Orthogonal Complement of Even Functions.

August 5, 2012

Question:  Consider the subspace E of L^{2}([-1,1]) consisting of even functions (that is, functions with f(x) = f(-x)).  Find the orthogonal complement of E.

 

One Solution.  It’s easy to prove E is a subspace.  Then, there is a representation of any function in this space by adding odd and even functions together; more precisely, given f\in L^{2}([-1,1]) we have that \frac{f(x) + f(-x)}{2} is even and \frac{f(x) - f(-x)}{2} is odd and f(x) = \frac{f(x) + f(-x)}{2} + \frac{f(x) - f(-x)}{2}.  For uniqueness, note that if f(x) = f(-x) = -f(x), then f(x) = -f(x) for each x, giving us that f(x) = 0.  Hence, the orthogonal complement of E is the set of odd functions.  \diamond

 

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

 

Another Solution.  We want to find all g\in L^{2}([-1,1]) such that \langle f, g \rangle_{2} = 0 for every even function f\in L^{2}([0,1]).  This is equivalent to wanting to find all such g with \displaystyle \int_{-1}^{1}f(x)g(x)\, dx = 0.  Assume g is in the orthogonal complement.  That is,

\displaystyle 0 = \int_{-1}^{1}f(x)g(x)\, dx = \int_{-1}^{0}f(x)g(x)\, dx + \int_{0}^{1}f(x)g(x)\, dx

\displaystyle = -\int_{1}^{0}f(-x)g(-x)\, dx + \int_{0}^{1}f(x)g(x)\, dx

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

\displaystyle = \int_{0}^{1}f(-x)g(-x)\, dx + \int_{0}^{1}f(x)g(x)\, dx

\displaystyle = \int_{0}^{1}f(x)g(-x)\, dx + \int_{0}^{1}f(x)g(x)\, dx

\displaystyle = \int_{0}^{1}f(x)g(-x) + f(x)g(x)\, dx = \int_{0}^{1}f(x)(g(x) + g(-x))\, dx.

We may choose f(x) = g(x) + g(-x) since this is an even function, and we note that this gives us

\displaystyle 0 = \int_{0}^{1}(g(x) + g(-x))^{2}\, dx.

Since (g(x) + g(-x))^{2}\geq 0, it must be the case that (g(x) + g(-x))^{2} = 0.  [Note: The fact that this is only true "almost everywhere" is implicit in the definition of L^{2}.]  Hence, g(x) + g(-x) = 0, giving us that -g(x) = g(-x)

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

\displaystyle \langle f, g \rangle_{2} = \int_{-1}^{1}f(x)g(x)\, dx = \int_{0}^{1}f(x)(g(x) + g(-x))\, dx = 0

where the last equality comes from the fact that g is odd.  \diamond

Advertisements

2 Responses to “Orthogonal Complement of Even Functions.”

  1. Silas said

    Hello! I know this is kind of off topic but I was wondering if
    you knew where I could locate a captcha plugin for my comment form?
    I’m using the same blog platform as yours and I’m having trouble finding one?
    Thanks a lot!

  2. Bobbye said

    Thank you for the good writeup. It in truth was a amusement account it.
    Glance advanced to far delivered agreeable from you! However, how could we keep up a correspondence?

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: