I'm trying to follow a result in a paper and I can't get it to work out.
The result requires deriving the bound
\begin{equation} \int_{\mathbb{R}^N\times\mathbb{R}_+} \left|\frac{\mathrm{d}\psi}{\mathrm{d}t}\right|^p \psi^{1-p}\mathrm{d}x\mathrm{d}t \leq CR^{N+k-pk}, \end{equation}
where we have:
- $\psi:\mathbb{R}^N\times\mathbb{R}_+ \to \mathbb{R}$,
- $\psi(x,t) = \phi(\frac{t^{2/k} + |x|^2}{R^2})$,
- $\phi \in C^{\infty}_0(\mathbb{R})$,
- $\phi(s) = 1$ for $s<1$ and $\phi(s) = 0$ for $s>2$,
- $|\phi'(s)| \leq C\phi^{1/q}(s)$ with $q$ such that $p = q/(q-1)$,
For $C>0$ denoting any (not necessarily the same) constant, and $R, k >0$ and $p, q>1$ fixed positive constants.
It is straightforward to get the original integral is equal to
\begin{equation} \int_{\mathbb{R}^N\times\mathbb{R}_+} |\phi'|^p |\frac{2}{kR^2}t^{(2/k)-1}|^p\phi^{1-p}\mathrm{d}x\mathrm{d}t \leq CR^{-2p}\int_{\mathbb{R}^N\times\mathbb{R}_+} \phi^{p/q}\phi^{1-p} t^{(2p/k)-p}\mathrm{d}x\mathrm{d}t. \end{equation}
Now, $\phi^{p/q}\phi^{1-p} = \phi^0$ and so is equal to zero for $|x|^2 > 2R^2 - t^{2/k}$, and equal to one otherwise. This allows us to evaluate the spatial part of the latter integral in radial coordinates as \begin{equation} C\int_0^{\sqrt{2R^2 - t^{2/k}}} r^{N-1}\mathrm{d}r = C(2R^2 - t^{2/k})^{N/2} \end{equation}
and so we're left with a time integral of the form \begin{equation} CR^{-2p}\int_0^{(\sqrt{2}R)^k} (2R^2 - t^{2/k})^{N/2}t^{(2p/k)-p}\mathrm{d}t. \end{equation}
This is where I get stuck - the obvious substitution would work were it not for that pesky factor of $p$ and that's where I run out of ideas. I feel like given the rather simple form of the given bound it shouldn't be that hard and there may be a flaw in my reasoning somewhere, most likely in the way I've treated the $\phi^{p/q}\phi^{1-p}$ and ensuing spatial integral, but I can't see it at the moment. Alternatively I'm missing an obvious trick and the integral can be easily evaluated or at least bounded in some way. Any input would be appreciated.
Well, I figured it out in case anyone is interested. Shouldn't really have taken me as long as it did, probably, given where I was.
It turns out the 'pesky' factor of $p$ preventing a nice clean integral evaluation can be dealt with using good old Cauchy-Schwarz. The trick is to rewrite the last integral as
\begin{equation} CR^{-2p}\int_0^{(\sqrt{2}R)^k} (2R^2 - t^{2/k})^{N/2}t^{(1/k)-(1/2)}t^{(2p-1)((1/k)-(1/2))}\mathrm{d}t, \end{equation}
and then applying Cauchy-Schwarz separating the terms like $(2R^2 - t^{2/k})^{N/2}t^{(1/k)-(1/2)}$ and $t^{(2p-1)((1/k)-(1/2))}$ we can bound this by
\begin{equation} CR^{-2p}\left(\int_0^{(\sqrt{2}R)^k} (2R^2 - t^{2/k})^{N}t^{(2/k)-1}\mathrm{d}t\right)^{1/2}\left(\int_0^{(\sqrt{2}R)^k} t^{(2p-1)((2/k)-1)}\mathrm{d}t\right)^{1/2}. \end{equation}
We are now able to evaluate the first of the terms in parentheses with a simple substitution to get, on taking the square root, $CR^{N+1}$and the second directly to get (again after taking the square root) $CR^{2p-1-pk + k}$. The result then follows immediately.