Quick aside. Does anyone know how to allow WordPress to understand user-defined Latex commands? [Edit: Professor Tao informed me how]

My hang up is everything looks better fixed-width, but WordPress cuts off some of the equations when I have a fixed-width theme. I don’t know if that problem was particular to the theme I chose, or whatever, but because of it, I’ve been forced into this less nice flexible-width theme.

[Edit: I have finally fixed the known errors in this post. I actually made the corrections at least a month ago, but out of sheer laziness, did not repost until just now.]

Exercise 7

We already know that is a vector space (with the conjugate complex structure). So we need to show that it’s Hausdorff and that vector addition and scalar multiplication are continuous. The proof for each of these statements hinge on being Hausdorff and that we endowed with the weak- topology.

Let be a convergent net in . Suppose that our sequence converges to two distributions, and . Then for each test function , , and in . with its usual topology is Hausdorff, so convergent sequences have only one limit. Since converges to both and , that must mean that they are equal. Therefore, for all test functions . Thus . Every convergent sequence in has only one limit point. Therefore is Hausdorff when endowed with weak- topology.

Now suppose that the net converges to in with the product topology. So we’re given that , and , we need to show that . For any test function , we know that , and . Since for each , , we therefore see that , which is . Since this holds for each test function , we see that in the sense of distributions. Vector addition is therefore continuous.

Now suppose that the net converges to in with the product topology. So we’re given that , and , we need to show that . For any test function , we know that , and . Since for each , , we therefore see that , which is . Since this holds for each test function , we see that in the sense of distributions. Scalar multiplication is therefore continuous.

is a Hausdorff topological vector space.

Exercise 8

Let be a locally integrable function, if is the Lebesgue measure, then is a regular measure and thus a Radon measure on . If we can show the identification of a complex Radon measure with distributions is injective, then since the identification of locally integrable functions with its induced regular measure is injective, we have shown that the identification of locally integrable functions with distributions is also injective.

Let be a Radon measure. induces a distribution defined on each test function by . Suppose we have two Radon measures such that for all test functions . By the Reisz Representation Theorem, for each (the space of all Radon measures on ), if we let , then the map is an isometric isomorphism between and . Since is a subspace of , we have on .

Now according to Part iv. of Exercise 1, is a dense subset of in the uniform topology. Thus and are continuous linear functionals on that agree on the dense subset . Therefore on all of .

Our map between and is an isomorphism, and thus . We see . The identification of Radon measures with their induced distributions is injective, and therefore, the identification of locally integrable functions with their distributions is also injective.

Exercise 10

Let be a sequence of approximations to the identity. Then each induces a distribution (which we’ll also denote , defined on each test function by . Note that since each is non-negative (and hence real), for each .

Fix any test function , and . Now is continuous, particularly at 0. Thus, there exists a such that whenever .

is bounded. Now use the same as above. Then for any ,

is a sequence of approximations to the identity, and therefore by definition, . Thus, there exists an such that if , then we have .

Then given any greater than the above , we see

Therefore, for any test function , . Thus, for each test function . converges in the sense of distributions to the Dirac distribution .

Exercise 16

The first part is trivial. For any test function , we see

Therefore, is identically 0.

The converse is not as straight forward. Fix a test function such that . Now for any test function , consider . It’s immediate that , and if , then

Therefore .

Now pick any test function .

Consider . Let . is a test function, and from our previous calculation, . Therefore

We combine our two calculations, and we see

Let , then since the above equality holds for each test function . We have that .

Exercise 18

We’re working in . Pick any distribution . For each positive integer , let , and let . By Part iii. of Exercise 1, there exists a test function such that is 1 on and vanishes outside . We claim that for each , the distribution is compactly supported, and the converge to in the sense of distributions.

Consider . Suppose the test function vanishes on . Now . vanishes on , so must as well. vanishes outside , so must vanish outside as well. So is identically 0 on all of . Distributions are linear functionals, and thus . for each that vanishes on . The distribution is supported on . The support of is thus a closed subset of the compact set , and therefore compact. Each is a compactly supported distribution.

Now given any test function , we claim . is compactly supported. So there exists a such that contains the support of . Then for any , consider . If lies in the support of , then also lies in , and therefore . If lies outside the support of , then . Therefore for all . Then for any greater than that same , we see . Therefore , and since our choice of was arbitrary, this must hold for all test functions. The converge to in the sense of distributions.

