Let $p, q\geq 2$, $s\geq p$ and $f,g$ be non-negative smooth enough functions. Then why does the following inequality hold: $$-f^{q-2}g^{s}|\nabla f|^{p}+f^{q-1}g^{s-1}|\nabla f|^{p-1}|\nabla g|\leq C(s, q)(-|\nabla (f^{\frac{p+q-2}{p}})|^{p}g^{s}+|\nabla g|^{p}g^{s-p}f^{p+q-2}),$$ for some constant depending $C(s, q)$ on $q$ and $s$?

In the source [1] (Proof of Lemma 3.1) I have, it is said that this follows from Young's inequality, but I do not know with which exponents and applied to which function.

Any help is appreciated!

[1] (S.A.J. Dekkers "Finite propagation speed for solutions of the parabolic p-Laplace equation on manifolds" Comm. in Analysis and Geometry, 13 (2005), no.4, 741-768)


1 Answer 1


This inequality cannot hold in general. Indeed, \begin{equation*} |\nabla(f^{\frac{p+q-2}{p}})|^p=k^pf^{q-2}|\nabla f|^p, \end{equation*} where \begin{equation*} k:=\frac{p+q-2}p=1+\frac{q-2}p\ge1. \end{equation*} So, at all points where $g>0$, $|\nabla g|>0$, and $|\nabla f|>0$, we can rewrite the inequality in question as \begin{equation*} r-1\le C(s,q)(r^p-k^p) \tag{1}\label{1} \end{equation*} where $r:=a/b$, $a:=f/g$, and $b:=|\nabla f|/|\nabla g|$.
In general, $r$ can take any nonnegative real value.

Letting now $r\to\infty$ in \eqref{1}, we get $C(s,q)>0$. Letting then $r=1$, we get a contradiction: $0\le C(s,q)(1-k)<0$ if $k>1$, that is, if $q>2$.

If, finally, $q=2$, then $k=1$ and the only value of $C(s,q)$ such that \eqref{1} holds for all real $r\ge0$ is $1/p$ -- so that $C(s,2)$ must depend, not on $s$, but on $p$.

  • $\begingroup$ Thank you for your answer! I added a source for this. Could you imagine what is meant with "Young's inequality" here? $\endgroup$
    – Shaq155
    May 4, 2022 at 9:32
  • 2
    $\begingroup$ I have looked at the paper. In your question, you omitted positive constant coefficients of $f^{q-2}g^{s}|\nabla f|^{p}$ and $f^{q-1}g^{s-1}|\nabla f|^{p-1}|\nabla g|$. Yet, since the ratio of those coefficients can be any positive real number, including $1$, what is said in my answer still holds. The inequality in question can still be rewritten as (1) in my answer. As was shown, this inequality cannot be generally true under the specified conditions. (I cannot imagine what Young's inequality may have to do with this.) $\endgroup$ May 4, 2022 at 13:56
  • $\begingroup$ Ok, thank you!! $\endgroup$
    – Shaq155
    May 4, 2022 at 19:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.