in . Assume we have Harnack inequality (Indeed we do have due to Moser), i.e. for any positive weak solution we will have

where is a constant independent of .

First one will know , see this note.

If is a weak solution, so are and . And they are strictly positive in due to the strong maximum principle, otherwise will be a constant. By Harnack inequality, we will have

Adding these two inequalities up we will have

hence

Denote , and choose such that , by interation we know

Then for any , choose such that , we will have

So Holder continuity follows.

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Definition 1Let be a bounded open set, we define the Newtonian potential of is the function on bywhere is the fundamental solution of Laplace’s equation.

We need to check if is well-defined. If , then one can easily see . We can further relax to be in , where . Since the Holder conjugate , simply by Holder’s inequality, for any

Therefore is well-defined in the whole space. If we only require to be defined in , we only need to be integrable plus a little bit local boundedness near the singularity , for example, local Holder continuity or so. In this case, we can see is also well defined outside , but not defined on . So this case will become less interesting when we try to solve Dirichlet problem through Newtonian potential, since the function on boundary needs to be defined.

**— 2. Regularity of Newtonian Potential —**

Theorem 2 (Representation for first derivative)Let be bounded and integrable in . Then and for any ,

*Proof:* Define a function

Since , we know is well-defined. We want to show .We fix a function in satisfying for , for . And define for small,

where . Clearly since we make the singularity vanished. And we have

Therefore uniformly in compact subsets of as . One can also easily see that uniformly as . Therefore, and

Theorem 3 (Representation for second derivative)Let , with , and also bounded. Then , in , and for any ,where is any domain containing for which the divergence theorem holds and is extended to be 0 outside .

*Proof:* The proof follows from the same strategy as the previous theorem. We define to be the RHS of the equation stated in theorem. Since is locally Holder continuous, the integrand is integrable near the singularity, hence is well-defined. Let be the same as defined in last theorem, then define

We will have

provided . Hence, by subtraction

Therefore converges to uniformly on any compact subset of as . One can also easily see that converges uniformly to in . Therefore and .Finally, setting , we have for sufficiently large ,

But we can estimate

for any . Therefore and hence .

Remark 1If the local Holder continuity of is replaced by Dini continuity, we still have the same result. And this theorem suggests that we can use Newtonian potential to get a solution of Poisson equation.

Theorem 4 (interior Schauder estimate)If , then for any . Furthermore,

*Proof:* Without loss of generality we can assume and . Take , from the representation we have, we know

If we set , We can split the RHS into 6 parts, namely,

where

We will estimate all these 6 terms. Since , we have

And should have a same estimate by this argument. To estimate , notice that if , , and if , . And we will have by divergent theorem

To estimate , note that is smooth when , by mean value theorem, we know for some between and . And for , we always have . Then we will have

Since on , we have trivial estimate for :

To estimate , we again have mean value theorem as in estimate of , and since and are bounded, we have

Combining these 6 estimates we can conclude . From the previous theorems we can bounded the norm of and its gradient by . Therefore we have the inequality

This tells us if and , then . And the following example tells us if only continuous, then is not necessarily ( cannot be 0).

Remark 2If is only continuous, is , but not necessarily .

An example: For with , let be a homogeneous harmonic polynomial of degree 2 with . Choose with when . Define Notice that , when or . Therefore

And since vanishes on the boundary of each annulus and as , we can easily see that is continuous.

\noindent Now for such that define . Then by a similar argument we can see is well-defined and smooth away from . For any , we can find an such that , and Notice that

Since is not , as , we can see is not around . Noticing that

since is homogeneous. We can see as , Also as , therefore stays bounded when . Now suppose is a solution of . Then for . Since is bounded around 0, is a removable singularity for , which makes it a harmonic function. But this implies , which leads to a contradiction. So there is no solutions in any neighborhood of the origin to .

**— 3. Solution to Dirichlet problem —**

Theorem 5Let be a bounded domain and suppose each point of is regular (in the sense that Perron’s method will give a harmonic function prescribed continuous boundary value, e.g. is .) Then if is bounded and locally Holder continuous, the classical Dirichlet problem : in , on is uniquely solvable for any continuous boundary values .

