Category Archives: Real Analysis

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.

Sharpness of Morrey’s inequality

Thm: Suppose {n<p\leq \infty}, then there exists a constant {C} such that

\displaystyle ||u||_{C^{0,\gamma}(\mathbb{R}^n)}\leq C(n,p)||u||_{W^{1,p}(\mathbb{R}^n)}

for all {u\in C^1(\mathbb{R}^n)}, where {\gamma=1-n/p}. For functions in {W^{1,p}(\Omega)}, we have

Thm: Suppose {\Omega} is bounded domain in {\mathbb{R}^n}, and {\partial\Omega\in C^1}. Assume {n<p\leq \infty} and {u\in W^{1,p}(\Omega)}, there exists one version {u^*\in C^{0,\gamma}(\bar{\Omega})}, for {\gamma=1-n/p}, such that

\displaystyle ||u^*||_{C^{0,\gamma}(\mathbb{R}^n)}\leq C(n,p,\Omega)||u||_{W^{1,p}(\mathbb{R}^n)}

Remark: This {\gamma} is critical number. For {\beta\in (\gamma,1]}, we can choose {\alpha\in (\gamma,\beta)}. Consider the function {u(x)=|x|^\alpha} on the unit ball {B_1}. Then

\displaystyle D_iu=\alpha x_i|x|^{\alpha-2}

{u\in W^{1,p}(B_1)} if and only if {(1-\alpha)p<n}, which is {\alpha>\gamma}.

However consider the {[u]_\beta} which is

\displaystyle [u]_{\beta;B_1}=\sup_{x,y\in B_1}\frac{\big||x|^\alpha-|y|^\beta\big|}{|x-y|^\beta}\geq \sup_{x\in B_1}\frac{|x|^\alpha}{|x|^\beta}=+\infty

So {u\not\in C^{0,\beta}(B_1)}.

Sharpness of Sobolev embedding

Consider the Sobolev embedding

Thm 1: Suppose {u\in W^{1,p}(\Omega)}, {\Omega} is a bounded domain in {\mathbb{R}^n} with {C^1} boundary, {p<n}, then

\displaystyle ||u||_{L^q(\Omega)}\leq C(n,p,q,\Omega)||u||_{W^{1,p}(\Omega)}

for any {1\leq q\leq \frac{np}{n-p}}.

Consider the sharpness of Sobolev embedding, which means {q} can not be bigger than {\frac{np}{n-p}}.

WLOG assume {\Omega=B_1(0)}. Choose a particular function {u(x)=\frac{1}{|x|^\alpha}}, then {u\in W^{1,p}(B_1)} when {\alpha<\frac{n-p}{p}}. The reason is the following

{u} is smooth away from {0} with

\displaystyle D_iu=\frac{-\alpha x_i}{|x|^{\alpha+2}}

For any {\phi\in C_c^\infty(B_1)}

\displaystyle \int_{B_1-B_\epsilon}uD_i\phi=-\int_{B_1-B_\epsilon}D_iu\phi+\int_{\partial B_\epsilon}u\phi\nu_ids

Let {\epsilon \rightarrow 0}, we have

\displaystyle \left|\int_{\partial B_\epsilon}u\phi\nu_ids\right|\leq C\int_{\partial B_\epsilon}\epsilon^{-\alpha}ds\leq \epsilon^{-\alpha+n-1}\rightarrow 0

Then

\displaystyle \int_B uD_i\phi=-\int_BD_iu\phi

for any {\phi\in C^\infty_c(B)} when {\alpha<n-1}. {u} has weak derivative {D_iu}, {Du\in L^p(B)} only when {\alpha<\frac{n-p}{p}}.

Let us calculate

\displaystyle ||u||_{L^q}=C\left(\int_0^1r^{-\alpha q+n-1}dr\right)^{1/q}

{u\in L^q(B_1)} if and only if {\alpha<\frac{n}{q}}.

So if {q>\frac{np}{n-p}}, we can find one {\alpha} such that {\alpha\in (\frac{n}{q}, \frac{n-p}{p})}. By the above analysis, such {u\in W^{1,p}(B_1)} but {u\not\in L^q(B_1)}.

 

 

Thm 2: Suppose {u\in W^{1,p}(\Omega)}, {\Omega} is a bounded domain in {\mathbb{R}^n} with {C^1} boundary, {p<n}, then the embedding {W^{1,p}(\Omega)\rightarrow L^q(\Omega)} is compact when {q\in [1,\frac{n-p}{np})}.

Consider the sharpness of compact embedding, {q} must be strictly less than {\frac{n-p}{np}}. Actually we can find a sequence of {u_n\in W^{1,p}(\Omega)} but {u_n} does not have convergent subsequence in {L^{p*}(\Omega)}.

