**Theorem 1** Let with , such that , where , if ; for all , if . If satisfies

then .

Before proving this, let’s recall the definition of (Hardy – 1 space).

Definition 2 (Hardy space)Fix a smooth function with . Set . We say a function if and .

*Proof:* It is obvious that by Holder’s inequality. It remains to show .

First notice that in . To see this, one can mollify and get a class of divergent free smooth vector fields . For any , we have

Then we have

where is the average value of in . Now choose such that

By Holder’s inequality, we have

We further employ Poincare-Sobolev inequality, and adopt the notation for maximal function, we have

Recall that is a strong- operator for , by Holder’s inequality, we have

since .

Here is an application. Let’s consider a stationary Euler system in :

Let for , and the pair satisfies the system. We claim that is continuous.

From the system, we know satisfies the equation weakly. By using a similar mollifying argument as above, one can show that

We can apply the theorem above since and is divergent free in distribution sense. Therefore . Note that we are in 2 dimension, can be rewritten as up to some harmonic function. Recall that the dual space of is , and , we can conclude is continuous since translation is continuous in norm.

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Theorem 1 (Brouwer fixed point theorem)Let be a convex compact set, for any continuous function , there exists a point such that .

There are couple of ways to extend this theorem to Banach spaces. First recall that a mapping between two Banach spaces is called **compact**, if the mapping is continuous (not necessarily linear) and the images of bounded sets are pre-compact.

Lemma 2Let be a Banach space, be a bounded nonempty set, and be a compact operator. Then there exists a sequence of continuous operator such that

*Proof:* Since is relatively compact, there exist balls of radius that cover , with for all . Consider defined by

One can easily check the denominator has a positive lower bound, and both numerator and denominator are continuous (since is continuous and the composition of two continuous functions is continuous). Hence is continuous. One can also easily check that the denominator is positive only when for some , i.e. . Therefore

Theorem 3 (Schauder fixed point theorem)Let be a Banach space, be a bounded closed convex set. If is a compact operator, then there exists an such that .

*Proof:* Let be the sequence of continuous operator from the previous lemma. Let be the convex hull of , the center of covering balls. Set , then by the definition of , one can see maps into itself. Notice that is finite-dimensional, convex and compact, we can apply Brouwer fixed point theorem to obtain a fixed point for , say .

Since and is pre-compact, there exists a subsequence, still denote by , that converges to . Then

Remark 1From the proof we can see the boundedness of and compactness of have only been used to deliver the pre-compactness of . So if we know is pre-compact, we can drop those 2 conditions, and of course, still needs to be continuous.

Theorem 4 (Leray-Schauder fixed point theorem)Let be a Banach space, be a compact operator from into itself, and suppose there is an a-priori boundfor all and satisfying . Then has a fixed point.

*Proof:* Define a mapping by

Then is a continuous mapping. Since is compact, so is (one can test against a bounded sequence to see this). By Schauder fixed point theorem, there exists a fixed point for . It remains to show is also a fixed point for . Assume , then , with . And we will have , which contradicts the a-priori estimate. Therefore and .

Remark 2None of the above fixed point theorems give uniqueness of fixed point, which is different from Banach (contracting) mapping theorem.

Remark 3If one can reduce an equation to a fixed point problem of a compact operator (especially for integral equation), an a-priori estimate of the type above will give the existence of solution through Leray-Schauder fixed point theorem, but not the uniqueness!

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**Definition 1** Let . Define

where .

Then is called d-dimensional Hausdorff measure in .

Remark 1Hausdorff measure is a Borel outer measure, it is not Radon for because is not finite with respect to . And it is a measure if restricted on Lebesgue measurable sets (by Caratheodory condition: We say a set satisfies Caratheodory condition, if for any set , ). Moreover, note that gives the volume of unit ball in dimension if is an integer, we naturally have (not trivially) , where is the n-dimensional Lebesgue measure.

Here are some basic properties of Hausdorff measures:

Theorem 2on , for all . And hence on .

*Proof:* Note that , we have for .

Denote by for . Then

So for all , hence on .

Theorem 3on for .

*Proof:* Let be a unit cube in , we can divide it into cubes with side-length , diameter . So

Send we will have , hence on .

Theorem 4Let , , then

- If , then .
- If , then .

*Proof:* Note that the second statement is the contraposition of the first one. To prove the first statement, fix a , there exists a cover of , with , such that

Then we have

Send we will have .