Every distribution is the limit of a sequence of compactly supported distributions.

Exercise 20

Without loss of generality, we’ll differentiate with respect to the -th variable, . Just apply the definition of the derivative to the left-hand side, and we see for each test function that

Now apply the definition to each of the summands on the left-hand side. So for any test function , we see

And for the other summand we see

Adding the last two sums we obtain

Since this equality holds for any test function we obtain our desired result

Since we didn’t use any property special to the -th variable, the above equality holds for each .

Exercise 22

i.Just do the calculations. For any test function ,

Thus .

ii. From Exercise 16, , the derivative of 0 is 0, thus . We can then apply the chain rule from the previous exercise, and see that . Therefore, .

iii. Let be a sequence of approximations to the identity. We want to show for any test function , that . Well, and each are smooth functions and therefore integration by parts is fair game, and we see that for each , . Now is a also a test function, and by Exercise 10, . Therefore .

By Exercise 10, the sequence converges in the sense of distributions to , and as we’ve just shown, the sequence converges in the sense of distributions to . Now differentiation is a continuous operation on the space of distribution. Thus, is the derivative of of .

Exercise 23

i. Given any test function , consider the following calculation, which makes use that the fundamental Theorem of Calculus applies to , since is differentiable in the classical sense. The calculation makes use that is compactly supported, and thus is 0 for all sufficiently large values.

Since this holds for all test functions, the derivative of is .

ii. Again, given any test function , and utilizing the same properties of as before.

Since this holds for any test function, the derivative of is .

iii.

iv.

v. For the last, time chose an test function , then

Now both and are differentiable, so integration by parts shows us

n The other thing we should note is that has compact support and thus vanishes outside a bounded area. Therefore

The derivative of as distribution is .

Exercise 24

Choose any distribution , and any test function . We immediately see that for any

and similarly

is smooth, and therefore . Thus we see

for each test function .

We therefore have . Weak derivatives commute.

Exercise 29

We need to show that if in , then . Actually, we’ll simplify our lives by showing that each element in the Schwartz class is absolutely integrable with respect to the measure . By doing that, then is automatically a linear map from to . If we show that, then we can assume without less of generality, that our sequence converges to 0.

Let lie in the Schwartz class. Let . Then . Now is rapidly decreasing, since it's Schwartz. So , thus , for all .

Therefore, given any .

Thus the series is term-wise dominated by the series , which converges. Since each term of the first series is non-negative, that must mean that the first series converges. And since , we see that . is absolutely integrable with respect to . Since our pick of was arbitrary, each Schwartz function is absolutely integrable with respect to . For the reasons we stated before, the map is linear.

So assume that in . By definition, that means that for each non-negative integers , we have . Thus, given any , there exists an such that if , then ,

Since is just some constant that doesn’t depend on , that means .

Similarly, tends towards 0 as goes to . And since

we see that is 0.

Now since for each , we can split into the sum , and the limit of each summand as goes to infinity is 0, we see . Since we defined to be , we see that whenever in the Schwartz space. Thus the map is continuous at 0, and since the map is linear, it’s continuous everywhere. Any Radon measure of polynomial growth induces a tempered distribution.

Exercise 31

Remember, for a tempered distribution , that we define for each function in the Schwartz class. We define similarly. Also note that on the Schwartz space, that . These exercises are very quick when one is familiar with Notes 2.

i. We simply do the calculation. Given a tempered distribution , then for any function in the Schwartz space, we see

Thus . An identical calculation shows . Therefore, for each tempered distribution , we see .

ii. Calculating the left-hand side gives us

For any Schwartz function , let . The right-hand side gives us

Now we have to deal with that messy expression. We can be clever and avoid having to see any messy integrals First we note that

Next thing we note, using the trick that complex conjugation and integration commutes. That is, for any two Schwartz functions and , we see

The last few things we note is that for two Schwartz function, , and that for any Schwartz function, we have .

We have all we need

Thus we see that for all Schwartz functions ,

Therefore, .

iii. Well, for any Schwartz function ,

For that Schwartz function ,

and thus,

Therefore, for each Schwartz function ,

Now we note that given any ,

Thus, for any Schwartz function ,

Thus, for any Schwartz function , we see that

Therefore, .

