Suppose you have got a relentless operate f(x) = c. What’s the Fourier remodel of f?
We are going to present why the direct strategy doesn’t work, give two hand-wavy approaches, and a rigorous definition.
Direct strategy
Sadly there are a number of conventions for outlining the Fourier remodel.
For this submit, we are going to outline the Fourier remodel of a operate f to be
If f(x) = c then the integral diverges until c = 0.
Heuristic strategy
The extra concentrated a operate is within the time area, the extra it spreads out within the frequency area. And the extra unfold out a operate is within the time area, the extra concentrated it’s within the frequency area. For those who assume this sounds just like the Heisenberg uncertainty precept, you’re proper: there’s a connection.
A relentless operate is as unfold out as potential, so evidently its Fourier remodel must be as concentrated as potential, i.e. a delta operate. The delta operate isn’t actually a operate, however it may be made rigorous. Extra on that beneath.
Gaussian density strategy
The Fourier remodel of the Gaussian operate exp(−x²/2) is identical operate, i.e. the Gaussian operate is a set level of the Fourier remodel. Extra usually, the Fourier remodel of the density operate for a traditional random variable with commonplace deviation σ is the density operate for a traditional random variable with commonplace deviation 1/σ.
As σ will get bigger, the density turns into flatter. So we may consider our operate f(x) = c as some a number of of a Gaussian density within the restrict as σ goes to infinity. The Fourier remodel is then some a number of of a Gaussian density with σ = 0, i.e. some extent mass or delta operate.
Rigorous strategy
If f and φ are two well-behaved capabilities then
In different phrases, we will transfer the “hat” representing the Fourier remodel from one operate to the opposite. The equation above is a theorem when f and φ are good capabilities. We are able to use it to inspire a definition when the operate f is just not so good however the operate φ may be very good. Particularly, we are going to assume φ is an infinitely differentiable operate that goes to zero at infinity quicker than any polynomial.
Given a Lebesgue integrable operate f, we will consider f as a linear operator by way of the map
Extra usually, we will outline a distribution to be any steady [1] linear operator from the area of check capabilities to the advanced numbers. A distribution that may be outlined by integral as above is known as a common distribution. After we say we’re taking the Fourier remodel of the fixed operate f(x) = c, we’re really taking the Fourier remodel of the common distribution related to f. [2]
Not all distributions are common. The delta “operate” δ(x) is a distribution that acts on check capabilities by evaluating them at 0.
We outline the Fourier remodel of (the common distribution related to) a operate f to be the distribution whose motion on a check operate φ equals the integral of the product of f and the Fourier remodel of φ. When a operate is Lebesgue integrable, this definition matches the classical definition.
With this definition, we will calculate that the Fourier remodel of a relentless operate c equals
Observe that with a distinct conference for outlining the Fourier remodel, you would possibly get 2π c δ or simply c δ.
A bonus of the conference that we’re utilizing is that the Fourier remodel of the Fourier remodel of f(x) is f(−x) and never some a number of of f(−x). This suggests that the Fourier remodel of √2π δ is 1 and so the Fourier remodel of δ is 1/√2π.
Associated posts
[1] To outline continuity we have to put a topology on the area of check capabilities. That’s an excessive amount of for this submit.
[2] The fixed operate doesn’t have a finite integral, however its product with a check operate does as a result of check capabilities decay quickly. In truth, even the product of a polynomial with a check operate is integrable