As before assume {\Omega=B_1}. Choose {u\in C^1_0(B_1)}, define {u_{\lambda}(x)=\lambda^\alpha u(\lambda x)}, for {\lambda\geq 1}, then {u_\lambda\in C^1_0(B_{1/\lambda})}

\displaystyle ||u_\lambda||_{L^{p*}(B_1)}=||u_\lambda||_{L^{p*}(\mathbb{R}^n)}=\lambda^{\alpha-n/p*}||u||_{L^{p*}(B_1)}

\displaystyle ||Du_\lambda||_{L^p(B_1)}=\lambda^{\alpha+1-n/p}||Du||_{L^p(B_1)}

Since {u_\lambda} has compact support, then

\displaystyle ||u||_{W^{1,p}}\leq C||Du||_{L^p}\leq C\lambda^{\alpha+1-n/p}||Du||_{L^p(B_1)}

Choose {\alpha=n/p*=n/p-1}, then {||u||_{W^{1,p}}} is bounded. However, {u_\lambda} has no convergent subsequence. Otherwise as {\lambda\rightarrow \infty}, {u_\lambda(x)\rightarrow 0} when {x\neq 0}, such subsequence must converge to 0 in {L^{p*}(\Omega)}. Since we have {||u_\lambda||_{L^{p*}}=||u||_{L^{p*}}}, apparently this can not be true.

 

Small things about Hausdorff measure

Suppose {X} is a metric space, {\mathscr{F}} is a families of subsets of {X}, and {\phi} is a function on {\mathscr{F}} such that

\displaystyle 0\leq \phi(A)\leq \infty\quad\text{ whenever }A\in \mathscr{F}

we can obtain a measure through a prodecure called Caratheodory’s construction. Define

\displaystyle \tilde{\phi}_\delta(E)=\inf\left\{\sum_i\phi(A_i)\big| E\subset\cup_i A_i\text{ and diam}A_i<\delta\right\}

for any set {E\subset X}. Then

\displaystyle \tilde{\phi}_\delta\geq\tilde{\phi}_\sigma \text{ if }0<\delta<\sigma<\infty

\displaystyle \tilde{\phi}_\delta(E\cup F)\geq \tilde{\phi}_\delta(E)+\tilde{\phi}_\delta(F)\text{ if dist}(E,F)>\delta>0

This means {\tilde{\phi}_\delta(E)} is monotone, then it is reasonable to define

\displaystyle \lambda(E)=\lim\limits_{\delta\rightarrow 0+}\tilde{\phi}_\delta(E)=\sup_{\delta>0}\tilde{\phi}_\delta(E)

{\tilde{\phi}_\delta} and {\lambda} are (outer) measures. Moreover

\displaystyle \lambda(E\cup F)\geq \lambda(E)+\lambda(F)\text{ if dist}(E,F)>0

This means all open set are measurable in {\lambda}. If {\mathscr{F}} contains Borel sets of {X}, then {\lambda} is Borel regular measure.

Suppose {X=\mathbb{R}^n}, {\mathscr{F}} is the Borel set of {\mathbb{R}^n}, {m\geq0}

\displaystyle \phi(B)=w_m\left(\frac{\text{diam }B}{2}\right)^m, \quad B\in \mathscr{F}

Then the Caratheodory’s construction will yield the {m-}dimensional Hausdorff measure.

Lemma: {X} is a (separable?) metric space, {\mu} is a measure on {Y}, {f:X\rightarrow Y}, {f(A)} is measurable whenever {A} is a Borel subset of {X}, define

\displaystyle \phi(A)=\mu(f(A))

{\lambda} is obtained from the Caratheodory’s construction from {\phi} on all Borel subsets of {X}. Then

\displaystyle \lambda(A)=\int_YN(f|A,y)d\mu(y)\text{ whenever }A\text{ is a Borel set }

where {N(f|A,y)=card\{x\in A|f(x)=y\}}. Proof: Find Borel partitions {\mathscr{P}_i} of {A} such that {\mathscr{P}_{i+1}} is a refinement of {\mathscr{P}_i}, or all subsets in {\mathscr{P}_i} are unions of that of {\mathscr{P}_{i+1}}. And also assume

\displaystyle \lim_{i\rightarrow \infty} \sup_{E\subset \mathscr{P}_i} \text{diam}(E)=0

Let {\chi^i_{f(E)}} be the characteristic function of {f(E)}, {E} is from partition {\mathscr{P}_i}.

\displaystyle \sum_{E\in \mathscr{P}_i}\chi^i_{f(E)}(y)\nearrow N(f|A,y)

From Levy’s monotone convergence thm,

\displaystyle \int_Y N(f|A,y)d\mu(y)=\lim_{i\rightarrow \infty}\int_Y \sum_{E\in \mathscr{P}_i}\chi^i_{f(E)}(y) d\mu(y)=\lim_{i\rightarrow \infty}\sum_{E\in \mathscr{P}_i}\mu(f(E))

\displaystyle =\lim_{i\rightarrow \infty}\sum_{E\in \mathscr{P}_i}\phi(E)\geq \lambda(A)

by the defintion of {\lambda(A)}.

