## Category Archives: Inequalities

### BV function and its property involves translation

Theorem 1 Suppose ${u\in L^1(\mathbb{R})}$, then ${u\in \text{BV}}$ if and only if ${\exists\, C}$ such that

$\displaystyle ||\tau_hu-u||_{L^1(\mathbb{R})}\leq C|h|,\quad \forall\, h$

Moreover, one can take ${C=|u|_{BV}}$. Here ${\tau_hu(\cdot)=u(\cdot+h)}$ is the tanslation operator.

Proof: Firstly suppose ${u\in \text{BV}}$. Let us prove

$\displaystyle \left|\int_{\mathbb{R}}(\tau_hu(x)-u(x))\phi(x)dx\right|\leq |u|_{BV}|\phi|_{L^\infty}|h|,\quad \forall \phi\in C^\infty_c(\mathbb{R}) \ \ \ \ \ (1)$

To show that

$\displaystyle LHS=\left|\int_{\mathbb{R}}u(x)(\phi(x-h)-\phi(x))dx\right|$

$\displaystyle =\left|\int_{\mathbb{R}}u(x)\psi(x)hdx\right|$

$\displaystyle \leq |u|_{BV}|\psi|_{L^\infty}|h|$

where

$\displaystyle \psi(x)=\int^x_{-\infty}\frac{\phi(s-h)-\phi(s)}{h}ds\in C_c^\infty(\mathbb{R})$

it is easy to verify ${|\psi|_\infty=|\phi|_\infty}$ therefore (1) is proved. Next one can choose such ${\phi_n\rightarrow sign(\tau_hu-u)\in L^1}$ with ${|\phi_n|\leq 1}$(it is easy to show by mollification). By dominating theorem, one get

$\displaystyle \int_{\mathbb{R}}|\tau_hu(x)-u(x)|dx\leq |u|_{BV}|h|.$

The other direction need more analysis. $\Box$

### Sobolev functions on puncture ball

Let us first see the prove of sobolev embedding of ${W^{1,1}\rightarrow L^2}$ on the plane.

Lemma: Suppose ${f\in W^{1,1}(\mathbb{R}^2)}$ with compact support. Then

$\displaystyle ||f||_{L^2}\leq ||\nabla f||_{L^1}$

Proof: Let us suppose ${f\in C^\infty_c(\mathbb{R}^2)}$, the general case can be proved by approximation. Since ${f}$ has compact support, then

$\displaystyle |f(x,y)|=\left|\int_{-\infty}^x\frac{\partial f}{\partial x}(t,y)dt\right|\leq \int_{-\infty}^{\infty}|\nabla f|(t,y)dt=F(y)$

$\displaystyle |f(x,y)|=\left|\int_{-\infty}^y\frac{\partial f}{\partial y}(x,s)ds\right|\leq \int_{-\infty}^{\infty}|\nabla f|(x,s)ds=G(x)$

Then

$\displaystyle \int_{\mathbb{R}^2}f^2(x,y)dxdy\leq \int_{-\infty}^\infty\int_{-\infty}^\infty F(y)G(x)dxdy$

$\displaystyle =\int_{-\infty}^\infty F(y)dy\int_{-\infty}^\infty G(x)dx\\ =\left(\int_{-\infty}^\infty\int_{-\infty}^\infty|\nabla f|dxdy\right)^2$

$\Box$

Suppose we have a function ${u\in W^{1,1}_{loc}(D\backslash\{0\})}$, where ${D}$ is the unit disc in ${\mathbb{R}^2}$, ${u}$ can blow up wildly near the origin. However if we know ${\nabla u\in L^1(D)}$, then actually ${u\in L^2(D)}$ and ${u\in W^{1,1}(D)}$.