Similarly, for any Schwartz function ,

On the other hand, for any Schwartz function ,

For any Schwartz function ,

Thus, . So for any Schwartz function ,

Therefore .

iv. First note, that is a linear operator on , and as a result, the adjoint is simply the transpose of . In general, our life is greatly simplified, since is real, and for real invertible matrices, taking the adjoint and taking the inverse are operations that commute. Also note that the determinant of the transpose is the same as the determinant of the original operator. Well, for any Schwartz function ,

Computing form the other direction, we see

We should first note, that direct calculation shows .

From Exercise from Notes 2, we know how the Fourier Transform under changes of variables. In particular, if is an invertible linear transformation, and is a Schwartz function, then

We again use the identity for each Schwartz function to see for any

So returning to our first calculation, we see

We have obtained our desired result.

v. Exercise 23 and 24 of Notes 2 are key here. The former says that for any Schwartz function ,

Similarly, the latter exercise says that for any Schwartz function ,

Using the identity, , we see that

and

So given any Schwartz function and distribution , we see

Therefore .

Similarly, given any Schwartz function , we see

Therefore, .

I’ll try and continue to add to this solutions list. I make no claims of accuracy, and any and all helpful comments are welcome.

I found out that doing all the exercises a very time consuming task, and this is a significant lowering of the bar, but I’ll proably have to resign to just giving partial solutions or outlines for most exercises.

Exercise 1

Let be a locally compact Hausdorff space that isn’t compact. Let be the collection of subsets of such that if and only if is an open subset of or is a compact subset of . The empty-set is an open subset of and therefore in . It is also a compact subset of and therefore is in .

Pick two sets . If both are open relative to , then is open relative to and thus lies in . If both and are compact subsets of , then so is , and therefore is in . If is an open set of , and is a compact subset of . Then since and does not contain . Well, is Hausdorff and therefore is closed relative to , and thus is open relative to . must then be open relative to and therefore lies in .

Let be a collection of subsets of . Then where if is open relative to , and if is a compact subset of . Let , then is an open subset of . Let , then . Each is closed relative to since the space is Hausdorff. Thus is closed, and is a subset of a compact set, and therefore compact. So we know that lie in . since does not contain . is a closed subset of , and therefore is a closed subset of the compact subset . is a compact subset of , and therefore is in . contains the empty-set and , it is closed under countable intersections and arbitrary unions. is a topology on .

Consider the map where for all . Let be open in . Suppose is an open subset of , then is open relative to in the subspace topology inherited from . , and is open. Now suppose that is a compact subset of . Then is an open set relative to as a subspace of . Then and is therefore closed, thus , is open. Since there are only two types of open sets in , there are only two types of open sets in when considered as a subspace of , and for both types in , the inverse image under is open. is continuous. Now suppose is open in , then by construction, is open in , and , so is continuous. is an embedding.

is an open subset of . Now here’s where we use that isn’t compact. Suppose that is an open set in , then since isn’t an open subset of , the complement of is a compact subset of , but the complement of is , and by assumption, isn’t compact. Therefore every open set containing contains a point in it that isn’t , that said point must lie in . is a limit point of , as is every point in . is a dense open subset of . If is compact, then when we consider , is an open neighborhood around , and therefore isn’t dense in . The one point compactification process fails if is compact.

We’re almost there, we just need to show that is Hausdorff and compact. We already know we can separate any two point in with disjoint open sets because any set that’s open in is open in . Pick some . Since is locally compact, there exists a compact set such that . Then the complement of in , , is an open set in that contains , and is empty. We were able to separate and any point in by disjoint open sets. is Hausdorff. Now let be an open cover in . lies in some . Remove from our collection of open sets, and let . Then our modified collection is an open cover of the . Now is a compact subset of . consider the collection of sets where each is equal to . We’ve already shown in this proof that each is open in . Therefore the is an open cover of the compact set . There exists a finite sub-cover . Then is a finite sub-cover of . Every open cover of gives arise to a finite sub-cover. is compact.

We’ve shown an embedding where is a compact Hausdorff space that has as a dense open sub-set. is a compactification.

Let be some compactification, and let be the one-point compactification. We want to show that is finer than . We need to find a continuous map such that . We’ll define as follows. If , then let where is the element in such that . And if is not in , let . Then .

