# The Unapologetic Mathematician

## Products of integrable functions

From the linearity of the Riemann-Stieltjes integral in the integrand, we know that the collection of functions that are integrable with respect to a given integrator over a given interval form a real vector space. That is, we can add and subtract them and multiply by real number scalars. It turns out that if the integrator is of bounded variation, then they actually form a real algebra — we can multiply them too.

First of all, let’s show that we can square a function. Specifically, if $\alpha$ is a function of bounded variation on $\left[a,b\right]$, and of $f$ is bounded and integrable with respect to $\alpha$ on this interval, then so is $f^2$. We know that we can specialize right away to an increasing integrator $\alpha$. This will work here (unlike for the order properties) because nothing in sight gets broken by subtraction.

Okay, first off we notice that $f(x)^2$ is the same thing as $|f(x)|^2$, and so they have the same supremum in any subinterval of a partition Then the supremum of $|f(x)|^2$ is the square of the supremum of $|f(x)|$ because squaring is an increasing operation that preserves suprema (and, incidentally, infima). The upshot is that $M_i(f^2)=M_i(|f|)^2$. Similarly we can show that $m_i(f^2)=m_i(|f|)^2$. This lets us write

$\displaystyle M_i(f^2)-m_i(f^2)=M_i(|f|)^2-m_i(|f|)^2=\left(M_i(|f|)+m_i(|f|)\right)\left(M_i(|f|)-m_i(|f|)\right)$
$\leq2M\left(M_i(|f|)-m_i(|f|)\right)$

where $M$ is an upper bound for $|f|$ on $\left[a,b\right]$. Riemann’s condition then tells us that $f^2$ is integrable.

Now let’s take two bounded integrable functions $f$ and $g$. We’ll write

$f(x)g(x)=\frac{1}{2}\left(\left(f(x)+g(x)\right)^2-f(x)^2-g(x)^2\right)$

and then invoke the previous result and the linearity of integration to show that the product $fg$ is integrable.

March 20, 2008 Posted by | Analysis, Calculus | 8 Comments