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Suppose we operate on the unit simplex $\Delta \subset \mathbb{R}^d$ with $0$ as a corner point.

Define the integral

$$ Iu(x):=\int_0^1 t^{|\beta|-1}x^\beta u(tx)\mathrm{d}t,\quad x\in \Delta $$ and $\beta \in \mathbb{N}^d$ a multiindex.

Question: Do we have the inequality \begin{equation} \left\|\frac{\partial}{\partial x_i } Iu \right\| \le C \|u\\| \end{equation} for some constant $C>0$ and $\|\cdot \|$ meaning the $L^2(\Delta)$-norm?

The integral $I$ arises for instance as remainder term in a Taylor expansion. I found (1) easy to prove for $d=1$ (simply substitute $\tau = tx$), for higher dimensions I tried polar coordinates but to no avail.

Edit My guess is that the answer is no, since $I$ only smoothes along rays emanating from zero...

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  • $\begingroup$ Could you edit the tex? Is $u$ an arbitrary $L^2$ function? $\endgroup$
    – András Bátkai
    Commented Mar 30, 2012 at 9:26
  • $\begingroup$ I edited the tex. There seems to be something wrong with the interpreter. Yes, $u$ can be arbitrary in $L^2$. $\endgroup$
    – pil
    Commented Mar 30, 2012 at 9:35
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    $\begingroup$ Your suspicion is right, take any function which depends only on $x/|x|$, it comes out of the integral $\endgroup$
    – Piero D'Ancona
    Commented Mar 30, 2012 at 11:50
  • $\begingroup$ yes, sorry for the stupid question... $\endgroup$
    – pil
    Commented Mar 31, 2012 at 11:15

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