Now to show that it is continuous. Let be an open set in . Suppose is not in . Then . , and is therefore open relative to . is a homeomorphism between and , and thus maps open sets to open sets. Therefore, is open relative to , and since is open relative to , is open in . Now suppose . Now is closed in , and does not contain . By the definition of the one-point compactification of , is a compact subset of the space . Now . is continuous, and thus maps compact sets to compact sets. is a compact subset of the Hausdorff space , and therefore closed. is closed. Therefore is open, and that’s equal to . We’ve exhausted our cases. is open whenever is open in . We have found our desired continuous function from to . is courser than , and since was an arbitrary compactification, we see that the one-point compactification is courser than any other compactification.

Exercise 2

Consider the continuous function , and suppose that there exists a continuous extension of to a function . Let and from to be continuous extensions of . Pick some . can be considered an open dense subset of , so pick some net in that converges to . a continuous function on , and since we have . Similarly, we have . and agree on , and therefore we have the net converging to and . is a Hausdorff space and thus every net converges to at most one point. So we see that . and agree everywhere. If an extension of exists, it is unique.

Now to show that an extension always exists. Consider the graph . consider the closure of in . Note that is a dense open set in . Let be a limit point in , that is, there exists a net that converges to . Similarly, let be another limit point of , and let be a net in that converges to . NEED TO SHOW THAT AND THUS THE CLOSURE OF IS A FUNCTION, AND THEN THAT IS CONTINUOUS.

Now to show the converse. Let be a compactification of such that every continuous map where is compact can be extended continuously to . We can identify as a subspace of without loss of generality. Let be any other compactification of . We need to show that finer than , that is, that there exists a continuous function such that . Well, is a continuous map from to a compact set, namely , so there exists an continuous extension . Since we’re considering to be a subset of , for all . , therefore . is finer than , and since was arbitrary, that must mean that is the Stone-Cech compactification.

Exercise 3

Bullet Point One

is the direct product of compact spaces with the product topology, and therefore compact by Tychonoff’s Theorem. is a closed subset of and therefore compact, now we need to show that is a compactification. We also need to note that since is the product of Hausdorff spaces, is Hausdorff and sine we know it’s compact, is normal. We need to show that is a homeomorphism between and and that is an open dense subset of .

First let’s show that is injective, and therefore bijective between and . Pick two points that aren’t equal. is Hausdorff and therefore and are closed subsets in . By Urysohn’s Lemma, there exists a such that and . Then and , these two tuples disagree at the -th coordinate. Therefore . is injective, and therefore a bijection between and .

We consider to have the subspace topology as a subset of , and we consider to have the subspace topology as a subset of . This is exactly the same topology as the subspace topology of when considered as a subset of . Thus, inherits the topology of point-wise-convergence. Now to show that is continuous. Suppose we’re given any net in in that converges to some point . Consider in , we need to show that this net converges to . converges to if and only if converges to for all . For each , we have that the net converges to since each is continuous. The net converges point-wise to , and therefore converges to in the product topology inherits from . is continuous at , and since our choice of was arbitrary, is a continuous on .

Now let’s show that is continuous. Suppose that the net in converges to some point , then we need to show that the net converges to in . Let be an open set in containing . is a locally compact Hausdorff space, and we can appeal to the generalized version of the Urysohn’s Lemma we proved in Exercise 6 in Notes 12. So we can find some such that and vanishes outside a compact set that is contained in . By assumption, , and thus there exists a such that whenever . The support of is contained in , and therefore whenever . We’ve shown directly that the net converges to . Therefore, for any , we see that whenever . is continuous.

By construction, is dense in . So all we need to do is show that is open in . Pick an . Since is a locally compact Hausdorff space, we know there exists a function if compact support (say the support is ) that is 1 at . Consider . Consider all such that , this is clearly an open set in that contains . Suppose that we’re given a that is not in . If we show that , then can’t be in the set of all such that . Thus, the contrapositive would be true, and we’ve found an open set in around that is contained in . Let be a net that converges to . Then there must be a such that lies in for all , or we could construct a convergent subset that lies entirely within . But since is compact, that net would converge to a point in , contradicting our choice of . If , then choose the open set in consisting of all points such that intersect . Then this is an open set around , but whenever , which contradicts that converges to . must be 0, and therefore we’ve shown that any point in is an interior point of . is open relative to . is indeed a compactification of .

Bullet Point Two

Let be the compactification defined above, and let be some other compactification, and identify as a subspace of . Given a , then the restriction map is a function from , and since is dense, if , then . So our restriction map is injective, and therefore we can identify as a subspace of .

