# Is $(1-x)^{\alpha} \log(1-x)$ a Sobolev function?

Let $$f(x) = (1-x)^{\alpha} \log(1-x)$$ be defined on $$[0,1]$$ with $$\alpha > 0$$ some real exponent.

Does such a function belong to the Sobolev (or Bessel potential) space $$H^{\beta}((0,1))$$ with $$\beta < \alpha + 1/2$$ ?

If $$\alpha, \beta$$ are non-negative integers, then $$H^{\beta}((0,1)) = W^{\beta,2}((0,1))$$ and using Leibniz rule, I find that

$$f^{(k)}(x) \approx (1-x)^{\alpha-k}\log(1-x) + (1-x)^{\alpha-k+1}$$

A primitive of $$f^{(k)}(x)^2$$ is found using integration by parts :

$$\int f^{(k)}(x)^2 dx \approx (1-x)^{2\alpha-2k+3} + (1-x)^{2\alpha-2k+1} \left( \log(1-x)^2 + \log(1-x) + 1\right)$$

so integrating on $$(0,1)$$, this will be finite given $$2\alpha-2k+1 > 0$$ and this must be true for $$k = 0, 1, ..., \beta$$. This is satisfied for $$\beta < \alpha + 1/2$$.

Is there an easy way to generalize this for real-valued $$\alpha$$ and $$\beta$$ ? Extension operators and Fourier transforms will be involved and it seems a bit tricky. Maybe there is some way using some embeddings or interpolation results ?

• An idea, which does not directly use Besov spaces as in the book cited in the answer below, is to write $f(x) = \sum_{j \leq 0} \phi(2^{-j}(1-x)) f(x)$, where $\phi( \cdot )$ defines a dyadic partition of unity (as in the Littlewood-Paley decomposition). The sum is locally finite, so we can interchange derivative and infinite sum. Each term in the sum is smooth. After differentiating, we can square-integrate and estimate this. To get estimates on the $L^2$ norms of fractional derivatives, we use interpolation. Nov 28, 2021 at 13:13

They do the case when the singularity is at $$x=0$$ and not at $$x=1$$, but you can reduce to that case with a change of variables.