Need help understanding proof for a Poincaré inequality.

Question

I need help understanding the first part of the proof for a Poincaré inequality taken from Evans p. 141. Because I couldn't find a proper way to display the average integral symbol, I will be using "⨍" instead.

For each $1 \leq p < n$ there exists a constant $C_2$, depending only on $p$ and $n$ such that

$$⨍_{B(x,r)} |f - (f)_{x,r} |^p dy \leq C_2 r ⨍_{B(x,r)}|Df|^pdy$$ for all $B(x,r) \subset \mathbb{R}^n, f \in W^{1,p}(U(x,r))$ where $(f)_{x,r} = ⨍_{B(x,r)} f dy$.

Proof:

We may assume $f \in C^1(B(x,r))$.

$$⨍_{B(x,r)}|f-(f)_{x,r}|^pdy = ⨍_{B(x,r)}|⨍_{B(x,r)}f(y) - f(z) dz|^pdy$$

$$\leq ⨍_{B(x,r)}⨍_{B(x,r)}|f(y)-d(z)|^pdz\ dy$$

$$\leq C⨍_{B(x,r)}r^{p-1} \int_{B(x,r)} |Df(z)|^p |y-z|^{1-n} dz\ dy$$

$$\leq Cr^p⨍_{B(x,r)}|Df|^p dz$$

(The second inequality uses a Theorem proven before.)

I have several questions: How does one get the first equality? The first equality only holds iff $f - (f)_{x,r} = ⨍_{B(x,r)} f(y)-f(z)dz$ but writing the average integrals out gives a contradiction.

Furthermore for the first inequality: I suspected the proof might be using Holder's inequality since $ \lvert\lvert f \rvert\rvert_{L^1}^p \leq \lvert\lvert f \rvert\rvert_{L^p}^p \lvert\lvert 1 \rvert\rvert_{L^q}^p $, but where is the last term in the proof?

And lastly in the last inequality: Why does the integral $\int ... dy$ as well as $|y-z|^{1-n}$ vanish?

I would appreciate any help.

Answer 1

The first equality is just plugging in the definition of the average (for $(f)_{x,r}$ on the left hand side) and using the fact that $⨍f(y) dz = f(y)$. (Note the integration variable)

The first inequality is in fact just the Hölder inequality in the form $$|\int\limits_{B_r} f(y)-f(z)\,dz| \le \left( \int\limits_{B_r} 1\, dz \right)^{\frac{1}{q}} ||f(y)-f||_{L^p(B_r)}$$ Just write out the average definitions and rearrange. Note $\frac{1}{q}=\frac{p-1}{p}$

The last one can be obtained by noting that $\int\limits_{B_r(x)}|z-y|^{n-1}dy$ can be estimated from above by (e.g.) $C\int_{B_{2r}(z)} |z-y|^{n-1}\,dy\le C\int_0^{2r} dt = Cr$ (using $n-$dimensional polar coordinates).