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

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