Proof: Because the bad thing happened only at origin, we can suppose ${u}$ has spt inside ${\frac{1}{4}D}$ or ${D_{1/4}}$. Put a substantially large square box ${B_\epsilon}$ with length ${\frac 12}$ inside the left half of the disc ${D^-}$whose distance to the origin is ${\epsilon}$ see the picture.

puncture disc

Then on the three sides, ${l_1}$, ${l_2}$, ${l_3}$, ${u=0}$. Using the proof of the above, one can prove

$\displaystyle ||u||_{L^2(B_\epsilon)}\leq ||\nabla u||_{L^1(B_\epsilon)}\leq ||\nabla u||_{L^1(D)}$

Letting ${\epsilon\rightarrow 0}$, we get ${u\in L^2(D^-)}$. The same proof works for the right part ${D^+}$. Finally ${u\in L^2(D)}$. Choose a cut off function ${\zeta_\epsilon=\zeta(x/\epsilon)}$. Then

$\displaystyle u\zeta_\epsilon\rightarrow u\text{ in }L^1$

$\displaystyle \nabla(u\zeta_\epsilon)\rightarrow \nabla u\text{ in }L^1$

So ${u\in W^{1,1}(D)}$. $\Box$

Remark: This is called the removable singularity. There is a more general theorem related to this. Assume ${n\geq 2}$, ${K\subset\subset\Omega}$ such that ${\mathcal{H}^{n-2}(K)=0}$. Suppose ${u\in W^{1,1}_{loc}(\Omega\backslash K)}$ and ${\int_{\Omega\backslash K}|\nabla u|dx<\infty}$. Then ${u\in W^{1,1}(\Omega)}$.

I learned this from Prof. Brezis’s class. Also see his book: Sobolev maps with values into the circle.

### Hardy inequality in dimesion 2

Suppose ${u}$ is a smooth function defined in ${B^c=\{|x|>1\}}$ in the plane ${\mathbb{R}^2}$, assume ${u=0}$ on ${\partial B^c}$ and also has compact support, then

$\displaystyle \int_{|x|> 1}\frac{u(x)^2}{|x|^2\ln^2|x|}dx\leq 4\int_{|x|>1}|\nabla u|^2dx$

There is a way presented by my advisor to explain why the strange function ${|x|^2\ln^2|x|}$ pops up here. First we transform LHS to polar coordinates

$\displaystyle \int_{|x|> 1}\frac{u(x)^2}{|x|^2\ln^2|x|}dx=\int_{0}^{2\pi}\int_{1}^\infty\frac{u(r,\theta)}{r^2\ln^2 r}rdrd\theta$

which leads us to start with a very general ${f(r)}$

$\displaystyle \int_{0}^{2\pi}\int_{1}^\infty\frac{u(r,\theta)}{f(r)}rdrd\theta$

Suppose ${F'(r)=r/f(r)}$, then

$\displaystyle \int_{0}^{2\pi}\int_{1}^\infty\frac{u(r,\theta)}{f(r)}rdrd\theta=\int_{0}^{2\pi}\int_{1}^\infty u(r,\theta)dF(r)d\theta$

$\displaystyle =\int_{0}^{2\pi}\left(u(r,\theta)F(r)|_1^\infty-\int_{1}^\infty F(r)\partial_r u(r,\theta)dr\right)d\theta$

$\displaystyle =-\int_{0}^{2\pi}\int_{1}^\infty F(r)\partial_r u(r,\theta)drd\theta=-\int_{|x|>1} \frac{F(r)\partial_r u(x)}{r}dx$

$\displaystyle \leq 2\left(\int_{|x|>1}\frac{u(x)^2}{f(r)}dx\right)^{1/2}\left(\int_{|x|>1}\frac{f(r)F^2(r)}{r^2}|\nabla u|^2dx\right)^{1/2}$

So we only need to find ${f(r)}$ and ${F(r)}$ such that

$\displaystyle \frac{f(r)F^2(r)}{r^2}\leq C$

Actually one can solve the ODE

$\displaystyle \frac{f(r)F^2(r)}{r^2}=1, \quad F'(r)=\frac{r}{f(r)}$

to get ${F(r)=\frac{-1}{\ln r}}$, ${f(r)=r^2\ln^2 r}$. Plugging in this function back to the above proof gives you the desired inequality.

