## 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.  $\diamond$

### 2 Responses to “Orthogonal Complement of Even Functions.”

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