Consider the natural projection from to where we consider as a subset of . Let be a basic open set in . That is, where each is open in all but finitely many ‘s are equal to . Let be the subsets of that aren’t equal to . Then where for each and for all other , thus . is open whenever is a basic open set, and therefore is open whenever is open. is continuous. Let be the restriction of to . The image of is the image of inside by the Gelfand transformation. So we can consider to be a continuous function from to . We need to show that . Pick a , then since we’re considering to be a subset of . , which is the image of under the Gelfand transformation. Thus, when we associate with it’s image under the Gelfand transformation, . Therefore is finer than , and since was an arbitrary compactification, must be the Stone-Cech compactification.

Exercise 4

Bullet Point One

Let be a space with discrete topology, and let be the Boolean algebra of all subsets of (this is also the topology on ). By Stone’s Representation Theorem, is isomorphic to the clopen algebra of a Stone space, a totally disconnected compact Hausdorff space. Denote the clopen algebra by . Denote the isomorphism between and by .

For each , , and consider . Fix . In order to find a embedding from to , we need to show that . Suppose there is a strictly smaller clopen set that contains , denote it by . But is an isomorphism, so exists and preserves orderings. Thus . If , then which contradicts our assumption that is strictly smaller than . If , then , which contradicts our assumption that contains . We’ve exhausted our cases, is the smallest clopen set that contains . We haven’t yet shown that is the singleton set that contains , we have one more step. The proof of Lemma 1 in Notes 4 shows that the intersection of all clopen sets that contain is . is a subset of any clopen set that contains , and thus lies in the intersection of all clopen sets that contain . Therefore .

Now we can define our compactification. Consider the map where where . is an injection because if then . Therefore if and , then . is injective. Therefore is a bijection between and .

Let be open in , then is some subset of . But any subset is open in the discrete topology, therefore is open. is continuous. Now let be an open set in , then . Then . Each is a clopen set in and a subset of and therefore clopen relative to , and thus is the union of open sets and therefore open relative to . is continuous. is a homeomorphism between and . To finish our proof we need to show that is a dense open subset of .

, and as we said in the previous paragraph each is open in , and thus is open in . Finally, to show that is dense in . Pick a point in , and pick an open neighborhood around . By Lemma 1 in Notes 4, the clopen algebra of forms an open base for the topology, so there exists a clopen set such that . Pick some , then let , then . is dense in . We have finished our proof. is a compactification.

Bullet Point Two

Bullet Point Three

Exercise 5

The first bullet point is a consequence of the second one. So we’ll just prove the second one.

Bullet Point Two

is an normed complex algebra under the sup-norm. The space is commutative, and unital since it contains the constant function . The sup-norm is sub-multiplicative. Consider the map that maps . This map is anti-linear, and since it is norm-preserving. Also, . So our space satisfies the -identity.

Let be the Stone-Cech compactification of . Then is a unital -algebra when equipped with the sup-norm and with the anti-linear map . Consider the map where . Since continuous functions on compact sets are bounded for each . Consider as a map between and . is a homomorphism and it even preserve the anti-linear map. We need to show that is a bijection and an isometry.

Since is dense in , if we have and then . Thus, is injective. Now if . Let , then . Then by Exercise 2 in these notes, there exists a unique extension (since is already compact) such that . Then and . is surjective, and thus bijective.

Pick a . Pick an . is dense in and therefore there exists a net in that converges to . Pick an . is continuous at and therefore there exists a such that whenever . Then if we pick an , then we see that , we see that . And since is a subset of , . Thus, preserves norms.

We can now use this fact to show that is complete, and therefore a -algebra. Let be a Cauchy sequence in . Consider the sequence in where . Since is a bijection that preserves norms, is a Cauchy sequence in the complete space and therefore converges to some . Let . Since , and is a homomorphism that is also an isometry, . Every Cauchy sequence in is a convergent sequence. is complete, and therefore we have our last criteria met. is a -algebra and our map is our -isomorphism between and .

We’ve simultaneously shown points one and two.

Bullet Point Three

Let be a compactification of . Consider the map that takes a function in and returns the function . From the previous bullet point, is an injective homomorphism from the commutative unital -algebra to the commutative unital -algebra that preserves the anti-linear transformation on both -algebras. We don’t have to change anything in our argument in the previous part to show that this is also an isometry. Now to show that the image of contains .