### Hardy’s inequality in higher dimension

Thm: Supppose ${u\in H^1(\mathbb{R}^n)}$ with ${n\geq 3}$ prove that

$\displaystyle \int_{\mathbb{R}^n}\frac{u^2}{|x|^2}dx\leq C(n)\int_{\mathbb{R}^n}|\nabla u|^2dx$

Proof: Suppose ${u\in C^\infty_c(\mathbb{R}^n)}$ firstly. We can find ball ${B_R(0)}$ large enough such that ${Supp. u\subset B_R}$, then we only need to prove the inequality within ${B_R}$.

Let

$\displaystyle I=\int_{B_R}\frac{u^2}{|x|^2}dx=\frac{-1}{2}\int_{B_R}x\cdot D\left(\frac{1}{|x|^2}\right)u^2dx$

Apply the Green identity on ${B_R\backslash B_\epsilon(0)}$,

$\displaystyle \int_{B_R\backslash B_\epsilon(0)}x\cdot D\left(\frac{1}{|x|^2}\right)u^2dx+\int_{B_R\backslash B_\epsilon(0)}\frac{1}{|x|^2}D(xu^2)dx=\int_{\partial B_\epsilon}\frac{1}{|x|^2}xu^2\cdot\nu dx$

where ${\nu}$ is the unit normal vector of ${\partial B_\epsilon}$

$\displaystyle \left|\int_{\partial B_\epsilon}\frac{1}{|x|^2}xu^2\nu dx\right|\leq \int_{B_\epsilon}\frac{u^2}{|x|^2}dx\leq \frac{1}{\epsilon}\leq \frac{1}{\epsilon}\sup u^2n w_n\epsilon^{n-1}$

Let ${\epsilon\rightarrow 0}$, we have

$\displaystyle -2I+\int_{B_R}\frac{1}{|x|^2}D(xu^2)dx=0$

$\displaystyle -2I+\int_{B_R}\frac{nu^2}{|x|^2}+\frac{1}{|x|^2}2uDu\cdot x=0$

$\displaystyle -2I+nI+2\int_{B_R}\frac{u}{|x|^2}Du\cdot xdx=0$

$\displaystyle I=\frac{2}{2-n}\int_{B_R}\frac{u}{|x|^2}Du\cdot xdx$

From Holder inequality, we have

$\displaystyle I\leq \left(\frac{2}{n-2}\right)^2\int_{B_R}|\nabla u|^2dx$

Next suppose ${u\in H^1(\mathbb{R}^n)}$. Since ${C^\infty_c(\mathbb{R}^n)}$ is dense in ${H^1(\mathbb{R}^n)}$, there exist ${u_k\subset C^\infty_c(\mathbb{R}^n)\rightarrow u}$ in ${H^1(\mathbb{R}^n)}$. From Fatou’s lemma

$\displaystyle \int_{\mathbb{R}^n}\frac{u^2}{|x|^2}dx\leq \lim\inf_{k\rightarrow \infty}\int_{\mathbb{R}^n}\frac{u_k^2}{|x|^2}dx\leq C(n)\lim\inf_{n\rightarrow \infty}\int_{B}|Du_k|^2=C(n)\int_{\mathbb{R}^n}|\nabla u|^2dx$

Remark: If ${n=2}$, we should prove that if ${u\in C^\infty_c(\Omega)}$, where ${\Omega=\{x|1\leq |x|<\infty\}}$

$\displaystyle \int_{\Omega}\frac{u^2}{|x|^2\ln^2|x|}dx\leq 4\int_{\Omega}|Du|^2dx$

### Pointwise gradient estimate of function

$\mathbf{Problem:}$ Suppose $f\in C^\infty_c(\mathbb{R})$, and $f\geq 0$, then

$\displaystyle |f'(x)|^2\leq 2f(x)\max|f''|$, $\forall\, x\in\mathbb{R}$

$\mathbf{Proof:}$ Let $f_\epsilon=f+\epsilon>0$, then $f'_\epsilon=f'$ and $f''_\epsilon=f''$.