Note that the defition of {\phi(A)} implies {\tilde{\phi}_\delta\geq \phi(A)} for any {\delta>0}, hence {\lambda(A)\geq \phi(A)} for very Borel set {A}. Since all Borel set are {\lambda} measurable, for any partition {\mathscr{P}}

\displaystyle \lambda(A)=\sum_{E\in \mathscr{P}}\lambda(E)\geq \sum_{E\in \mathscr{P}}\phi(E)=\sum_{E\in \mathscr{P}}\mu(f(E))

combining with the result before,

\displaystyle \lambda(A)=\int_YN(f|A,y)d\mu(y)

\Box

Theorem: Suppose {f:\mathbb{R}^l\rightarrow \mathbb{R}^n} is a Lipschitzian map. {m} is nonegative number, then

\displaystyle \int_{\mathbb{R}^n}N(f|A,y)d\mathscr{H}^m(y)\leq (Lip f)^m\mathscr{H}^m(A)

whenever {A\subset\mathbb{R}^l} is a Borel subset. Proof: Let us apply the above lemma with {\mu=\mathscr{H}^m} on {\mathbb{R}^n}, then

\displaystyle \lambda(A)=\int_{\mathbb{R}^n}N(f|A,y)d\mathscr{H}^m(y)

For any Borel set {B}, any covering of {B} by sets diameter {\leq \delta} yields a covering of {f(B)} of diameter {\leq (Lip f)\delta}, then

\displaystyle \phi(B)=\mathscr{H}^m(f(B))\leq (Lip f)^m\mathscr{H}^m(B)

\displaystyle \tilde{\phi}_\delta(A)\leq \sum_{\text{diam}B_i<\delta, B_i\in \mathscr{P}}\phi(B_i)\leq (Lip f)^m\mathscr{H}^m(A)

By taking the limit, {\lambda(A)\leq (Lip f)^m\mathscr{H}^m(A)}. \Box

Corollary: For any connected set {E\subset \mathbb{R}^n}

\displaystyle \mathscr{H}^1(E)\geq \text{diam}\,E

Proof: Since {\mathscr{H}^1} is a regular measure on {\mathbb{R}^n}, we can find Borel set {B} containing {E} with the same {\mathscr{H}^1} measure. Choose {a\in B}, define {f:\mathbb{R}^n\rightarrow \mathbb{R}^1}, {f(x)=|x-a|}. Then {f} is Lipschitzian map and {Lip\, f=1}. By the previous theorem

\displaystyle \mathscr{H}^1(B)\geq\int_{\mathbb{R}^1}N(f|B,y)d\mathscr{H}^1(y)\geq \mathscr{H}^1(f(E))=\mathscr{L}^1(f(E))\geq\text{diam}\,E

\Box

 

Product rule or Product formula for weak derivative

Problem: Derive the product formula for weak derivative

\displaystyle D(uv)=uDv+vDu

holds for all {u,v\in W^1(\Omega)} such that {uv} and {uDv+vDu\in L^1_{loc}(\Omega)}.

Proof: Step 1: prove the case {u\in W^1(\Omega)}, {v\in C^1(\Omega)}

Step 2: prove the case {u,v\in W^{1}(\Omega)\cap L^\infty_{loc}(\Omega)} by step 1. Readers can also see the proof on page269 of Functional Analysis, Sobolev Spaces and Partial Differential Equations (Haim Brezis)

Step 3: Define {f\in C^0(\mathbb{R}^n)}

\displaystyle f_n(t)=\begin{cases}n,\quad t>n\\ t,\quad |t|\leq n\\ -n,\quad t<-n\end{cases}

then {f_n} is piecewise smooth in {\mathbb{R}} and {f_n'\in L^\infty (\mathbb{R})}. So {f_n(u)\in W^1(\Omega)} by lemma 7.8 in Gilbarg and Trudinger’s bk and

\displaystyle D(f_n(u))(x)=\begin{cases}Du(x),\quad |u(x)|\leq n,\\ 0,\quad\quad |u(x)|>n.\end{cases}

Denote {u_n=f_n(u)\in W^1(\Omega)\cap L^\infty_{loc}(\Omega)} and {v_n=f_n(v)\in W^1(\Omega)\cap L^\infty_{loc}(\Omega)}. By step 2 we have {u_nv_n\in W^1(\Omega)} and

\displaystyle \int_{\Omega} u_nv_n D\phi dx=-\int_{\Omega}(u_nDv_n+v_n Du_n)\phi dx\quad \forall\, \phi\in C^\infty_0(\Omega)

Note that by assumption

\displaystyle |u_nv_n|\leq |uv|\in L^1_{loc}(\Omega)

\displaystyle |u_nDv_n|\leq |uDv|\in L^1_{loc}(\Omega)

\displaystyle |v_nDu_n|\leq |vDu|\in L^1_{loc}(\Omega)

By dominating convergence theorem, letting {n\rightarrow \infty}

\displaystyle \int_{\Omega} uv D\phi dx=-\int_{\Omega}(uDv+v Du)\phi dx\quad \forall\, \phi\in C^\infty_0(\Omega)

Step 4:Consider the problem with additional assumption {u\geq 0} and {v\geq 0}.

