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Silly Mistake August 15, 2009

Posted by putnam120 in Math Related.
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I was on a math help forum and was trying to help someone with the following problem.

Given two disjoint closed sets A and B in a metric space S, prove that there exists disjoint open sets U and V such that A\subset{U} and B\subset{V}.

In my proof I made the following mistake: Let r=\inf d(x,y) for x\in{A}, y\in{B}. Then r>0.

I should have stopped at this time, since I remember talking about something very similar to this in my analysis class.  It turns out that to gaurantee r>0 you need that at least one of A,B to be compact.

Infinity Norm June 18, 2009

Posted by putnam120 in Math Related.
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Well here is a question that I used to wonder about for a while.

If f is integrable over U, then

\displaystyle\lim_{n\to\infty}\left\{\int_U|f|^nd\mu\right\}^{\frac 1n}=\sup_U|f|.

Well here is a proof of a simplified version.  I only assume that M=\sup|f|<\infty, and \mu(U)<\infty

proof: By “the world’s most obvious integral inequality” \int_U|f|^nd\mu\le M^n\mu(U). So obviously ||f||_\infty\le M.

For the reverse direction let \epsilon>0 then |f|>M-\epsilon on some set K\subset U. Therefore \int_U|f|^n\ge (M-\epsilon)^n\mu(K). To get the result take n\to\infty then \epsilon\to 0. \mathbb{Q.E.D.}

The only part that bothers me is making sure that K is measurable, but that shoudn’t be too difficult to fix.

Solutions to IVT Problems November 26, 2008

Posted by putnam120 in Math Related, Uncategorized.
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(a) Since f is continuous on a compact set (\left[a,b\right] it attains both its maximum and minimum values, so m\le f(x)\le M for all x\in\left[a,b\right]. Now using these poor estimates we have that m\int_a^b g(x)dx\le\int_a^b f(x)g(x)dx\le M\int_a^b g(x)dx. So if we consider the function V(x)=f(x)\int_a^b g(x)dx we have that V is continuous on the same interval as f. Thus we just apply the intermediate value theorem to V and the desired result follows.

(b) WLOG assume that f(0)=f(1)=0. Let n be given and define new function by L(x)=n\left(f(x+1/n)-f(x)\right). This is just the equation for the slope of the line connecting the points f(x) and f(x+1/n) and is continuous because f is continuous. Now consider the set of points x_i=\frac{i}{n} for i=0,1,\dots,n-1. Now look at the values L takes at these points. If L(x_i)=0 for any i then we have found a solution, so assume that L(x_i)\neq 0 for any chose of i. So WLOG assume that L(x_0)>0, since f(0)=f(1) we know that there must be some i such that L(x_i)<0, call this x_k. Then we there is a x\in\left(x_0,x_k\right) such that L(x)=0.

(c) For this one I will prove something a little stronger. All that we need to assume about f is that it is integrable, and has what I shall call the extreme value property on our interval. Basically this property says that on the interval there exists a point c such that f(c)=\sup f(x)=M and a point d such that f(d)=\inf f(x)=m, where x ranges over all the values in our interval. So clearly an increasing function satisfies these conditions. Now define a function by

V(x)=m\int_a^xg(x)dx+M\int_x^bg(x)dx.

Now V is continuous and V(b)\le\int_a^bf(x)g(x)dx\le V(a), thus by the intermediate value theorem we are done.

Some after thoughts: Problem (a) was pretty straight forward and didn’t take too much insight to actually solve. However, problem (b) was a little more challenging until I visualized what I was being asked to prove. This led me to the idea to look at the slope of the connecting line segment. Finally problem (c), I at first wondered why you were given that f was increasing and not continuous, or even just integrable. As can be seen in my proof continuity imposes more conditions than are necessary, while integrability does not provide you with enough. This led me to think about what makes increasing functions special (well one of the things that makes them special).

Some Problems Involving I.V.T. November 25, 2008

Posted by putnam120 in Math Related, Uncategorized.
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Well I am in a big Analysis mood. I guess that it has something to do with the fact that at this point in my life Analysis is the topic in math that spikes my interest the most. This can be accredited to my Calculus BC teacher from high school (Mrs. Johnson) and my undergraduate Advanced Calculus professor (Dr. Shen). They both did a wonderful job of presenting material and showing applications and fascinating consequences.