Consider $\displaystyle \frac{|f'_\epsilon|^2}{f_\epsilon}\in C^\infty_c(\mathbb{R}^n)$, it must has maximum at some point $x_0$,

$\displaystyle \max\frac{|f'_\epsilon|^2}{f_\epsilon}=\frac{|f'_\epsilon|^2}{f_\epsilon}(x_0)$.

Then at $x=x_0$, $\displaystyle \left(\frac{|f'_\epsilon|^2}{f_\epsilon}\right)'(x_0)=0$. Since we have

$\displaystyle \left(\frac{|f'_\epsilon|^2}{f_\epsilon}\right)'(x_0)=\frac{f'_\epsilon(2f_\epsilon f_\epsilon''-f_\epsilon'^2)}{f_\epsilon^2}(x_0)=0$

If $\displaystyle f'(x_0)=0$, then obviously $\displaystyle |f_\epsilon'(x)|^2\leq 2f_\epsilon(x)\max|f_\epsilon''|$.

Otherwise we have $2f_\epsilon(x_0) f_\epsilon''(x_0)-f_\epsilon'^2(x_0)=0$, that is

$\displaystyle \max\frac{|f'_\epsilon|^2}{f_\epsilon}=\frac{|f'_\epsilon|^2}{f_\epsilon}(x_0)=f''_\epsilon(x_0)$

This also means $\displaystyle |f_\epsilon'(x)|^2\leq 2f_\epsilon(x)\max|f_\epsilon''|$.

Letting $\epsilon\to 0$, we have $\displaystyle |f'(x)|^2\leq 2f(x)\max|f''|$, $\forall\, x\in\mathbb{R}$

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$ This proof is due to Ming Xiao.

### Poincare inequality from different approach

$\mathbf{Problem:}$ Let $\Omega$ be convex and $u\in C^1(\Omega)$. Suppose $S\subset\Omega$ with $\displaystyle \frac{|S|}{|\Omega|}>0$,

$\displaystyle |u(x)-u_S|\leq \frac{d^n}{n|S|}\int_{\Omega}\frac{|\nabla u(y)|}{|x-y|^{n-1}}dy$

where $d$ is the diameter of $\Omega$ and $\displaystyle u_S=\frac{1}{|S|}\int_S u(y)dy$.

$\mathbf{Proof\,1: }$

$\displaystyle u(y)-u(x)=\int_0^1\frac{d}{dt}\left(u(ty+(1-t)x)\right)dt=\int_0^1\nabla u(ty+(1-t)x)\cdot(x-y)dt$

$\displaystyle |u(x)-u_S|\leq\frac{1}{|S|}\int_S|u(x)-u(y)|dy\leq \frac{1}{|S|}\int_S\int_0^1|\nabla u(ty+(1-t)x)||x-y|dtdy$

Set $z=ty+(1-t)x$, then $t|x-y|=|z-x|$. Since $|x-y|\leq d$, $\displaystyle t\geq \frac{|z-x|}{d}$. We also have $t^ndy=dz$

Then

$\displaystyle |u(x)-u_S|\leq \frac{d}{|S|}\int_\Omega\int_\frac{|z-x|}{d}^\infty \frac{|\nabla u(z)|}{t^n}\,dtdz=\frac{d^n}{(n-1)|S|}\int_\Omega\frac{|\nabla u(z)|}{|z-x|^{n-1}}dz$.

$\mathbf{Proof\,2: }$ Using

$\displaystyle u(x)-u(y)=\int_{0}^{|x-y|}\frac{d}{dt}u(x+tw)dt=\int_0^{|x-y|}\nabla u(x+tw)\cdot wdt$

where $\displaystyle w=\frac{x-y}{|x-y|}$. Then

$\displaystyle |u(x)-u_S|\leq\frac{1}{|S|}\int_S |u(x)-u(y)|dy\leq \frac{1}{|S|}\int_S\int_0^{|x-y|}|\nabla u(x+tw)|dtdy$

$\displaystyle \leq \frac{1}{|S|}\int_\Omega\int_0^{\infty}|\nabla u(x+tw)|\chi_{\{t<|x-y|\}}dtdy$

