# Monthly Archives: October 2017

## Dual norm in R^n

Using the separating hyperplane theorem, we prove the dual norm of the dual norm is the original norm.

### 1. Extension Theorem

Let be a norm on . The dual norm of a norm is defined by

For any , we show

By definition, we have

So

It remains to prove the reverse inequality. We will show that there exists such that and

(1)

Theorem 1. Let be a subspace of and let be a linear functional on satisfying

Then there exists satisfying

and

One can prove this theorem directly. See [1] or [3].However, we prove this theorem using the following lemma:

Lemma 1. Let be an affine set in and let be a non-empty convex subset of , not intersecting . Then there exists a hyperplane in containing and not intersecting .

Proof. By the separating hyperplane theorem, there exist and such that

Define

Then it is a hyperplane in . There are two cases to consider. First, suppose . Then there exists with . Then for , note that since is an affine set. So

On the other hand,

Thus,

This implies

which proves .

Suppose . After a translation, we may assume has so that is a subspace of . Then since . Note that for all . If it were not, then there exists satisfying . If , then for , . Letting , then there exists with , which contradicts . Similarly, leads a contradiction. Hence for all .

Note . So for all and for all . This shows that is a hyperplane in containing and not intersecting .

Thus, both cases gives a desired result. This completes the proof.

Using this lemma, we are ready to prove the theorem.

Proof of Theorem 1 Define

Then is convex, affine subspace of . Consider

Then it is also convex. Then is a hyperplane containing . Note that since

Hence by Lemma 1, there exists a hyperplane containing and . So there exist and such that

with for all . Since , . So . Hence we may assume

Then for all .

Note that if , then for all . Since is linear, is the zero map. So there is nothing to prove. So we assume . Then there exists with .

For nonzero vector , . So

and this shows

If with , then

and so

If for some , then

and so

Since , . So

and

This completes the proof of theorem.

### 2. Proof of the duality

Now we are ready to prove the claim (1). If , there is nothing to prove. If , consider . Define by

(2)

Clearly it is linear, and

Also,

for all . Hence by Theorem 1, there exists with

with

Since

we conclude that

From (2), if we take , then and , which proves the claim.

Therefore,

### 3. Some remarks

The motivation of the proof is the philosophy that the Hahn-Banach theorem and the separating hyperplane theorem are closely related. We can prove the separating hyperplane theorem using an extension of Theorem 1. One may ask whether the converse is true. In fact, both statements are equivalent.

The proof of Theorem 1 using Lemma 1 is given in [2]. We modify the lemma and theorem and their proof so that it is suitable to our setting.

References

1. H. Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations, Springer, 2011.
2. H. H. Schaefer and M. P. Wolff, Topological vector spaces, Springer, 1966.
3. E. Stein and R. Sharkarchi, Functional Analysis: Introduction to Further Topics in Analysis, Princeton University Press, 2011.

## Local Existence and Uniqueness of IVP

In this note, we prove the local existence and uniqueness of initial value problem of  ODEs

where satisfies Lipschitz condition.

### 1. Function spaces

To discuss our main topic, we need to extend some concepts that we already studied. In linear algebra, we studied vector spaces, which satisfies nine axioms. Such concepts give our concept of `vectors’ in high school as an example. One can also check that the space of real-valued continuous function on is a vector space with usual addition and scalar multiplication. We denote this space by .

In the Euclidean space , we can measure a distance of two vectors and by

Likewise, as is a vector space, one tries to measure a distance of two vectors. In this case, we want to measure a distance of two functions in . There is some issues to be considered. First, how can we define a distance on ? Next, the distance is useful?

There are many ways to define a distance on . Here we pick some special distance. Consider with , where

We check induces a distance on . Define by . To say is a distance, it should satisfies basic properties which the Euclidean distance
has.

• is nonnegative. Indeed, .
• if and only if . Indeed,  implies for all . So . The converse is obvious.
• . It is also easy since
• for all : Indeed, this follows from the triangle inequality of Euclidean space:

So

This shows is a distance. So induces a distance .

