The Unapologetic Mathematician

Mathematics for the interested outsider

Two Mean Value Theorems

We’ve got two different analogues of the integral mean value theorem for the Riemann-Stieltjes integral.

The first one says that if \alpha is increasing on \left[a,b\right] and f is integrable with respect to \alpha, with supremum M and infimum m in the interval, then there is some “average value” c between m and M. This satisfies

\displaystyle\int\limits_{\left[a,b\right]}fd\alpha=c\int\limits_{\left[a,b\right]}d\alpha=c\left(\alpha(b)-\alpha(a)\right)

In particular, we should note that if f is continuous then the intermediate value theorem tells us that there is some x_0 with f(x_0)=c. That is, there is some x_0 such that

\displaystyle f(x_0)=\frac{1}{\alpha(b)-\alpha(a)}\int\limits_{\left[a,b\right]}fd\alpha

When \alpha(x)=x this gives us the old integral mean value theorem back again.

So why does this work? Well, if \alpha(a)=\alpha(b) then both sides are zero and the theorem is trivially true. Now, the lowest lower sum is L_{\alpha,\{a,b\}}(f)=m\left(\alpha(b)-\alpha(a)\right), while the highest upper sum is U_{\alpha,\{a,b\}}(f)=M\left(\alpha(b)-\alpha(a)\right). The integral itself, which we’re assuming to exist, lies between these bounds:

\displaystyle m\left(\alpha(b)-\alpha(a)\right)\leq\int\limits_{\left[a,b\right]}fd\alpha\leq M\left(\alpha(b)-\alpha(a)\right)

So we can divide through by \int_{\left[a,b\right]}d\alpha=\alpha(b)-\alpha(a) to get the result we seek.

We can get a similar result which focuses on the integrator by using integration by parts. Let’s assume \alpha is continuous and f is increasing on \left[a,b\right]. Our sufficient conditions tell us that the integral of f with respect to \alpha exists, and the integration by parts formula says

\displaystyle\int\limits_{\left[a,b\right]}fd\alpha=f(b)\alpha(b)-f(a)\alpha(a)-\int\limits_{\left[a,b\right]}\alpha df

But the first integral mean value theorem tells us that the integral on the right is equal to \alpha(x_0)\left(f(b)-f(a)\right) for some x_0. Then we can rearrange the above formula to read

\displaystyle\int\limits_{\left[a,b\right]}fd\alpha=f(b)\alpha(b)-f(a)\alpha(a)-\alpha(x_0)\left(f(b)-f(a)\right)
\displaystyle=f(a)\left(\alpha(x_0)-\alpha(a)\right)+f(b)\left(\alpha(b)-\alpha(x_0)\right)
\displaystyle=f(a)\int_{\left[a,x_0\right]}d\alpha+f(b)\int_{\left[x_0,b\right]}d\alpha

So there is some point x_0 so that the integral of f is the same as the integral of the step function taking the value f(a) until x_0 and the value f(b) after it.

March 28, 2008 Posted by | Analysis, Calculus | 4 Comments

   

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