*Proof:* From Theorem 2 and 3, we know that , in . Consider , then the Dirichlet problem is equivalent to in , on , which has a unique solution by Perron’s method.

The regularity of can be relaxed as following theorem.

Theorem 6Let be as described above, for some and is local Holder continuous then the classical Dirichlet problem can by uniquely solved.

*Proof:* From section 1 we know is well-defined on . And we need to check . We fix a function as in Theorem 2. And define for small,

where . Clearly . Extending to be 0 outside , then

Hence uniformly as , and thus . Then follow the proof of Theorem 2 one can see . Note that we might not have anymore since we lose boundedness of near the boundary of . Then we can follow the exact same proof of Theorem 3 and Theorem 8 to get the result.

Remark 3In order to solve Dirichlet problem, we can’t assume for some , otherwise might not be well-defined on boundary.

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**Theorem (Analytic Fredholm Alternative)** Let be a connected open subset of and be a separable Hilbert space. Suppose that is an analytic operator-valued function such that is compact . Then either

- does not exist .
- exists , where is discrete.

*Proof:* We will prove it in 6 steps. First we will prove the result holds in the neighborhood of any , then by a connectedness argument to get the result on all .

Step 1:

Let , there exists such that if , then . Since is compact, there exists a finite rank operator such that . This implies for all . So exists and is bounded and analytic on .

Step 2 (Relation between and ):

Let , then one has

- is invertible iff is invertible.
- has non-zero solution iff has non-zero solution.

Step 3 (Analysis of ):

Since is finite rank, there exists orthonormal vectors such that where . By Riesz representation, there exists vectors such that . Therefore . And we have

Denoting , we get that .

Step 4:

If , then and for , and some . This is equivalent to say

Since is analytic on , is also analytic. This implies the set is either all of or a discrete set in . If , then the first alternative in the theorem holds.

Step 5:

Suppose , we will show that is invertible. Indeed, is injective by step 4. By Inverse Mapping Theorem it remains to show it is also onto. Let , we need to solve the equation for some . Let , then has to solve

But ran . So for some . Substitute back and use the formula for , we get . Therefore

Since , this inhomogeneous system has a solution. This implies is invertible if .

Step 6:

We have proved for every , there is a neighborhood where the Analytic Fredholm Alternative holds. Define the sets there exists a neighborhood around where the first alternative holds\}, there exists a neighborhood around where the second alternative holds\}. Therefore and . By the definition of and , we know both sets are open. Since is connected, we know either or . This concludes the proof.

Corollary 1 (Fredholm Alternative)If is compact, nonzero, then either exists or has a nonzero solution.

*Proof:* Apply the theorem to with at

Corollary 2 (Riesz Schauder Theorem)Let be a compact operator on a separable Hilbert space , then the spectrum is a discrete having no limit points except perhaps at . Furthermore, any nonzero is an eigenvalue with finite multiplicity.

*Proof:* Let , define the set . Since , by the Analytic Fredholm Alternative, is discrete. If , then exists so exists. Hence . Therefore .

The fact that nonzero eigenvalues have finite multiplicity follows from compactness. Otherwise we get infinite orthonormal set of eigenvectors with eigenvalue , and it have no convergence subsequence.

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**Theorem** Suppose is a subsolution of , . Then for all , ,

Remark 1Without loss of generality one can assume . Since if is a subsolution, is also a subsolution. Since the result is scaling invariant, one can assume . Further more one can assume , if we have , for any , any , we can take a point such that . ThenLet , we can derive the inequality for any .

LemmaLet be a nonnegative bounded function on , if there exist , nonnegative constant such thatfor all . Then

*Proof:* Set , for some to be chosen later. Then . So

After iterating we will have

Now we can choose such that . Let ,

Remark 2We only need to prove for case , and use the above lemma to recover the cases when . That is because for any , ,Set and . Apply the above lemma we will have

*Proof of the theorem:* By the above remarks, it suffices to prove the case when . Since

for all . Take , where is a cut off function. Then we have

which implies

By ellipticity, we have

By Holder inequality we will have

Since , we can derive , which implies

By Sobolev inequality, we have

Now let , so and as . Take , , and on . Then . Therefore

Take , Then

Therefore

After iteration, we will have

Since we can take and get

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**Theorem** Let , satisfies

If there exists such that

If as , then in .

