## Pointwise Convergent Sequence of Polynomials.

### August 7, 2012

An interesting question came up during my studying today which made me think about some of the different ways we can think about polynomials. The solution to this problem was not immediately obvious to me, and, in fact, it wasn’t until I looked up a completely unrelated problem (in a numerical methods book!) that some solution became clear.

**Question: **Suppose that is a sequence of polynomials with each of degree and suppose pointwise. Show that is also a polynomial of degree no more than .

Some interesting points come up here. First is that we only have pointwise convergence — it wasn’t even immediately obvious to me how to prove the resulting limit was *continuous*, let alone a polynomial of some degree. Second, we know very little about the polynomials except for what degree they are. This should be an indication that we need to characterize them with respect to something degree-related.

Indeed, polynomials can be represented in a few nice ways. Among these are:

- In the form where it is usually stated that .
- In terms of their coefficients. That is, if we have a list of polynomials of degree 3 to store on a computer, we could create an array where the first column is the constant, the second is the linear term, and so forth. This is sort of like decimal expansion.
- Where they send each point. That is, if we know what is equal to for each , we could recover it.
- If a polynomial is of degree then, somewhat surprisingly, we can improve upon the previous statement: if we know the value of for distinct points, then we can find , and is the unique such polynomial of that degree which has those values. (Note that if we were to have points and a polynomial of degree , then
*many*polynomials of this degree could fit the points. Consider, for example, and . Then we have one points and we want to fit line through it. Clearly this can be done in infinitely many ways.)

This last one is going to be useful for us. So much so that it might be a good idea to prove it.

**Lemma. **Let be distinct points in , and let be distinct points in . Then there is a *unique *polynomial of degree at most such that for each considered here.

*Proof. *This is an exercise in linear algebra. We need to solve the system of linear equations

where spans , for the constants . Notice that this is simply plugging into a general polynomial of degree . Notice that the matrix that this forms will be a Vandermonde matrix. Since each is distinct, the determinant of this matrix is nonzero, which implies that there is a unique solution. This gives us our coefficients, and note that this is a polynomial not necessarily of degree exactly , since some coefficients may be 0, but it is at most .

[Note: For those of you who forgot your linear algebra, the end of this goes like this: if we let our coefficients be denoted by the column matrix and our Vandermonde matrix is denoted by , then we want to solve where is the column vector with entries . If has non-zero determinant, then it is invertible, and so we have that gives us our coefficients.]

Neato. But now we need to specialize this somewhat for our proof.

**Corollary. **Let the notation and assumptions be as in the last lemma. For , let be the unique polynomial of degree at most with (where if and ). Then every polynomial of degree at most is of the form for each .

This might be a bit more cryptic, so let’s do an example. Let’s let so that we have two points. Let’s say and . Then we have is the unique polynomial of degree at most 1 such that and . Of course, this function will be . Now and ; this gives us that . The theorem now states that any polynomial of degree at most can be written in the form

.

For example, let . Then the lemma says , as we’d expect. The power of this lemma will become clear when we use this in the solution. The proof of this corollary is just a specialization of the previous lemma, so we exclude it.

**Solution.** Recall, just for notation, that our sequence pointwise. Let’s let be our distinct points, as usual. In addition, let’s let be defined as in the corollary above. Represent each as follows:

for each . Here comes the magic: let and note that at every point, so, in particular, on each and on . We obtain

for each . But this is the sum of polynomials of degrees at most , which gives us that is itself a polynomial of degree at most .

I’ll admit, I did a bit of digging around after finding the lemma above; in particular, this corollary representation of polynomials seems to be a nice way to represent a polynomial if we do not know the coefficients but do know the values at a certain number of points and have that its degree is bounded below that number of points.

**Exercise: **Try to write out this representation for and . If you’re a programmer, why not make a program that allows you to input some points and some values and spits out the polynomial from the corollary above?

## Orthogonal Complement of Even Functions.

### August 5, 2012

**Question:** Consider the subspace of consisting of even functions (that is, functions with ). Find the orthogonal complement of .

**One Solution. **It’s easy to prove is a subspace. Then, there is a representation of any function in this space by adding odd and even functions together; more precisely, given we have that is even and is odd and . For uniqueness, note that if , then for each , giving us that . Hence, the orthogonal complement of is the set of odd functions.

Here’s another solution that "gets your hands dirty" by manipulating the integral.

**Another Solution. **We want to find all such that for every even function . This is equivalent to wanting to find all such with . Assume is in the orthogonal complement. That is,

The last equality here re-parameterizes the first integral by letting , but note that our new gives us the negative sign.

.

We may choose since this is an even function, and we note that this gives us

.

Since , it must be the case that . [Note: The fact that this is only true "almost everywhere" is implicit in the definition of .] Hence, , giving us that .

We now have one direction: that *if * is in the orthogonal complement, *then *it will be odd. Now we need to show that if is any odd function, it is in the orthogonal complement. To this end, suppose is an odd function. Then by the above, we have

where the last equality comes from the fact that is odd.