TITLE: Properties of Riesz transform
QUESTION [1 upvotes]: I am studying something about the Riesz transform. The wikipedia page about Riesz transform https://en.wikipedia.org/wiki/Riesz_transform has something not clear for me. In particular, i refer to the section titled "Relationship with the Laplacian".
The first one thing that is not clear for me is why it says that the Riesz transform of $ f $ gives the first partial derivatives of a solution of the equation
\begin{equation}
\left( -\Delta\right)^{\frac{1}{2}} u = f.
\end{equation}
I know that the fractional laplacian can be defined in this way
$$\left(-\Delta\right)^{s} u\left(x\right) = c_{n} P.V. \int_{R^{n}} \dfrac{u\left(x\right) - u\left(y\right)}{\vert x -y\vert^{n + 2s}} dy $$
and the Riesz transform of a function $ f $ in the Schwartz space is defined as
$$ R_{j} f\left(x\right) = C_{n} P.V. \int_{R^{n}} \dfrac{y_{j}}{\vert y\vert^{n+1}} f\left(x-y\right) dy $$
but i'm not able to justify the equation
$$\left( -\Delta\right)^{\frac{1}{2}} u = f.$$
The second thing that is not clear for me is the relation
\begin{equation}
R_{i}R_{j} \Delta u = -\dfrac{\partial^{2} u}{\partial x_{i}\partial x_{j}}.
\end{equation}
I have no idea how to verify it.
Could anyone please help me?
Thank you!

REPLY [1 votes]: One usually gets this via the $L^2$ results.  Setting all constants to be 1, and allowing all functions  to be Schwartz, the sketch is as follows.
First, you can check that we have the equalities for the Fourier Transforms,
$$\mathcal F (-\Delta)^{1/2} u = |\xi|\hat u,\\
\mathcal F R u = \frac{-i\xi}{|\xi|} \hat u $$
i.e. these two operators are multiplier operators. Such operators automatically commute with derivatives(and other multipliers), and in fact
$$ \mathcal F (R_j (-\Delta)^{1/2} u)  = -i\xi \hat u = \mathcal F(\partial _j u).$$
This implies the first result you're asking about, and by using it twice, also the second result.
The first Fourier transform above can be found in this well-written expository paper. The second is similarly "well-known" and can be found in e.g. Stein's book "Singular Integrals and Differentiability Properties of Functions", or for instance, this PDF file.


TITLE: How would you show that a non-cyclic group of order 10 must contain an element of order 5?
QUESTION [2 upvotes]: How would you show that a non-cyclic group of order 10 must contain an element $r$ of order 5?
Also, is every pair of groups homomorphic?
Thanks

REPLY [5 votes]: Let $G$ be a group of order $10$ with no element of order $5$. By Lagrange's theorem, all elements would have order $2$ (except $e$) and so $G$ would be abelian. Let $a, b \in G$ of order $2$. Then $\{e,a,b,ab\}$ would be a subgroup of order $4$, which cannot happen because $4$ does not divide $10$.