*Proof:* Without loss of generality assume on , or we can add a large constant on to make it positive on boundary. , there exists such that

Since , we have

Also on since is non-negative on boundary. Then maximum principle tells in , where . Then we can first take and then take to get in .

RemarkThe condition is essential, otherwise one can take plus a large constant as a counterexample.

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**Theorem** Let be a closed ideal of a C* algebra , then for any self-adjoint element , there exists an such that

*Proof:* Since is a closed ideal, is also a C* algebra and one can define a natural projection homomorphism , with . Fix an element , define a continuous function as follow:

Let , then is identical on . By continuous functional calculus, . And since can be approximated by polynomials, we know . The spectral radius , so we know is identical on the spectrum of , which implies . Therefore is in the kernel of , and hence in the ideal . Since and by the Gelfand Naimark isomorphism theorem, we know and hence . So is the we are looking for.

Note: this is a homework exercise given by Dr. Carlen.

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DefinitionA function , if , and the nth order distributional derivative is a finite Radon measure.

TheoremIf , then has a continuous representative.

*Proof:* Note that is dense in , i.e. for any , there exists a sequence such that in , and . (See for example, book by Evans and Gariepy). So it suffice to show for all ,

Indeed,

This gives us the desire inequality. Then by density result we know that admits continuous representative for every element.

Corollary.

RemarkIf the dimension is greater than the order of BV function, then the above theorem will fail. For example, let be defined in a neighborhood around 0, with smooth boundary in . and . So . Then we can extend on , since is smooth (See for example, PDE book by Evans). But then as , which doesn’t admit a continuous representative.

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**Lemma **In a metric space , if is a sequence of compact sets such that for and diam. Then , where is a point in .

*Proof:* First note that cannot contain more than 1 points. If , then . Since diam=0, it forces .

It remains to show is nonempty. Assume it is empty, then . Since metric space is Hausdorff, is closed for all and is an open cover for . Since is compact, there exists finite subcover such that , which contradicts .

TheoremIn a metric space (no matter it has countable basis or not), let be a subset in . The following statement are equivalent by assuming axiom of choice:

- is compact;
- is sequentially compact;
- is complete and totally bounded.

I will prove , and then .

*Proof:* :

Let be a sequence in , we want to show that it has convergent subsequence. Denote by the ball centered at of radius . Then , by compactness . Therefore there must exist a for some that contains infinite elements of . Denote by the intersection of that ball and . Use balls of radius to cover , again there exists a such that contains infinite elements of . Repeating this process we will have a sequence of sets . Pick an element from in each , we will have a cauchy subsequence, still denote by . Since for all , and is compact, then is compact for all . By the above lemma, for some . Therefore we know the cauchy subsequence converges to .

:

If is sequentially compact, every Cauchy sequence in converges to a point in , which means is complete. To show is also totally bounded, assume it is not, then there exists an such that cannot be covered by finitely many open balls of radius . Pick an , and pick but not in . And then pick that is not in . Since cannot be covered by finitely many balls of radius , we can find a sequence of open balls . Notice that

the sequence will not have convergent subsequence. So must be totally bounded.

:

If is complete and totally bounded. Given any sequence in , first we cover the set by finite open balls of radius , there must be one ball, say , contains infinite elements of the sequence. Then we cover the set by balls of radius , again there must be one ball, say , such that contains infinite elements of the sequence. Keep doing the process we will have a Cauchy subsequence of , and it converges to some point in since is complete. Hence is sequentially compact.

:

Assume we have an open cover for , we want to show there exists a finite subcover. Since is totally bounded, it suffices to show there exists an such that for some . This is because can be covered by finitely many , and hence can be covered by finitely many . Now assume for all , there exists an such that is not contained in any . Take for , we then have a sequence such that is not contained in any . By sequential compactness, there exists a convergent subsequence converges to a point . Since is an open cover, for some . Since is open, if we take large such that , , which contradicts our assumption. Therefore there exists an such that for some , and hence there is a finite subcover for .

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DefinitionIn a metric space, a set is totally bounded, if for any fixed , the set can be covered by finitely many open balls of radius .

