Lighthill derivative of $f(x)=|x|^\alpha$

Question

I am reading Lighthill's book 'An introduction to Fourier analysis and generalized functions' and it is not clear to me what the author writes on the first page of chapter 3:

$$ \frac{\mathrm{d}}{\mathrm{d}x}|x|^\alpha = |x|^{\alpha-1} \text{sgn}(x) $$

The function $f(x)=|x|^\alpha$ ($\alpha>-1$), understood in the classical sense, is not differentiable in 0 when $\alpha <1$. If we understand it as a generalized function according to Lighthill's approach (definition 7 of his book), then its derivative exists and must be calculated in general by deriving the sequence of 'good functions' that defines the generalized function $f(x)$ (which I indicate with the same symbol). However he does a different thing, stating that the result of the derivative written above ($f'(x)=|x|^{\alpha-1} \text{sgn}(x)$) can be obtained by directly deriving the function $f(x)$ understood in the classical sense, and then using theorem 10.

My doubt is: the hypotheses of Theorem 10 do not hold in this case, since the derivative in the classical sense $f'(x)$ does not exist in 0, and therefore the function is not derivable in all points of its domain of definition , which is necessary for the application of the theorem.

Answer 1

The theorem in question reads as follows:

Theorem (Theorem 10, Lighthill). If $f(x)$ is an ordinary differentiable function such that both $f(x)$ and $f'(x)$ satisfy the condition of definition 7, then the derivative of the generalised function formed from $f(x)$ is the generalised function formed from $f'(x)$.

The definition used here is as follows:

Definition (Definition 7, Lighthill). If $f(x)$ is a function of $x$ in the ordinary sense, such that $(1 + x^2)^{-N} f(x)$ is absolutely integrable from $-\infty$ to $\infty$ for some $N$, then the generalised function $f(x)$ is defined by a sequence $f_n(x)$ such that for any good function $F(x)$ $$\lim_{n \to \infty} \int_{-\infty}^{\infty} f_n(x) F(x) dx = \int_{-\infty}^{\infty} f(x) F(x) dx .$$

It seems that Lighthill assumes knowledge about "ordinary functions" and "ordinary differentiable functions", but does not prescribe exact definitions to these. Given some of the examples and discussions in the text around and before the quoted material, it does seem that Lighthill allows for "ordinary differentiable functions" to be functions that have points of discontinuity, but whose derivatives can nevertheless be defined by differentiating from the left and right separately. The application of Theorem 10 is then to make rigorous sense of this "pasted" derivative (so to speak) as a generalised function. The notion of generalised function is made precise by Definition 7.

Therefore, while it is true that $f(x) := |x|^{\alpha}$ may be discontinuous at $x = 0$ for $\alpha > 0$, it seems that Lighthill is allowing $f'(x) := \alpha |x|^{\alpha - 1} \operatorname{sgn}(x)$ to be used as its derivative. Note that both, $f$ and $f'$ are well-defined functions on $\mathbb{R}$ and satisfy the condition that their respective products with $(1 + x^2)^{-N}$ for some $N \in \mathbb{N}$ gives an absolutely integrable function, so in this sense, they do satisfy the conditions of the quoted theorem (Theorem 10).