Firstly we assume {v\geq 1}. Define {\tilde{u}_n=\min\{u,\frac{n}{v}\}=u+(\frac{n}{v}-u)^+}. Since {u,\frac{n}{v} \in W^1(\Omega)}, then {\tilde{u}_n\in W^1(\Omega)}, satisfies {0\leq \tilde{u}_n v\leq n}, moreover

\displaystyle D\tilde{u}_n=\begin{cases}Du,\quad uv<n\\-n\frac{Dv}{v^2},\quad uv\geq n\end{cases}\in L^1_{loc}(\Omega)

\displaystyle vD\tilde{u}_n+\tilde{u}_n Dv=\begin{cases}vDu+uDv,\quad uv<n\,\quad uv\geq n\end{cases}\in L^1_{loc}(\Omega)

Suppose {\displaystyle f_\epsilon(t)=\frac{t}{1+\epsilon t}} defined on {\mathbb{R}_+^1}, where {\epsilon>0}. Then {f_\epsilon} is a piecewise smooth function in {\mathbb{R}_+^1} and {f'_\epsilon\in L^\infty(\mathbb{R}_+^1)}. By lemma 7.8 on Gilbarg and trudinger’s book, {f_\epsilon(\tilde{u}_n)\in W^1(\Omega)}. Note that {f_\epsilon(u)\in L^\infty(\Omega)} by the conclusion of step 3,

\displaystyle \int_{\Omega}f_\epsilon(\tilde{u}_n) vD\phi=-\int_{\Omega}[Df_\epsilon(\tilde{u}_n)v+f_\epsilon(\tilde{u}_n)Dv]\phi

that is

\displaystyle \int_{\Omega}\frac{\tilde{u}_n}{1+\epsilon \tilde{u}_n}vD\phi=-\int_{\Omega}\frac{\tilde{u}_nDv+vD\tilde{u}_n}{1+\epsilon \tilde{u}_n}\phi+\int_{\Omega}\frac{\epsilon \tilde{u}_nvD\tilde{u}_n}{(1+\epsilon \tilde{u}_n)^2}\phi\quad (1)

By dominating convergence theorem, as {\epsilon\rightarrow 0}

\displaystyle \int_{\Omega}f_\epsilon(\tilde{u}_n) vD\phi\rightarrow \int_{\Omega}\tilde{u}_nvD\phi

\displaystyle \int_{\Omega}\frac{\tilde{u}_nDv+vD\tilde{u}_n}{1+\epsilon\tilde{u}_n}\phi\rightarrow \int_{\Omega}(\tilde{u}_nDv+vD\tilde{u}_n)\phi

And

\displaystyle \left|\int_{\Omega}\frac{\epsilon \tilde{u}_nvD\tilde{u}_n}{(1+\epsilon\tilde{u}_n)^2}\phi\right|\leq \epsilon n\int_{\Omega}|D\tilde{u}_n||\phi|\rightarrow 0\text{ as }\epsilon\rightarrow 0

Letting {\epsilon\rightarrow 0}, {(1)} implies

\displaystyle \int_{\Omega}\tilde{u}_nvD\phi=-\int_{\Omega}[vD\tilde{u}_n+\tilde{u}_nDv]\phi

Since for any {n>0}

\displaystyle |\tilde{u}_nv|\leq |uv|

\displaystyle |vD\tilde{u}_n+\tilde{u}_nDv|\leq |vDu+uDv|

by dominating convergence theorem, letting {n\rightarrow \infty}

\displaystyle \int_{\Omega}uvD\phi=-\int_{\Omega}[vDu+uDv]\phi

If we only know {v\geq 0}, consider {u} and {v+1}, repeat the above proof

\displaystyle \int_{\Omega}u(v+1)D\phi=-\int_{\Omega}[(v+1)Du+uD(v+1)]\phi

note that {u\in W^1(\Omega)}, this is equivalent to

\displaystyle \int_{\Omega}uvD\phi=-\int_{\Omega}[vDu+uDv]\phi

Step 5: Consider the most general case with no extra assumption. Since

\displaystyle u^+v^+, u^+v^-, u^-v^+,u^-v^-\in L^\infty_{loc}(\Omega)

\displaystyle v^{\pm}Du^\pm+u^\pm Dv^\pm\in L^\infty_{loc}(\Omega)\text{ respectively }

step 4 will imply

\displaystyle u^\pm v^\pm\in W^1(\Omega),\quad D(u^\pm v^\pm)=v^{\pm}Du^\pm+u^\pm Dv^\pm

So

\displaystyle uv= u^+v^+-u^+v^-- u^-v^++u^-v^-\in W^1(\Omega)

and

\displaystyle D(uv)=uDv+vDu

\Box

 

Remark: Who can simplify this proof? It is ugly.

