Estimate of supremum of a function by an integral Let $\Omega\subset\mathbb{R}^n$ be a bounded open set, $n\geq 2$. For $r>0$, denote by $B_r(x_0)=\{x\in\mathbb{R}^n:|x-x_0|<r\}$ whose closure is a proper subset of $\Omega$. Let $u\in W^{1,p}(\Omega)$ (the standard Sobolev space) for $1<p<n$ be a nonnegative, bounded function such that for every $\frac{1}{2}\leq\sigma^{'}<\sigma\leq 1$, we have
\begin{equation}
\sup_{B_{\sigma^{'}r}(x_0)}\,u\leq \frac{1}{2}\sup_{B_{\sigma r}(x_0)}\,u+\frac{c}{(\sigma-\sigma^{'})^{\frac{n}{q}}}\left(\frac{1}{|B_r(x_0)|}\int_{B_r(x_0)}u^q\,dx\right)^\frac{1}{q}\quad\forall q\in(0,p^{*}),
\end{equation}
where $c$ is some fixed positive constant,independent of $x_0,r$, $p^{*}=\frac{np}{n-p}$ and $|B_r(x_0)|$ denote the Lebsegue measure of the ball $B_r(x_0)$. Then by the iteration lemma stated below, we have
\begin{equation}
\sup_{B_{\frac{r}{2}}(x_0)}\,u\leq c\left(\frac{1}{|B_r(x_0)|}\int_{B_r(x_0)}u^q\,dx\right)^\frac{1}{q}\quad\forall q\in(0,p^{*}),
\end{equation}
where $c$ is some fixed positive constant,independent of $x_0,r$.
Iteration lemma: Let $f=f(t)$ be a nonnegative bounded function defined for $0\leq T_0\leq t\leq T_1$. Suppose that for $T_0\leq t<\tau\leq T_1$ we have
$$
f(t)\leq c_1(\tau-t)^{-\theta}+c_2+\xi f(\tau),
$$
where $c_1,c_2,\theta,\xi$ are nonnegative constants and $\xi<1$. Then there exists a constant $c$ depending only on $\theta,\xi$ such that for every $\rho, R$, $T_0\leq \rho<R\leq T_1$, we have
$$
f(\rho)\leq c[c_1(R-\rho)^{-\theta}+c_2].
$$
Applying the iteration lemma with $f(t)=\sup_{B_t(x_0)}\,u$, $\tau=\sigma r$, $t=\sigma^{'}r$, $\theta=\frac{n}{q}$ in the given estimate on $u$ above, the second estimate on $u$ above follows. My question is can we obtain the following estimate
\begin{equation}
\sup_{B_r(x_0)}\,u\leq c\left(\frac{1}{|B_r(x_0)|}\int_{B_r(x_0)}u^q\,dx\right)^\frac{1}{q}\quad\forall q\in(0,p^{*}),
\end{equation}
which estimates the supremum of $u$ over the whole ball $B_r(x_0)$? If it is possible, does it follow by a covering argument? Here $c$ is some fixed positive constant, independent of $x_0,r$.
Thank you very mych.
 A: No: let $u(x) = |x|^a$ and $x_0 = 0$. I do this with $q = 1$ to simplify computation, but you can repeat the argument with any (or all) $q$. The first inequality (up to modifying the constant, and assuming say $a \geq 1$) reads
$$
(\sigma' r)^a \leq \frac{1}{2} (\sigma r)^a + \frac{c_1}{a(\sigma - \sigma')^{n}}r^a.
$$
When $(\sigma'/\sigma)^a \leq \frac{1}{2}$ this is true just by looking at the first term, and if not then
$$
\frac{c_1}{a(\sigma - \sigma')^{n}} \geq \frac{c_1}{a\sigma(1 - (1/2)^{1/a})^{n}} \geq \frac{c_1}{a(1 - (1/2)^{1/a})^{n}} \geq \frac{c_1}{A}
$$
uniformly for large $a$, where $A>0$ is some upper bound on the denominator. Then the inequality is satisfied with $c_1 = A$ for all $a$ large, and all $\sigma, \sigma', r$.
On the other hand, the desired inequality (with $r = 1$) reads
$$
1 \leq \frac{c_2}{a}.
$$
This is false for any given $c_2$ if $a = a(c_2)$ is taken large enough.
Let me make a further comment, based on you tagging this question under partial differential equations and regualrity. The fact that this inequality is false for some simple example is accidental and not particularly interesting. More importantly:

*

*You should not expect inequalities like this to be true. You will almost always lose some room between nested balls when you use cutoff functions, local estimates, etc. Maybe once in a blue moon you can "recover" the loss by some argument, but only with enormous effort which hinges on much finer analysis of situations unrelated to what you are mostly interested in.

*The regularity arguments we develop are OK with needing some room in estimates.

*Covering arguments let you modify the amount of room you need between sets at the expense of larger constants (e.g. replacing $\frac{1}{2}$ by $\frac{999}{1000}$ here), not remove all the room entirely.

A: @ user378654 the below limit
$$
\lim_{a\to\infty} a(1-(\frac{1}{2})^\frac{1}{a})^n=0,
$$
but I hope you have used the above limit to be non-zero to give a lower bound in your argument, which is independent of $a$.
The above limit is zero, you may see the answer from here: To find the limit of a sequence
Also, it seems to give a lower bound independent of $a$, the whole sequence should be considered. Can you please see it. Thanks.
