$\DeclareMathOperator{\diam}{diam}\newcommand{\norm}[1]{\lVert#1\rVert}\newcommand{\abs}[1]{\lvert#1\rvert}$For $u \in C^{1}(\overline{\Omega})$, for $\Omega\subset \subset \mathbb{R^{n}}$ a bounded convex set, and for any $1 \leq p \leq q$ such that $\frac{1}{p}-\frac{1}{q}<\frac{1}{n}$, I need to show that

$$ \norm{u-\overline{u}_{\Omega}}_{L^{q}} \leq c_{n} \left[ \frac{1+\frac{1}{q} - \frac{1}{p}}{\frac{1}{n}+\frac{1}{q}-\frac{1}{p}}\right]^{1+\frac{1}{q}-\frac{1}{p}} \cdot \frac{(\diam \Omega)^{n}}{\abs\Omega^{1-\frac{1}{n}+\frac{1}{p}-\frac{1}{q}}}\norm{Du}_{L^{p}(\Omega)}, $$

For context, this is a part (c) to a problem where in part (a), we were asked to show that

$$ \abs{u(x)-\overline{u}_{\Omega}} \leq \frac{(\diam \Omega)^{n}}{n \abs{\Omega}}\int_{\Omega}\frac{\abs{Du(y)}}{\abs{x-y}^{n-1}}dy $$

for any function $u \in C^{1}(\overline{\Omega})$ and any point $x \in \Omega$, and in part (b), we were asked to show that if $u$ and $v$ were any two functions on $\Omega$ related as follows:

$$\abs{u(x)}\leq \int_{\Omega}K(x,y) \abs{v(y)} \, dy, \quad x\in \Omega$$

for some kernel $K(x,y)$, then for any $1 \leq p \leq q \leq \infty$, we have

$$\norm u_{L^{q}(\Omega)} \leq A \norm v_{L^{p}(\Omega)},$$

where $\displaystyle A = \max\left(\sup_{x}\norm{K(x,\cdot)}_{L^{\frac{pq}{pq+p-q}}}, \sup_{y}\norm{K(\cdot,y)}_{L^{\frac{pq}{pq+p-q}}}\right)$.

We are told to assume the integral estimate that

$$\int_{\Omega}\abs{x-y}^{-n\mu}dy \leq c_{n}\frac{1}{1-\mu}\abs{\Omega}^{1-\mu}$$

for any $0 \leq \mu < 1$, where $c_{n}$ is a constant dependent only on the dimension $n$.

Here is how I did the problem:

By part (a), we have that $\displaystyle \abs{u(x)-\overline{u}_{\Omega}}\leq \frac{(\diam \Omega)^{n}}{n\abs{\Omega}}\int_{\Omega}\frac{\abs{Du(y)}}{\abs{x-y}^{n-1}}dy$, so we have the relationship required in part (b) between $\tilde{u}(x) = u(x)-\overline{u}_{\Omega}$ and $v(y)$, with kernel $K(x,y) = \frac{1}{\abs{x-y}^{n-1}}$ and $\displaystyle \abs{v(y)} = \frac{(\diam \Omega)^{n}}{n\abs{\Omega}}\abs{Du(y)}$.

Therefore, applying part (b), we obtain that

$$\begin{align} \norm{\tilde{u}}_{L^{q}(\Omega)} &= \norm{u(x) - \overline{u}_{\Omega}}_{L^{q}(\Omega)}\\ &\leq A \norm{v}_{L^{p}(\Omega)} = A\left( \int_{\Omega}\abs{v(y)}^{p} dy\right)^{1/p} \\ &= A \left( \int \left\lvert \frac{(\diam \Omega)^{n}}{n\abs\Omega}Du(y)\right\rvert^{p}dy\right)^{1/p} \\ &= A \frac{(\diam \Omega)^{n}}{n\abs\Omega}\left( \int |Du(y)|^{p}dy\right)^{1/p} \\ &= A \frac{(\diam \Omega)^{n}}{n\abs\Omega}\norm{Du}_{L^{P}(\Omega)}. \end{align}$$

Now, since by part (b), we are told that

$$A = \max\left(\sup_{x}\norm{K(x,\cdot)}_{L^{\frac{pq}{pq+p-q}}}, \sup_{y}\norm{K(\cdot,y)}_{L^{\frac{pq}{pq+p-q}}}\right),$$

we must show that

$$\displaystyle \sup_{x}\norm{K(x,\cdot)}_{L^{\frac{pq}{pq+p-q}}(\Omega)} \leq C_{n}\left[ \frac{1+\frac{1}{q} - \frac{1}{p}}{\frac{1}{n}+\frac{1}{q}-\frac{1}{p}} \right]^{1 + \frac{1}{q} - \frac{1}{p}}\abs\Omega^{\frac{1}{n}+\frac{1}{q}-\frac{1}{p}},$$

which I did. I'll spare you the gory details of this derivation, especially since it's not relevant to my question.

My question comes in the putting of it all together:

When we do put it all together, we have that

$$\begin{align} \norm{u-\overline{u}_{\Omega}}_{L^{q}(\Omega)} &\leq A \norm v_{L^{p}(\Omega)} \\ &= A \frac{(\diam \Omega)^{n}}{n\abs\Omega}\left( \int_{\Omega} \abs{Du(y)}^{p}dy \right)^{1/p} \\ &\leq c_{n} \left[ \frac{1+\frac{1}{q}-\frac{1}{p}}{\frac{1}{n}+\frac{1}{q}-\frac{1}{p}} \right]^{1+\frac{1}{q}-\frac{1}{p}}\abs\Omega^{\frac{1}{n} + \frac{1}{q} - \frac{1}{p}}\frac{(\diam \Omega)^{n}}{n\abs\Omega} \norm{Du}_{L^{p}(\Omega)} \end{align}$$

and eventually, we get the desired result.

From this point on, I was able to get the desired result just fine; however, my question is, in the last line of this last quoted formula, should I have a $\mathbf{\leq}$ sign as it is currently written, or should it be an $=$ sign, and why?

Thank you in advance. :)

  • $\begingroup$ What is $u_\Omega$? $\endgroup$ – Jochen Feb 3 '14 at 12:02
  • $\begingroup$ @Jochen, I have thoroughly edited the question so that now that is explained. $\endgroup$ – ALannister May 4 '14 at 4:49
  • $\begingroup$ @jessyCat, Are you sure that $u$ doesn't appear in the RHS of the inequality? $\endgroup$ – themaker May 4 '14 at 4:59
  • $\begingroup$ @themaker, which inequality are you talking about? $\endgroup$ – ALannister May 4 '14 at 5:02
  • $\begingroup$ @JessyCat, the first one $\endgroup$ – themaker May 4 '14 at 5:03

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