Regularity of Newtonian Potential

Suppose {\Gamma(\cdot)} is the fundamental solution of laplace equation.

\displaystyle \Gamma(x-y)=\begin{cases}\frac{1}{n(2-n)\omega_n}|x-y|^{2-n}, n>2\\\frac{1}{2\pi}\log|x-y|, n=2.\end{cases}

We know it has property

\displaystyle |D_{y_i}\Gamma(x-y)|\leq C\frac{1}{|x-y|^{n-1}}

We can define the Newtonianial potential of {f}

\displaystyle Nf(x)=\int_{\Omega}\Gamma(x-y)f(y)dy\quad x\in\mathbb{R}^n

{Nf} is well defined. We have the following propery of {Nf}

Thm: Suppose {\Omega\subset\mathbb{R}^n} is a bounded domain, {n\geq 2}. Assume {f} is bounded and local integrable in {\Omega}, then {Nf} is {C^1(\mathbb{R}^n)} and

\displaystyle D_{x_i}Nf=\int_{\Omega} D_{x_i}\Gamma(x-y)f(y)dy

Proof: Fix {0<\epsilon<1}, let {B_\epsilon(x)=\{y\in\Omega||x-y|<\epsilon\}} and {B_\epsilon^c(x)=\{y\in\Omega||x-y|\geq\epsilon\}}.

\displaystyle Nf=\int_{B_\epsilon(x)}\Gamma(x-y)f(y)dy+\int_{B^c_\epsilon(x)}\Gamma(x-y)f(y)dy=I+II

Since {D_{x_i}\Gamma(x-y)} is uniformly bounded in {B^c_\epsilon(x)}, by the Lebesgue dominating theorem, {II} is differentiable and

\displaystyle D_{x_i}II=\int_{B^c_\epsilon(x)}D_{x_i}\Gamma(x-y)f(y)dy

Considering {I}, we will prove if {|x-z|<\epsilon/2}

\displaystyle |I(x)-I(z)|\leq \begin{cases}C\epsilon|x-z|\quad\quad\text{ if } n>2\\ C\epsilon|\log \frac{\epsilon}{2}||x-z|\text{ if }n=2\end{cases}

where {C=C(||f||_{L^\infty},n,\Omega)}.

\displaystyle \int_{B_\epsilon(x)}\Gamma(x-y)f(y)dy-\int_{B_\epsilon(z)}\Gamma(z-y)f(y)dy=\int_{B_\epsilon(x)\cap B^c_\epsilon(z)}\Gamma(x-y)f(y)dy

\displaystyle +\int_{B_\epsilon(x)\cap B_\epsilon(z)}[\Gamma(x-y)-\Gamma(z-y)]f(y)dy-\int_{B^c_\epsilon(x)\cap B_\epsilon(z)}\Gamma(z-y)f(y)dy

{=\mathcal{A}+\mathcal{B}+\mathcal{C}}

For any {y\in B_\epsilon(x)\cap B^c_\epsilon(z)}, {|y-x|\geq |y-z|-|z-x|\geq \epsilon/2}. So

\displaystyle |\mathcal{A}|\leq C\frac{1}{\epsilon^{n-2}} |B_\epsilon(x)\cap B^c_\epsilon(z)|\leq C\frac{\epsilon^n-(\epsilon-|x-z|/2)^n}{\epsilon^{n-2}}\leq C\epsilon |x-z|\text{ if } n>2

\displaystyle |\mathcal{A}|\leq C|\log\frac{\epsilon}{2}| |B_\epsilon(x)\cap B^c_\epsilon(z)|\leq C\epsilon|\log\frac{\epsilon}{2}||x-z|\text{ if }n=2

Similarly for {C}, we also have

\displaystyle |\mathcal{C}|\leq\begin{cases}C\epsilon|x-z|\quad \quad \text{ if } n>2\\ C\epsilon|\log\frac{\epsilon}{2}||x-z|\text{ if }n=2\end{cases}

Considering {\mathcal{B}}, we have

\displaystyle |\mathcal{B}|\leq C\int_{B_\epsilon(x)\cap B_\epsilon(z)}\int_0^1|D_{x_i}\Gamma(tx+(1-t)z-y)||x-z|dtdy

\displaystyle =C|x-z|\int_0^1\int_{B_\epsilon(x)\cap B_\epsilon(z)}|D_{x_i}\Gamma(tx+(1-t)z-y)|dydt

\displaystyle \leq C|x-z|\int_0^1\int_{B_\epsilon(x)\cap B_\epsilon(z)} \frac{1}{|tx+(1-t)z-y|^{n-1}}dydt

