How is Young's inequality applied in this equation? In the book Superlinear Parabolic Problems Blow-up, Global Existence and Steady States, page 493 this equation appears in which the book says it uses Young's inequality
$$ |\Omega| u^p \leq ku^q +\epsilon(k)u $$
where $\epsilon(k) = C(|\Omega|,p,q) k^{{-(p-1)}/{(q-p)}}$.
The only things that are given are that $p<q$ ($p$ and $q$ are not conjugate exponents) and $k>0$ ($k$ can be taken large enough). I am failing to see how this is an application of classical Young inequality.
Is this really an application of the classic young inequality or is it an application of some modification of it?
attempt: The best I've achieved so far using Young's inequality was
$$ u^p = [\left( k \frac{(p-1)}{q} \right)^{-(p-1)/q}u][(\left( k \frac{(p-1)}{q} \right)^{(p-1)/q}u^{p-1}] \leq \dfrac{q}{(p-1)}[\left( k \frac{(p-1)}{q} \right)^{(p-1)/q}u^{p-1}]^{q/(p-1)} +   \dfrac{q}{(q-p+1)}[\left( k \frac{(p-1)}{q} \right)^{-(p-1)/q}u]^{q/(q-p+1)}  \leq ku^q+Ck^{-(p-1)/(q-p+1)}u^{q/(q-p+1)}.$$
 A: I assume that $1 \le p$ in the problem, as I believe the inequality is false without this assumption.  Let's assume this.
The estimate is trivial if $p=1$ since you're free to make the constant on the second term larger than $|\Omega|$.
Assume then that $1 < p < q$.  Then write
$$
u^{p} = u^{p-1/r} u^{1/r}
$$
for $1 < r < \infty$ and apply the standard Young inequality to bound
$$
u^{p-1/r} u^{1/r} \le \frac{1}{r'} u^{r'(p-1/r)} + \frac{1}{r} u.
$$
Now, $r' = r/(r-1)$, so
$$
r'(p-1/r) = \frac{rp-1}{r-1}.
$$
We now want to choose $1 < r < \infty$ such that $(rp-1)/(r-1) =q$.  Simple algebra shows this is equivalent to
$$
r = \frac{q-1}{q-p}
$$
and this satisfies $r>1$ precisely due to the assumptions that $1 < p < q$.  We have thus proved that
$$
u^p \le \frac{p-1}{q-1} u^q + \frac{q-p}{q-1} u.
$$
We're now most of the way there as we have our "basic" Young inequality.  We now apply this to $\varepsilon u$, and divide through by $\varepsilon^p$ to see that
$$
u^p \le \frac{p-1}{q-1} \varepsilon^{q-p} u^q + \frac{q-p}{q-1} \varepsilon^{1-p} u.
$$
We can now get your desired bound by choosing $\varepsilon$ appropriately.
A: The trick is to use the splitting
$$
u^p = u^ru^{p-r}
$$
with
$$
r = \frac{p-1}{q-1}q,
$$
and apply the Young inequality with the exponents
$$
\frac{q-1}{p-1} \qquad\textrm{and}\qquad \frac{q-1}{q-p}.
$$
