Mathematical Tools for Physics - Department of Physics - University

17—Densities and Distributions 413

I’ve arranged it so that the integral ofφnover allxis one. If I want the value offatx 0 I can

do an integral and take a limit.

F[φn]=

∫∞

−∞

dxf(x)φn(x) =

∫∞

−∞

dxf(x)

√

n

π

e−n(x−x^0 )

2

Asnincreases to infinity, all that matters forfis its value atx 0. The integral becomes

lim

n→∞

∫∞

−∞

dxf(x 0 )

√

n

π

e−n(x−x^0 )

2

=f(x 0 ) (17.12)

This means that I can reconstruct the functionfif I know everything about the functionalF. To get

the value of the derivativef′(x 0 )instead, simply use the function−φ′nand take the same limit. This

construction is another way to look at the functional derivative. The equations (17.10) and (17.12)
produce the same answer.
You say that this doesn’t sound very practical? That it is an awfully difficult and roundabout
way to do something? Not really. It goes back to the ideas of section17.1. To define a density you
have to know how much mass is contained in an arbitrarily specified volume. That’s a functional. It
didn’t look like a functional there, but it is. You just have to rephrase it to see that it’s the same thing.

∫ As before, do the case of linear mass density. λ(x)is the density, and the functionalF[φ]=

∞

−∞dxλ(x)φ(x). Then as in Eq. (17.4),m

(

[x 1 ,x 2 ]

)

is that mass contained in the interval fromx 1

tox 2 and you can in turn write it as a functional.

Letχbe the step function χ(x) =

{

1 (x 1 ≤x≤x 2 )

0 (otherwise)

}

then

∫x 2

x 1

dxλ(x) =F[χ]=m

(

[x 1 ,x 2 ]

)

What happens if the functionfis itself not differentiable? I can still define the original functional.

Then I’ll see what implications I can draw from the functional itself. The first and most important

example of this is forfa step function.

f(x) =θ(x) =

{

1 (x≥ 0 )

0 (x < 0 ) F[φ]=

∫∞

−∞

dxθ(x)φ(x) (17.13)

θhas a step at the origin, so of course it’s not differentiable there, but if it were possible in some way

to define its derivative, then when you look at its related functional, it should give the answer−F[φ′]

as in Eq. (17.11), just as for any other function. What then is−F[φ′]?

−F[φ′]=−

∫∞

−∞

dxθ(x)φ′(x) =−

∫∞

0

dxφ′(x) =−φ(x)

∣

∣∣∞

0

=φ(0) (17.14)

This defines a perfectly respectable linear functional. Input the functionφand output the value of the

function at zero. It easily satisfies Eq. (17.6), but there is no functionfthat when integrated against

φwill yield this result. Still, it is so useful to be able to do these manipulations as if such a function

exists that the notation of a “delta function” was invented. This is where the idea of a generalized
function enters.