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Since $u'\in L^1(0,1)$, you find from the Lebesgue differentiation theorem that $$ \int_{1/2}^x u'(t) dt=u(x)+Cst,\quad x\in(0,1). $$ As a result $u$ is a continuous function and the constant above is …

The function $\vert x\vert^{\alpha-n}$ is radial homogeneous of degree $\alpha-n$, so its Fourier transform is radial homogeneous of degree $-(\alpha-n)-n=-\alpha$ (both locally integrable since $\alp …

So $\sigma=\sum_{1\le j\le n}\sigma_j(x)\frac{\partial}{\partial x_j}$ is a vector field with distributions coefficients $\sigma_j$ and divergence in $L^2$: $$ \sum_{1\le j\le n}\frac{\partial \sigma …

The answer is negative: take $f$ smooth compactly supported in $(-1/4,1/4)$ equal to 1 in $(-1/8,1/8)$, take $g(x) =f(x+1)$ so that $g$ is supported where $-1/4<x+1<1/4,$ i.e. $-5/4<x<-3/4$ so tha …

More a comment than an answer, but too long for a comment. First a comment on Michael Renardy's remark: there is no homogeneous function in $L^1(\mathbb R^N)$ so the first assumption is not satisfied. …

Let $n\ge 1$ be an integer and $s>n/2$. Then you have $H^s(\mathbb R^n)\subset L^\infty(\mathbb R^n)$ and for $f,g\in H^s(\mathbb R^n)$, $$ \Vert fg \Vert_{H^s(\mathbb R^n)}\le c_n\bigl(\Vert f \Vert_ …

Continuity is a local property: functions which are in your $L^{2,s}$ are locally in $L^2$, so functions in $H^{2,s}$ are locally in the Sobolev space $H^2(\mathbb R^3)$, thus are continuous functions …

I believe that the answer is yes in dimension $d\ge 2$, from the following result. Theorem. Let $\Omega$ be an open subset of $\mathbb R^d$, let $X$ be a Lipschitz vector field on $\Omega$ and let $u …

For $\epsilon >0$, $u\in H^{\frac{n}{2}+2\epsilon}$, $N_0=-\Delta+1$, $$ \Vert N_0^{\frac{n}{4}+\epsilon} u\Vert_{L^2}=\Vert u\Vert_{H^{\frac{n}{2}+2\epsilon}}\ge c_{n,\epsilon} \Vert u\Vert_{L^\inf …

Maybe just a long comment: if you want this property for any $f$ polynomial (or any smooth $f$), you will need $H^1$ to be an algebra, which is true only in 1D ($N=1$). More generally, $H^s$ is an al …

The problem is indeed coming form the fact that singular integrals, such as the Hilbert transform, although bounded on $L^p$ for $1<p<+\infty$ are failing to be bounded on $L^1$ or $L^\infty$. Howev …

(1) Let me answer first to the last question: $\Delta \vert x\vert$ is homogeneous of degree $-1$ and radial. On $\mathbb R^d$ ($d\ge 2$) it is $$ (\partial_r^2+\frac{d-1}{r}\partial_r)(r)=\frac{d-1}{ …

Let $p\in (1,+\infty)$ and $s\in \mathbb R$. For $\xi \in \mathbb R^n$, we define $ \langle\xi \rangle=(1+\vert \xi\vert^2)^{1/2} $ and accordingly the Fourier multiplier $\langle D \rangle^s$ as $$ …

Since your manifold is compact, $H^s_{loc}$ regularity will be equivalent to $H^s$ regularity. To check the local regularity, you can use cutoff functions and the coordinate charts.

Up to some normalization, the Fourier transform of the characteristic function $\mathbf 1_I$ of a compact interval $I$ is $$ \widehat{\mathbf 1_I}(\xi)=\frac{\sin \xi}{\xi}. $$ Obviously the function …