LemmaIn a metric space, a set is compact if and only if is complete and totally bounded.

Proof of the lemma can be find here.

Theorem 1 (Kolmogorov Riesz Theorem (finite measure domain))Let be a bounded set of functions, Assume that , there exists a such thatwhenever , for all . Then has a compact closure in for any finite measure set .

*Proof:* Since is complete, by the above lemma we need to show is totally bounded. The idea is first find a bounded domain with almost same measure, use Arzela-Ascoli theorem to get compactness for , then we have finite open balls covering . Enlarge those balls a bit to get a finite cover for .

Let be a sequence of mollifiers with supp. Fix , also fix such that , we have

, for any . Direct verification shows for any ,

which implies

By (2) and the inner regularity of measurable sets, we can find a bounded measurable set such that . Therefore

By (2) and (3) we know the functions in the set are bounded and equicontinuous, so we can apply Arzela-Ascoli theorem and hence has a compact closure in . This means for every sequence in , there exists a uniformly convergent subsequence. Since the subsequence also converges in , we know that has a compact closure in . Therefore, can be covered by finite ball of radius in , say . i.e.

, where . Now define

Finally I claim that . For any ,

Choose such that , along with the estimate (1) and (4), we have

This proves that is totally bounded, hence has a compact closure in .

Theorem 2 (Kolmogorov Riesz Theorem (infinite measure domain))Let be a bounded set of functions, Assume that , there exists a such thatwhenever , for all . Furthermore, , there exists an with finite measure such that

Then has a compact closure in .

*Proof:* By the previous theorem, fix , . Define

Then by condition (5) and similar calculation in the previous theorem, .

RemarkThe condition (5) for infinite measure domain is essential. A counterexample without (5) would be as following.Let be any function, with supp. Let , and . So is a bounded set,whenever . However,

if . This implies does not have any convergent subsequence. Hence does not have compact closure in .

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**Definition **Let be a banach space, be the space of bounded linear operator from . If , then say a complex number is in the resolvent set of if has a bounded inverse. Then we define the resolvent of at to be . If , then is said to be in the spectrum of T.

We will learn some properties of the resolvent and the spectrum.

Lemma 1is a banach algebra under composition.

*Proof:* Let , then

Lemma 2Let be a banach algebra with identity . If with , then exists.

*Proof:* Since is a banach algebra, we have . Then is absolutely convergent in . Note that

Let we will have . Therefore .

Theorem (First resolvent identity)If , then and commute. Furthermore,

*Proof:*

Since we can interchange and , and commute.

Theoremis an open subset of . Fixed , is an analytic – valued function.

*Proof:* If , first define . From the previous lemma, we know that if , the series defines the inverse of . Therefore,

if . So is open. Since can be expressed as infinite series as , it is analytic.

Now we are going to show the spectrum of is non-empty (actually works for any complex banach algebra), which gives a nice corollary by Gelfand and Mazur. Before proving that, we need a banach-valued version of Liouville theorem:

Theorem (Liouville theorem for banach-valued function)Let be a banach space. If is a function from , which is entire and bounded, then is constant.

*Proof:* First note that if , then the function is a complex-valued analytic function.

Pick a and let . For any , we have is complex-valued analytic function and bounded in . Then by Liouville theorem is a constant, and , therefore for all .

Assume , by Hahn-Banach theorem there exists a nonzero linear bounded functional such that . By contraposition this implies , and hence is a constant.

**Remark**: There might be a proof without using axiom of choice. I just don’t know it yet.

Theoremis a non-empty compact subset in .

*Proof:* First define . By the previous lemma again, if , then defines the inverse of . Therefore is a bounded subset. Since is open, is closed. Hence is compact.

To see is non-empty. Assume it is, then is entire by the previous theorem. For , , and goes to 0 as goes to . Then by Liouville theorem, , which is not invertible. This gives a contradiction.

**Remark**: As mentioned before, this argument works for any complex banach algebra.

Corollary (Gelfand-Mazur)Every complex banach division algebra is an isometric isomorphic to .

*Proof:* Since is a division algebra, the only non-invertible element is 0. By the previous theorem, we know that for any , there exists a such that . Then . So the map gives an isometric isomorphism from to .

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