\displaystyle \leq C|x-z|\int^1_0 C\epsilon dt\leq C\epsilon |x-z|

Combing the fact about {\mathcal{A},\mathcal{B},\mathcal{C}}, we get

\displaystyle |I(x)-I(z)|\leq \begin{cases}C\epsilon|x-z|\quad\quad\text{ if } n>2\\ C\epsilon|\log \frac{\epsilon}{2}||x-z|\text{ if }n=2\end{cases}

So

\displaystyle \left|\frac{Nf(x)-Nf(z)}{x-z}-\int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy\right|

\displaystyle \leq \left|\frac{II(x)-II(z)}{x-z}-\int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy\right|+\left|\frac{I(x)-I(z)}{x-z}\right|

Applying the fact that {II} is differentiable and the fact about {I}

\displaystyle \overline{\lim\limits_{z\rightarrow x}}\left|\frac{Nf(x)-Nf(z)}{x-z}-\int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy\right|

\displaystyle \leq \left|\int_{B_\epsilon(x)}D_{x_i}\Gamma(x-y)f(y)dy\right|+\begin{cases}C\epsilon\quad\quad\text{ if } n>2\\ C\epsilon|\log \frac{\epsilon}{2}|\text{ if }n=2\end{cases}

\displaystyle \leq C\epsilon+\begin{cases}C\epsilon\quad\quad\text{ if } n>2\\ C\epsilon|\log \frac{\epsilon}{2}|\text{ if }n=2\end{cases}

Let {\epsilon\rightarrow 0}, we get {D_{x_i}Nf} exists and

\displaystyle D_{x_i}Nf=\int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy

Next we shall prove {D_{x_i}Nf} is continuous for {i=1,\cdots,n}, then {Nf\in C^1(\mathbb{R}^n)}

\displaystyle \left|\int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy-\int_{\Omega}D_{x_i}\Gamma(z-y)f(y)dy\right|

\displaystyle \leq\left|\int_{B^c_\epsilon(x)}D_{x_i}\Gamma(x-y)f(y)dy-\int_{B^c_{\epsilon}(z)}D_{x_i}\Gamma(z-y)f(y)dy\right|

\displaystyle +\left|\int_{B_\epsilon(x)}D_{x_i}\Gamma(x-y)f(y)dy\right|+\left|\int_{B_\epsilon(z)}D_{x_i}\Gamma(z-y)f(y)dy\right|

By the dominating theorem

\displaystyle \lim\limits_{x\rightarrow z}\left|\int_{B^c_\epsilon(x)}D_{x_i}\Gamma(x-y)f(y)dy-\int_{B^c_{\epsilon}(z)}D_{x_i}\Gamma(z-y)f(y)dy\right|=0

\displaystyle \left|\int_{B_\epsilon(x)}D_{x_i}\Gamma(x-y)f(y)dy\right|\leq C\epsilon

\displaystyle \left|\int_{B_\epsilon(z)}D_{x_i}\Gamma(z-y)f(y)dy\right|\leq C\epsilon

where {C=C(||f||_{L^\infty}, n)}. So {\displaystyle \int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy=D_{x_i}Nf} is continuous everywhere.

Remark: This revised version also have errors. It is a shame. What do you mean {II} is differentiable? Don’t you notice that the integral domain of {II} also has relation with {x}?

To prove this theorem rigorously and neatly, we need the following lemma from calculus

Lemma: Suppose {u_n{x}} is a sequence of differntiable function on {[a,b]\rightarrow \mathbb{R}}. If {u_n} converges to another function {u} uniformly({u} is finite somewhere) and {u'_n} converges uniformly to {v} in {\Omega} then {u} is differentiable and {u'=v}.

Proof: Let {\xi(r):C^1(\mathbb{R}_+^1)\rightarrow \mathbb{R}^1} and {0\leq\eta'\leq 2}, \xi(r)=0\text{ if } r\leq 1, \xi(r)=1\text{ if }r>2.

Define

{\displaystyle w_\epsilon=\int_{\Omega}\xi\left(\frac{|x-y|}{\epsilon}\right)\Gamma(x-y)f(y)dy}

Let {\xi_\epsilon(r)=\xi_{\epsilon}(r/\epsilon)} Since

\displaystyle \left|D_{x_i}\left(\xi_\epsilon\left(|x-y|\right)\Gamma(x-y)f(y)\right)\right|\leq \frac{C}{\epsilon^{n-1}}f(y)

which is in {L^1(\Omega)}. By the Lebesgue Differentiable theorem, {w_\epsilon} is differntiable and

\displaystyle D_{x_i}w_\epsilon=\int_{\Omega}D_{x_i}\left(\xi_\epsilon\left(|x-y|\right)\Gamma(x-y)\right)f(y)dy

\displaystyle \left|D_{x_i}w_\epsilon-\int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy\right|\leq \sup|f|\int_{\Omega}\left|D_{x_i}((1-\xi_\epsilon)\Gamma(x-y))\right|dy

