College Math Teaching

June 7, 2016

Infinite dimensional vector subspaces: an accessible example that W-perp-perp isn’t always W

Filed under: integrals, linear albegra — Tags: , — collegemathteaching @ 9:02 pm

This is based on a Mathematics Magazine article by Irving Katz: An Inequality of Orthogonal Complements found in Mathematics Magazine, Vol. 65, No. 4, October 1992 (258-259).

In finite dimensional inner product spaces, we often prove that $(W^{\perp})^{\perp} = W$ My favorite way to do this: I introduce Grahm-Schmidt early and find an orthogonal basis for $W$ and then extend it to an orthogonal basis for the whole space; the basis elements that are not basis elements are automatically the basis for $W^{\perp}$. Then one easily deduces that $(W^{\perp})^{\perp} = W$ (and that any vector can easily be broken into a projection onto $W, W^{\perp}$, etc.

But this sort of construction runs into difficulty when the space is infinite dimensional; one points out that the vector addition operation is defined only for the addition of a finite number of vectors. No, we don’t deal with Hilbert spaces in our first course. 🙂

So what is our example? I won’t belabor the details as they can make good exercises whose solution can be found in the paper I cited.

So here goes: let $V$ be the vector space of all polynomials. Let $W_0$ the subspace of even polynomials (all terms have even degree), $W_1$ the subspace of odd polynomials, and note that $V = W_0 \oplus W_1$

Let the inner product be $\langle p(x), q(x) \rangle = \int^1_{-1}p(x)q(x) dx$. Now it isn’t hard to see that $(W_0)^{\perp} = W_1$ and $(W_1)^{\perp} = W_0$.

Now let $U$ denote the subspace of polynomials whose terms all have degree that are multiples of 4 (e. g. $1 + 3x^4 - 2x^8$ and note that $U^{\perp} \subset W_1$.

To see the reverse inclusion, note that if $p(x) \in U^{\perp}$, $p(x) = p_0 + p_1$ where $p_0 \in W_0, p_1 \in W_1$ and then $\int^1_{-1} (p_1(x))x^{4k} dx = 0$ for any $k \in \{1, 2, ... \}$. So we see that it must be the case that $\int^1_{-1} (p_0(x))x^{4k} dx = 0 = 2\int^1_0 (p_0(x))x^{4k} dx$ as well.

Now we can write: $p_0(x) = c_0 + c_1 x^2 + ...c_n x^{2n}$ and therefore $\int^1_0 p_0(x) x^{4k} dx = c_0\frac{1}{4k+1} + c_1 \frac{1}{2 + 4k+1}...+c_n \frac{1}{2n + 4k+1} = 0$ for $k \in \{0, 1, 2, ...2n+1 \}$

Now I wish I had a more general proof of this. But these equations (for each $k$ leads a system of equations: $\left( \begin{array}{cccc} 1 & \frac{1}{3} & \frac{1}{5} & ...\frac{1}{2n+1} \\ \frac{1}{5} & \frac{1}{7} & \frac{1}{9}...&\frac{1}{2n+5} \\ ... & ... & ... & ... \\ \frac{1}{4k+1} & \frac{1}{4k+3} & ...& \frac{1}{10n+4} \end{array} \right) \left( \begin{array}{c} c_0 \\ c_1 \\ ... \\ c_n \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ ... \\ 0 \end{array} \right)$

It turns out that the given square matrix is non-singular (see page 92, no. 3 of Polya and Szego: Problems and Theorems in Analysis, Vol. 2, 1976) and so the $c_j = 0$. This means $p_0 = 0$ and so $U^{\perp} = W_1$

Anyway, the conclusion leaves me cold a bit. It seems as if I should be able to prove: let $f$ be some, say… $C^{\infty}$ function over $[0,1]$ where $\int^1_0 x^{2k} f(x) dx = 0$ for all $k \in \{0, 1, ....\}$ then $f = 0$. I haven’t found a proof as yet…perhaps it is false?