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From PDE Evans, 2nd edition, page 281:

Now \begin{align} \int_0^s \int_{\partial B(0,1)} |Du(x+tw)| \, dS(w) dt &=\int_0^s \int_{\partial B(x,t)} \frac{|Du(y)|}{t^{n-1}} \, dS(y) dt \\ &= \int_{B(x,s)} \frac{|Du(y)|}{|x-y|^{n-1}} \, dy \\ &\le \int_{B(x,r)} \frac{|Du(y)|}{|x-y|^{n-1}} \, dy, \end{align} where we put $y=x+tw,t=|x-y|$.

Can I ask where does the $\frac 1{t^{n-1}}$ factor come from in the first inequality? Somehow this results from transforming the variables by letting $y=x+tw$.


Also, from page 282 of the textbook (part of the same proof of Morrey's inequality),

Next, choose any two points $x,y \in \mathbb{R}^n$ and write $r :=|x-y|$. Let $W := B(x,r) \cap B(y,r)$. Then $$|u(x)-u(y)| \le \frac 1{|W|} \left[ \int_W |u(x)-u(z)| \, dz + \int_W |u(y)-u(z)| \, dz \right]\tag{24}$$ But inequality $\text{(21)}$ allows us to estimate \begin{align} \frac 1{|B(x,r)|} \int_W |u(x)-u(z)| \, dx &\le C \cdot \frac 1{|B(x,r)|} \int_{B(x,r)} |u(x)-u(z)| \, dz \\ &\le C \left(\int_{B(x,r)} |Du|^p \, dz \right)^{\frac 1p} \left(\int_{B(x,r)} \frac{dz}{|x-z|^{(n-1)\frac{p}{p-1}}} \right)^{\frac{p-1}p} \\ &\le C\left(r^{n-(n-1)\frac p{p-1}} \right)^{\frac{p-1}p} \|Du\|_{L_p(\mathbb{R}^n)} \\ &= Cr^{1-\frac np} \|Du\|_{L^p(\mathbb{R}^n)}. \end{align}

The only part I'm not understanding is how they were able to assert the following: $$\int_{B(x,r)} \frac{dz}{|x-z|^{(n-1)\frac{p}{p-1}}}=r^{n-(n-1)\frac p{p-1}}$$

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Integration in polar coordinates (or spherical, in higher dimensions). Write $z=x+\rho\theta$ where $\theta\in S^{n-1}$ is a unit vector and $\rho\in [0,r]$. Then for any integrable $f$ we have $$ \int_{B(x,r)} f(z)\,dz = \int_0^r \int_{S^{n-1}} f(x+\rho \theta) \rho^{n-1}\,d\theta \,d\rho \tag{1} $$ (Compare with $n=2$ case, when this is the usual polar coordinates.)

In your example the inner integral is easy because $f(x+\rho\theta) = \rho^{-(n-1)\frac{p}{p-1}}$. Integrating a constant over the unit sphere we pick up the factor of the surface area, which is a constant that depends only on $n$, traditionally denoted $\omega_{n-1}$. Proceed to integrate: $$ \int_0^r \omega_{n-1} \rho^{-(n-1)\frac{p}{p-1}} \rho^{n-1}\,d\rho =\omega_{n-1} \rho^{n-(n-1)\frac{p}{p-1}}\bigg|_{\rho=0}^{\rho=r} $$ which is the desired result.


Let me also bring this in line with "Calculus" Appendix in the book. Evans does not turn to unit sphere in (1), but rather presents the formula as $$ \int_{B(x,r)} f(z)\,dz = \int_0^r \int_{\rho S^{n-1}} f(x+\theta) \,d\theta \,d\rho $$ (his notation is different). This is the same as (1), because scaling an $(n-1)$-dimensional surface by factor of $\rho$ carries with it the Jacobian of $\rho^{n-1}$.

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