\displaystyle \leq \sup|f|\int_{|x-y|\leq 2\epsilon}|D_{x_i}\Gamma|+\frac{2}{\epsilon}|\Gamma|dy

\displaystyle \leq \sup|f|\begin{cases}C\epsilon\quad \quad \text{ if }n>2\\ C\epsilon (1+|\log\epsilon|)\text{ if }n=2\end{cases}

So {D_{x_i}w_\epsilon} converges to {\displaystyle \int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy} uniformly on any compact subset of {\mathbb{R}^n}. And it is easy to prove {w_\epsilon} converges to {Nf} uniformly on any compact subset of {\mathbb{R}^n}. So by the lemma, we have {Nf} is differentiable and has

\displaystyle D_{x_i}Nf(x)=\int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy,\quad i=1,2\cdots,n

Measure on sphere

Considering an n-sphere, which is the unit sphere in \mathbb{R}^n, we usually encounter this kind of integration

\displaystyle \int_{\mathbb{S}^n}f(x)dS_x

here dS_x is the sphere measure. For example find the value of \displaystyle \int_{\mathbb{S}^n}x^+_ndS_x.

To calculate this integral, we have to understand dS_x. What is the relation to Lebesgue measure in Euclidean space?

The key point is viewing \mathbb{S}^n as a manifold, then dS_x is the volume element of \mathbb{S}^n, which can be written as dS_x=\sqrt{g}du_1\wedge du_2\wedge \cdots \wedge du_{n-1} if u_1,u_2,\cdots,u_{n-1} is the local coordinates of \mathbb{S}^n.

\displaystyle \psi: B_1(0)\mapsto \mathbb{S}_+^n

\displaystyle (u_1,u_2,\cdots,u_{n-1})\to (u_1,u_2,\cdots,u_{n-1}, \gamma)

where B_1(0)\subset \mathbb{R}^{n-1} and \gamma=\sqrt{1-u^2_1-u^2_2-\cdots-u^2_{n-1}}.

Suppose the metric of \mathbb{R}^n is g_0, then it induces a metric \psi^*g_0 on \mathbb{S}^n.

\displaystyle \psi^*g_0\left(\frac{\partial }{\partial u_i}, \frac{\partial }{\partial u_j}\right)=\delta_{ij}+\frac{\partial \gamma}{\partial u_i}\frac{\partial \gamma}{\partial u_j}.

Since (u_1,u_2,\cdots,u_{n-1}) can serve as a local coordinates of \mathbb{S}_+^n, then

\displaystyle dS_x=\sqrt{\psi^*g_0}du_1\wedge du_2\wedge \cdots \wedge du_{n-1}=\sqrt{1+|D\gamma|^2}du_1\wedge du_2\wedge \cdots \wedge du_{n-1}.

Or we can get from the volume form

\displaystyle \omega = \sum_{j=1}^{n} (-1)^{j-1} x_j \,dx_1 \wedge \cdots \wedge dx_{j-1} \wedge dx_{j+1}\wedge \cdots \wedge dx_{n}

and substitute x_{n} by \gamma.

So

\displaystyle \int_{\mathbb{S}^n}x^+_ndS_x=\int_{B_1(0)}\gamma\sqrt{1+|D\gamma|^2}du_1\wedge du_2\wedge \cdots \wedge du_{n-1}

which can be calculated by the polar coordinates.

Bound of norm on gradient implies bound on function

\mathbf{Problem:} Suppose p\in L^2_{loc}(B_1), \nabla p\in L^2(B_1), prove that p\in L^2(B_1) where B_1\subset\mathbb{R}^n is a ball with radius 1.

We are going to use the following lemma.

Lemma 7.16 on GT’s book. p162

Let \Omega be convex and u\in W^{1,1}(\Omega). Then

\displaystyle |u(x)-u_S|\leq \frac{d^n}{n|S|}\int_{\Omega}\frac{|Du(y)|}{|x-y|^{n-1}}dy,   a.e. \Omega

where \displaystyle u_S=\frac{1}{|S|}\int_S udx, d=diam\,\Omega.

\mathbf{Proof:} For any B_r\subset B_1 concentric with B_1, p\in W^{1,1}(B_r), apply the lemma 7.6

\displaystyle |p(x)-p_S|\leq \frac{r^n}{n|S|}\int_{B_r}\frac{|\nabla p(y)|}{|x-y|^{n-1}}dy\leq \frac{1}{n|S|}\int_{B_1}\frac{|\nabla p(y)|^2}{|x-y|^{n-1}}dy\int_{B_1}\frac{1}{|x-y|^{n-1}}dy

Since \displaystyle \int_{B_1}\frac{1}{|x-y|^{n-1}}dy is uniformly bounded for \forall x\in\mathbb{R}^n, then the above inequality implies that

\displaystyle |p(x)-p_S|\leq C\int_{B_1}\frac{|\nabla p(y)|^2}{|x-y|^{n-1}}dy

\displaystyle \int_{B_r}|p(x)-p_S|^2dx\leq C\int_{B_1}\int_{B_1}\frac{|\nabla p(y)|^2}{|x-y|^{n-1}}dydx\leq C\int_{B_1}|\nabla p|^2dy

Let r\to 1^+, we get ||p||_{L^2}\leq C(p_S+||\nabla p||_{L^2}).

