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?


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