$\displaystyle \leq \frac{1}{|S|} \int_0^\infty\int_{|w|=1}\int_0^d|\nabla u(x+tw)|\chi_{\{t<|x-rw|\}}r^{n-1}dwdrdt$

$\displaystyle \leq \frac{d^n}{n|S|}\int_0^\infty\int_{|w|=1}|\nabla u(x+tw)|\chi_{\{t<|x-\xi(w)|\}}dwdt$

$\displaystyle =\frac{d^n}{n|S|}\int_\Omega \frac{|\nabla u(z)|}{|z-x|^{n-1}}dz$

where $\xi(w)$ is the intersection of ray $x+tw$ with $\partial \Omega$ (it is unique because $\Omega$ is convex).

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$

### Special inequality of O. A. Ladyzhenskaya

$\mathbf{Problem:}$ If $u=u(x,,y)$ is a smooth function of compact support in $\mathbb{R}^2$, then

$\displaystyle \int_{\mathbb{R}^2}u^4(x,y)dxdy\leq 2\int_{\mathbb{R}}u^2(x,y)dxdy\int_{\mathbb{R}^2}|\nabla u|^2dxdy$

$\mathbf{Proof:}$ Because $\displaystyle u^2(x,y)=2\int_{-\infty}^x uu_xdx=2\int_{-\infty}^y uu_ydy$

we have

$\displaystyle \max\limits_{x}u^2(x,y)\leq 2 \int_{-\infty}^\infty |uu_x|dx$

$\displaystyle \max\limits_{y}u^2(x,y)\leq 2 \int_{-\infty}^\infty |uu_y|dy$

Using the Schwarz’s inequality

$\displaystyle \int_{\mathbb{R}^2}u^4dxdy\leq \int_{-\infty}^{\infty}\max\limits_{x}u^2dy\int_{-\infty}^{\infty}\max\limits_{y}u^2dx$

$\displaystyle \leq 4\int_{\mathbb{R}^2}|uu_x|dxdy\int_{\mathbb{R}^2}|uu_y|dxdy$

$\displaystyle \leq 2\int_{\mathbb{R}^2}u^2dxdy\int_{\mathbb{R}^2}|\nabla u|^2dxdy$

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$ From this inequality and approximation argument, we know $W^{1,2}(\mathbb{R}^2)\hookrightarrow L^4(\mathbb{R}^2)$.

Excerpt from Mathematical theory of viscous incompressible flow. Ladyzhenskaya. Also it contains a similar inequality in $\mathbb{R}^3$.

### Integral inequality with f(x)/x and f'(x)

$\mathbf{Problem:}$ Suppose $f\in C^1[0,a]$, and $f(0)=0$, then

$\displaystyle \int_0^a\left(\frac{f(x)}{x}\right)^2dx\leq 4\int_0^a|f'(x)|^2dx$

$\mathbf{Proof:}$ From integration by parts, we get

$\displaystyle \int_\epsilon^a f^2(x)\left(\frac{-1}{x}\right)'dx=\frac{-1}{x}f^2(x)\vert_\epsilon^a+\int_0^a 2f'(x)\frac{f(x)}{x}dx$

letting $\epsilon \to 0$, note the fact that $\displaystyle \lim\limits_{x\to 0}\frac{f^2(x)}{x}=0$, we get

$\displaystyle \int_0^a \left(\frac{f(x)}{x}\right)^2dx=2\int_0^a f'(x)\frac{f(x)}{x}dx-\frac{f^2(a)}{a}$

So by Cauchy’s inequality

$\displaystyle \int_0^a \left(\frac{f(x)}{x}\right)^2dx\leq 2\int_0^a f'(x)\frac{f(x)}{x}dx\leq 2 \int_0^a \left(f'(x)\right)^2dx+\frac{1}{2}\int_0^a\left(\frac{f(x)}{x}\right)^2dx$

The inequality follows from this easily.

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$ I first saw this from Existence of solutions to the nonhomogeneous steady Navier-Stokes Equations. by Charles, J. Amick