Having this theorem, we know that for any set there exists only one such that is neither nor . It is natural to define this to be its dimension.

Definition 5We say the Hausdorff dimension of a set is

Note that the Hausdorff dimension need not be an integer. Even if is an integer, it doesn’t mean is a -dimensional surface in any sense.

Another important and related measure is called capacity, and it is defined as follows.

Definition 6Let , we define the p-capacity of by

Obviously, by density we can consider instead of .

It’s not hard to see is an outer measure, by know that given a sequence of functions , denote by and , then -a.e. As we know, given an outer measure, one can construct a -algebra by Caratheodory condition such that it is a measure restricted to this -algebra. But in this case, we are not that interested in it because the -algebra only contains sets of capacity 0 or . And here are some properties of capacities that one can easily prove.

Theorem 7 (Properties of )

- open,
- , for .
- .
- .

From the scaling, we can see that the capacity does not measure the “volume” of the set. And according this scaling, it is natural to look for the relations between and .

Theorem 8Assume , if , then .

Note that this is a refinement of the third statement of last theorem for $latex {1

*Proof:* First we claim that there exists a constant such that for any neighborhood containing , there exists an open set and such that

To proof the claim, let be a neighborhood of and set . Since , we can find a sequence of ball with such that , and

Set

Then . Set and , then and

Now we use the claim inductively. We find a sequence of , from the claim, and relate them by . Set and . Then and on . Therefore

since are discrete and .

The next important relation is that, given , we will have some information on its Hausdorff dimension. Before proving it, we need a Lemma.

Lemma 9Let , suppose and defineThen .

*Proof:* From Lebesgue differential theorem one can see . Then by the absolute continuity of Lebesgue integral, there exists a and such that whenever . Fix a and an , we define a set

Since , we can fix an open set with , and consider a family of balls

Then this family of balls cover , and by Vitali’s covering theorem, there exists a sequence of disjoint balls such that . Hence

Then we can take to yield the result.

Theorem 10Let and . If , then for all .

*Proof:* Since , we can choose a sequence of functions such that . Define , by GNS inequality we know is well defined in , and for any . So , the average of on , goes to 0 as if . Now I claim that

Assume by contradiction, there exists an such that , by Poincare inequality, we know that

for small enough. Then

for some . This means converges as , which leads to a contradiction. Therefore, , and by the previous lemma we know since .

For case, we can say more about the relation between capacity and Hausdorff measure since we will have more geometric insights. For instance, we can take advantage of isoperimetric inequality and co-area formula to prove the following result.

Theorem 11Let , then if and only if .

*Proof:* From Theorem 8 we know that implies . Assume , for any , we can find an such that and . By co-area formula, we know

This implies and for some . Fix this , by isoperimetric inequality, one can see . Obviously we know , then for each , we can find an such that

Since is a small portion in , by relative isoperimetric inequality we have

Since these balls cover , by Vitali’s covering theorem, there exists a sequence of disjoint balls such that . From the estimates we get above,

And we know that for each . Hence . Send we will get the result.

Remark 2In the previous theorem, we actually prove the smallness of is equivalent to the smallest of , which we don’t have for .

The next important result is fine property of Sobolev functions. Comparing to Lebesgue integrable functions, Sobolev functions have a smaller singular set, and their representatives enjoy a better continuity result. To show it, we first need a Chebyshev type inequality.

Lemma 12Let and , define the setThen

*Proof:* First note that, without loss of generality, we can assume by homogeneity. For each , we consider a ball such that . This collection of balls cover , and by Besicovitch covering theorem, there exist and countable collections of disjoint balls , such that

Denote by the element in for . For each , consider the function , which is in , on . By extension theorem we know this function can be extended to , say , such that

for all , where depends only on . Then we have in for all . Define , then since is open. Therefore

Now we are ready to prove the fine properties of Sobolev functions.

Theorem 13Let . Then there exists a Borel set such that , and exists for any . In addition,for all . Furthermore, given any , there exists an open set such that is continuous restricted on .

*Proof:* For each , choose an such that . Define the set

From Lemma 14, we know . Consider the set

then by Theorem 9 and Lemma 10, we have . For any , by Poincare’s inequality we have

Then for any , we have

Define , then for , we have . This implies converges uniformly to a continuous function in . So

as , which implies . Define , then

And we know that exists for all . In addition,

For the continuity statement, given any , we can find an such that . From theorem 8, we can find an open set such that . Then we are done since we know is continuous restricted on .

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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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