 x⇤[f(x) f(0)]dx. Z0 Conclude that we can write the result of Problem 2 as asserting the validity of (3) on the range Re ⇧ > 1.
⇤ 4. Obtain a formula for x+ which extends its range into the region Re ⇧ > n 1, ⇧ = 1, 2, , n ··· for any positive integer n, and obtain a simpler formula valid on the range n 1 < Re ⇧ < n to conclude the validity of (3) on the strip n