Anyway onto the post, I was reading through the intermediate real analysis section of Problem Solving Through Problems. Here are some of the problems I found interesting.

(a) Suppose that f:\left[a,b\right]\to\mathbb{R} is continuous and g:\left[a,b\right]\to\mathbb{R} is integrable and such that g(x)\ge{0} for all x\in\left[a,b\right]. Prove that there is a number c in \left[a,b\right] such that

\int_a^bf(x)g(x)dx=f(c)\int_a^bg(x)dx.

(b) Let f:\left[0,1\right]\to\mathbb{R} be continuous and suppose that f(0)=f(1). Prove that for each positive integer n there is an x in \left[0,1-\frac{1}{n}\right] such that f(x)=f(x+1/n).

(c) Assume the same conditions on f,g as in part (a) except that instead of being continuous f is now assumed to be increasing. Prove that there is a c\in\left[a,b\right] such that

\int_a^bf(x)g(x)dx=f(a)\int_a^cg(x)dx+f(b)\int_c^bg(x)dx.

My thoughts: The solution to (a) is a pretty straight forward application of the intermediate value theorem. For problem (b) I am considering looking at the slope of the line connecting f(x) and f(x+1/n). Finally for (c) I am going to try and generalize it in an appropriate way. My solutions should be posted in the near future.

Hölder’s Inequality November 23, 2008

Posted by putnam120 in Math Related.
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I am only going to do the case where f,g are real functions. The result however, still holds if they are complex.

Statement: Suppose that f,g are integrable functions with respect to \alpha on the interval \left[a,b\right]. Additionally p,q\in\mathbb{R^+} such that \frac 1p+\frac 1q=1. Then we have the following inequality:

\displaystyle\left|\int_a^bfgd\alpha\right|\le\left\{\int_a^b|f|^pd\alpha\right\}^{\frac 1p}\left\{\int_a^b|g|^qd\alpha\right\}^{\frac 1q}.

This can also be stated as ||fg||_1\le ||f||_p||g||_q.

Lemma: If u,v\ge{0} then uv\le\frac{u^p}{p}+\frac{v^q}{q}, where we have the same conditions on p,q as before.

Proof of lemma: Just apply Jensen’s Inequality to e^x and the fact that e^{\ln x}=x. \mathbb{Q.E.D.}

Proof: Without loss of generality we can assume that ||f||_p=||g||_q=1, if not we can just divide f,g by the appropriate constants and make it so. Now from the lemma we have that \forall x\in\left[a,b\right] \displaystyle |f(x)g(x)|\le \frac{|f(x)|^p}{p}+\frac{|g(x)|^q}{q} we then integrate both sides of the inequality and the result follows. \mathbb{Q.E.D.}

Integral Test November 22, 2008

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Well it’s been a while since I have posted a math related post. So I am going to do this one on one of the problems from our analysis homework.  Basically we were asked to prove the integral test, not too difficult but definitely something that should be done.

Statement: Assume that f(x)\ge 0 and that f decreases monotonically on \left[1,\infty\right). Then \displaystyle\int_1^\infty f(x)dx converges if and only if \displaystyle\sum_{n=1}^\infty f(n) converges.

Aside: When I submitted this to my professor for grading in proved the theorem in both direction. Here I am going to try and combine them, thus saving time on my part.

Proof: Consider the interval \left[m,n\right] where m,n are integers with m<n. Additionally let P be the partition \left\{m,m+1,\dots ,n\right\}. Now because f is monotonically decreasing and f\ge 0 we have

(1)   \displaystyle 0\le\sum_{k=m+1}^nf(k)=L(P,f)\le\sum_{k=m}^{n-1}f(k)=U(P,f)\le\sum_{k=m}^nf(k)

Let \epsilon>0. If \displaystyle\sum_{n=1}^\infty f(n) converges then there exists an $N$ such that \displaystyle\sum_{k=m}^n f(k)<\epsilon whenever m,n>N. Similarly if \int_1^\infty f(x)dx converges we have that there exists W such that \int_m^nf(x)dx<\epsilon whenever m,n>W. The theorem follows from combining these facts with (1). \mathbb{Q.E.D.}

I would like to mention that I have left out some of the small details, such as proving that the integral actually does exist if the sum converges.