To start an mathematical analysis, we need to ensure that with this kind of distance behaves likes One of the important property of is completeness. Here our completeness is Cauchy complete, i.e., for any Cauchy sequence in , there exists such that . We want to obtain such kind of completeness on with .

Now recall the definition of uniform convergence. Let be a sequence of real-valued functions on a set and a real-valued function on . We say that converges uniformly on if for every , there exists an such that

The function is called the uniform limit of the sequence . Note that

So the convergence in is a uniform convergence. Now considering Theorem 5.46 (Cauchy criterion for uniform convergence) in the textbook, for any Cauchy sequences in , converges uniformly on . Since is a sequence of continuous functions on and if we denote is a uniform limit of , then . This shows with is complete.

Complete space has various properties we can discuss in analysis. In the next section, we study one of the important property related to completeness.

### 2. Contraction Mapping Principle

Recall the Problem 1.14 in our textbook.

A sequence of real numbers is said to be \emph{contractive }if there exists a constant such that

Prove that for all . Also prove is convergent. Fianlly, if , then

In the proof of this problem, we used the Cauchy criterion to ensure is convergent. In the previous section, we already observed with is complete. So one can think that similar fact must be hold. Indeed, it is true. As an application, we study some convergence of sequences which is recursively defined. It is related to fixed points.

Theorem (Contraction mapping principle). Let be a contraction, i.e., there exists such that

Then there exists a unique function such that .

Proof. Uniqueness is easy. If and , then

which can only happen when . So .

Let . Define

Then . Iterate this so that

So if , then

For , choose so that . So implies . Hence is a Cauchy sequence in . So by the completeness of , there exists such that

Now from

we conclude that

in the uniform limit. This completes the proof.

Remark. The above theorem holds for any complete metric space, which will be studied in Topology I.

### 3. Existence and uniqueness of ODE

Now we give one application of the contraction mapping principle. We consider the following initival value problem of the first-order ODE:

(1)

where is an interval containing and is a Lipschitz function. Observe that is a solution of (1) if and only if

(2)

This follows from the fundamental theorem of calculus.

Theorem. Suppose that is a Lipschitz function. Then the equation has a unique -solution on for some .

Let be a Lipschitz function with constant , i.e., there exists a constant such that

for all .

Proof. It suffices to prove there exists saitsfying (2). Note that as we saw before is a complete with .

Define by

If we show has a fixed point , then this is a solution of (2), which is a solution of (1).

Observe that

Now choose so that . Hence by the contraction mapping principle, there exists a unique fixed point . This completes the proof.

Remark. If we try to use the contraction mapping principle on , we have a trouble since is not complete under the distance . Note that converges uniformly to on .
Also, , but is not differentiable at .

Remark. The th order ODE given by

can be reduced to the first order ODE. Define by

Then

where is the function

## Space-filling curve

Theorem. There exists a continuous curve in that passes through every point of the unit square .

Proof. Define by

Extend to all of by making periodic with period .

Define

By the Weierstrass M-test, both and converges uniformly on . Moreover, it is continuous on . Now define and let denote the image of the unit interval under . We show .

Observe and . Hence . Let . Write

with . Now let

Then since . Now we show and . We show for each … If we can show this, then we have , and this gives , .

Now write

where

Since has period 2, we have

If , then and hence . So in this case. If , then and hence . Therefore, . This proves , . So .

Meaning of this theorem. Observe that is continuous. has dimension 1 and has dimension 2. By the above theorem, the continuity does not guarantee the dimension of spaces. The above curve we constructed is nowhere differentiable. This is proved by Alsina.

It seems that this kind of curve is not useful, but it has quite a lot of applications. One can use this kind of fact to probability theory, topology, etc. It has an application to industry, e.g. Google map.

References

1.  J. Alsina,  The Peano curve of Schoenberg is nowhere differentiable, Journal of Approximation theory, Vol. 33 (1), 28– 42.
2. T. Apostol, Mathematical Analysis
3. C. S. Perone, Google’s S2, geometry on the sphere, cells and Hilbert curve