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

\mathbf{Remark:} For any bounded domain with locally lipschitz boundary, this proposition is true. Because the locally boundary enable the domain to have locally star-shaped property, we can apply this proposition locally.

One example related to weak derivative

\mathbf{Problem:} Suppose D=\{(x_1,x_2)||x_1|<1, |x_2|<1\} is the open square in \mathbb{R}^2. u is defined

\displaystyle u(x)=\begin{cases}1-x_1\quad if\quad x_1>0, |x_2|<x_1\\1+x_1\quad if\quad x_1<0, |x_2|<-x_1\\1-x_2\quad if\quad x_2>0, |x_1|<x_2\\1+x_2\quad if\quad x_2<0, |x_1|<-x_2.\end{cases}

Find the first weak derivative of u.

\mathbf{Proof:} Suppose v=\partial_\alpha u is the weak derivative, then for any \phi\in C^\infty_0(D), we have

\displaystyle \int_D v\phi dx=-\int_D u\partial_\alpha\phi dx.\quad \alpha=1,2\quad (1)

WLOG, assume \alpha=1. Let us denote the four domains in the definition of u as D_1,D_2,D_3,D_4 respectively. Note that

\displaystyle \int_D u\partial_1\phi dx=\int_{D_1} u\partial_1 \phi dx+\int_{D_2} u\partial_1\phi dx+\int_{D_3} u\partial_1\phi dx+\int_{D_4} u\partial_1\phi dx\quad(2)

While

\displaystyle \int_{D_1} u\partial_1\phi dx=\iint_{D_1}(1-x_1)\partial_1\phi dx_1dx_2=\int_{0}^1\int_{-x_1}^{x_1}(1-x_1)\partial_1 \phi dx_1dx_2

\quad                      \displaystyle =\int_{-1}^1 \int_{|x_2|}^1(1-x_1)\partial_1 \phi dx_1dx_2=\int_{-1}^1\left[ (1-x_1)\phi\big|^1_{|x_2|} +\int_{|x_2|}^1 \phi dx_1\right]dx_2

\quad                       \displaystyle =-\int_{-1}^1(1-|x_2|)\phi(x_2,x_2)dx_2-\int_{D_1}\partial_1u\,\phi dx\quad (3)

obtained from the integral by parts and the boundary behavior of \phi. Similarly for domain D_2

\displaystyle \int_{D_2} u\partial_1\phi dx=\iint_{D_2}(1+x_1)\partial_1\phi dx_1dx_2=\int_{-1}^0\int_{x_1}^{-x_1}(1+x_1)\partial_1 \phi dx_1dx_2

\quad                      \displaystyle =\int_{-1}^1 \int^{-|x_2|}_{-1}(1+x_1)\partial_1 \phi dx_1dx_2=\int_{-1}^1\left[ (1+x_1)\phi\big|_{-1}^{-|x_2|} -\int_{-1}^{-|x_2|} \phi dx_1\right]dx_2

\quad                      \displaystyle =\int_{-1}^1(1-|x_2|)\phi(-x_2,x_2)dx_2-\int_{D_2}\partial_1u\,\phi dx\quad (4)
As for the domain D_3 and D_4, we have
\displaystyle \int_{D_3} u\partial_1\phi dx=\iint_{D_3}(1-x_2)\partial_1\phi dx_1dx_2=\int_{0}^1(1-x_2)\int_{-x_2}^{x_2}\partial_1\phi dx_1dx_2

\quad                      \displaystyle =\int_{0}^1(1-x_2)[\phi(x_2,x_2)-\phi(-x_2,x_2)] dx_1dx_2\quad (5)

\displaystyle \int_{D_4} u\partial_1\phi dx=\iint_{D_4}(1+x_2)\partial_1\phi dx_1dx_2=\int_{-1}^0(1+x_2)\int_{x_2}^{-x_2}\partial_1\phi dx_1dx_2

\quad                      \displaystyle =\int_{-1}^0(1+x_2)[\phi(-x_2,x_2)-\phi(x_2,x_2)] dx_1dx_2\quad (6)
Substituting (3-6) to (2)

\displaystyle \int_D u\partial_1\phi dx=-\int_{D_1}\partial_1u\,\phi dx-\int_{D_2}\partial_1u\,\phi dx.

So from (1), we know

\displaystyle v=\partial_1 u=\begin{cases}-1, x\in D_1\\1,x\in D_2\,\\ 0,x\in D_3\cup D_4. \end{cases}

As for \partial_2u, one can get it in a similar way.

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

\mathbf{Remark:} Evans, partial differential equation. 5.3