IMPORTANT NEWS November 22, 2008

Posted by putnam120 in Life Events, Math Related.
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Well for unknown reasons (well not totally unknown) I can no longer use LaTex on this blog. I did look up ways to get this back but ran into a few problems. Most of the solutions only worked for Unix like systems, and I would like to be able to use LaTex on the blog even if I was on a Windows machine. Also most of these solutions would have required me to add another version of Tex to my linux system and I didn’t really want to do that. There was however, one solution that would work with all systems (supposedly). All I had to do was edit the source code for a GreaseMonkey script, but after doing that I was still running into the same issues as before so I quickly gave up on that dead end.

I will still be using this blog but whenever I want to post anything mathematically related I shall post a link to my WordPress blog in the post. The reason I am using WordPress is becasuse it has LaTex built in and thus less work on my part.

I Have Wireless September 21, 2008

Posted by putnam120 in Life Events, Math Related, Programming Related.
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So I have finally managed to get my wireless card working with Ubuntu 8.04. You have no idea how happy I am about this. I have been wanting to really give Ubuntu a try but have always been reluctant to because of the fact that it wouldn’t work with my wireless card and I would need an ethernet cable in order to connect to the internet. Now I know this happiness will not be too long lived because once I update to the next version of Ubuntu (which in my case will probably be 9.04 and not 8.10) I will most likely have to go through the same process again and hope that it still works on the new version.

On a more “depressing” note; I managed to finish only half of one my my 5 analysis homework problems. They all seem to be about connected sets. One of the funny things is that in Rundin for a set to be connected it must not be able to be written as the union of two separated sets. However, on Wikipedia, Mathworl, and one of my other analysis text they also say that the separated sets must be open sets. Now I know that if two open sets are disjoint they are separated so I don’t see the real need for this extra assumption. In any case the problem I managed to solve was as follows.

Let A and B be two connected subsets of a metric space X. Show that is A and B have nonempty intersection then their union is also connected.

The second half of the problem asked us to state and prove a generalized version of this for the union of arbitrary connected sets. I came up with:

Let {A} be collection of connected sets. If for each F in {A} there exist a G in {A} such that the intersection of F and G is nonempty then the union of all the members of {A} is a connected set.

solution August 2, 2008

Posted by putnam120 in Math Related.
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This is the solution for problem #1 in my previous post.

Let , since is a Cauchy sequence there exist an such that if if Similarly since a subsequence converges to we have that when . Now let be the larger of and .

Now .

Here is the solution for #2 in my previous post (this one took a little longer for me to think of).

Let be a sequence in such that . Now from properties of we have that for all Now let , and since we have that for large enough. So since is complete. Additionally, is a limit point of each , so .

From the Integers to Infinite Series July 30, 2008

Posted by putnam120 in Life Events, Math Related.
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In the upcoming fall semester I will be taking the first year graduate analysis class. Hopefully, this class will be more “rewarding” than the undergraduate analysis sequence I completed last year. The undergraduate sequence was not a total waste of time since I actually did learn some new things, but the class moved so slowly and the test were not all that difficult. I never really had to exert myself in order to obtain my A. As a result in the second semester I tried to get a 100% on every assignment and test. I almost achieved this goal, however, I did not make a small clarification on one of my proof thus resulting in me losing 1 or 2 point for the problem.

The graduate class will be using the same book as the undergraduate sequence. However, we will hopefully cover all of the sections. Last year the professor would pick and choose which topics he though were necessary and possible for the majority of the class to understand. As a result some of our proofs were longer than needed but I suppose knowing multiple ways to prove something isn’t all that bad.

Right now I am reading through as many chapters as I can and working all the exercises. My only requirement is that I can only be working on problems from at most two different chapters and that I can only be reading one chapter ahead of the problems I am solving. So basically right now I am working problems from chapter 2 and 3, and reading chapter 4. I am almost done with the problems from chapter 3, but the ones from chapter 2 are going to take a little more time. I suppose that it has to do with the fact that dealing with topology is still a little “new” to me.

Tomorrow I will hopefully be able to do the following problems:

1) Suppose is a Cauchy sequence in a metric space , and some subsequence converges to a point . Prove that the full sequence converges to .
2) Let be a sequence of closed, nonempty, bounded sets in a complete metric space . Also and . Prove that contains exactly one point.

I will eventually get around to making a post about my REU in Michigan. Sorry to disappoint, but I just need some more time to better collect my thought on this topic so that I will have a post worth reading.