7
$\begingroup$

If we consider the Cantor staircase function, let us say $f:[0,1]\to\mathbb{R}$, as a distribution, I was wondering whether there is an explicit way to express its generalised derivative as a distribution, which is defined by$$\varphi\mapsto-\int_{-\infty}^{+\infty}f(x)\varphi'(x)dx$$where $\varphi$ is a fundamental function, i.e. an infinitely differentiable function equal to 0 outside a finite interval.

I am not sure whether there is a "more direct" way to express it and I have not been able to find one -I already find it quite difficult to handle the definition of $f$, which I find quite clear here-, but I think it is an interesting issue. Thank you very much for any contribution!

$\endgroup$
1
  • $\begingroup$ The answer below is too confusing. Can anybody write a concrete formula for the functional? Is it a sum of the function values at the endpoints of the Cantor set taken with certain weights? $\endgroup$
    – Tyrell
    Feb 17, 2023 at 12:44

1 Answer 1

6
$\begingroup$

The distributional derivative of any increasing function is a positive measure on the real line. The Wikipedia article hints at what this measure is for the Cantor function $f$:

[the Cantor function] is the cumulative probability distribution function of a random variable that is uniformly distributed on the Cantor set. This distribution, called the Cantor distribution, has no discrete part. That is, the corresponding measure is atomless.

The measure $f'$ can be described as the pushforward of the Lebesgue measure under the map described in terms of binary and ternary expansions as: $$0.01100010111_2\mapsto 0.02200020222_3$$ Alternatively, the measure can be obtained as the limit of the natural measures on pre-Cantor sets: just restrict the Lebesgue measure to pre-Cantor set $C_n$ and normalize it so that the total measure is $1$. Let's call the resulting measure $\mu_n$. The weak* limit of $\mu_n$ exists and is the distributional derivative of the Cantor function.

And here is why. Let $f_n$ be the cumulative distribution function of $\mu_n$. Then $f_n$ is continuous and piecewise linear; in particular it's a Lipschitz function. Hence, its distributional derivative is its pointwise derivative, which is [the density function of] $\mu_n$. Observe that $f_n$ converge uniformly to the Cantor function $f$. Uniform convergence implies convergence in the sense of distributions. And when $f_n\to f$ distributionally, it follows that $f_n'\to f'$ distributionally.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .