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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 ?

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  • $\begingroup$ 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. $\endgroup$
    – Desura
    Nov 28, 2021 at 13:13

1 Answer 1

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Yes, you can find the proof in Lemma 1 on page 44 of the book Runst, Thomas and Sickel, Winfried. Sobolev Spaces of Fractional Order, Nemytskij Operators, and Nonlinear Partial Differential Equations, Berlin, New York: De Gruyter, 2011. https://doi.org/10.1515/9783110812411

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.

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