Let . So given an , there exists a compact set such that when . Consider the function where and on . Let be open in . If isn’t in , then . is open relative to , and is open relative to , and thus is open relative to . Now if is in , consider . Let . is open relative to , and therefore is open relative to . is an interior point. Suppose . . is an open set in that contains . Therefore there exists an such that , then . There exists a compact set such that for all . Now is compact relative to , and for all . Therefore . We’ve found an open neighborhood around that lies in . is an interior point. Therefore, is open. is a continuous function from to .

Therefore is in the image of under . Let denote the constant function , and let lie in . Now let . There exists a such that . Then . lies in the image of .

Exercise 6

Let be the Stone-Cech compactification of . The proof that is isomorphic to as Banach spaces was our proof that is isomorphic as a -algebra to with some steps omitted. Therefore is isomorphic to . Since is a compact Hausdorff space, is isomorphic to the space of real signed Radon measures on according to the Reisz Representation Theorem. The complex case is identical.

Any function on is continuous, therefore is the space of bounded continuous function on . Therefore .

The movie’s not an attack on Pentecostalism. Marjoe even gives them an honest endorsement as the Christian curch he’d belong to if he had to choose one. They never mention in the movie any doctronial isssues on which the Pentecost church differs from other churches. Sure there’s the shaking on the floor, the healing through prayer, and other shows, but the club handshakes don’t tell you a club’s beliefs and values.

I like Marjoe, it’s hard not to like him. He’s smart, charaismatic, and has a certain power. I must admit that I have bias towards people like that because I’m the polar opposite. Those people are strange to me. I don’t understand how they tick, from where they get that power. So I’m always interested when something takes a closer look at that type of person.

This movie is a behind the curtain look of a rouse and its con man. He explains some fake miracles to his camera crew, but that’s only a fraction of what he does. The con isn’t his parlor tricks but him. He has passion and energy. He speaks with conviction and power. He’s a skilled entertainer, and even when he’s not singing his voice has a musical quality to it. When you watch him preach there is no crack in the performance. There is no keyhole through which we see that jaded cynic that must be half-laughing half-crying at the power his performance has over his audience.

This isn’t harmless fun. He even recognizes the bad he’s doing. I’ve read accounts of James Randi doing cold-readings (for those of you that haven’t taken an interest in the battle between skepticism and psuedo-science/paranormal-activity, a cold-reading is the name of the parlor trick off of which John Edwards has based his entire), and he puts on a good show. But at the end, he always reveals that it was just that, a show. And at no points did he ever collect money from the people he was tricking. I don’t know if this was intenttional but the symbolism smacks you in the face. There is a scene where Marjoe’s audience line up and puts money in a waste bin. He realizes what he’s doing is wrong, he doens’t feel guilty enough to not take his victims’ money,  but the desire to end his sham is the impetus for the creation of the movie.

There isn’t really too much straight criticism found in the movie. The strongest piece would be a zoom-in on an elaborate broche pinned to a woman that asks the people in the pews to donate money to the cash-straped church, while reassuring them that the church doesn’t spend foolishly. Even when Marjoe’s father is presented, all Marjoe says is that the only thing they can really talk about is his father’s back-yard.

Probably the best scene is one towards the end that features Marjoe, his girl friend, and a dog. At this point he’s stated how he’s tired of his double-life, the lies, and he wants a new life with the girl he loves. It’s a huanting scene because he’s at full-force in it. It shows the incredible talent and personality that’s essentialy been wasted on a single trick. He’s not giving up just a rouse, but what’s been his life. It’s an increidbly interesting movie if only because it is the atonement of a fascianting person.

But I have to admit, if there is one child prohet I’m going to follow, it would be this one. Sorry Marjoe.

There is a huge backlog of problems. I plan on initially just having thirteen essentially blank posts (one for each of his notes), with each post just consisting of the exercise numbers, and I’ll fill the space next to each of those numbers with the corresponding proof as I get around to typing one up or posting a submitted answer. This way my posts are ordered the same way as Professor Tao’s.

I’ll post the problem lists soon, and I will fill in what I can when I have time, but as I said before, I do hope to receive help. If this blog is a sort of answering of a challenge from Professor Tao’s blog, then this blog is at a serious disadvantage in thus duel. As previously stated, there is a tremendous backlog of problems, intrinsically it takes less time read a problem than it is to answer it, and I am much more of a mere mortal than Professor Tao.

So here we go.