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Bound your integral by its base times its height

Quick description

This is an utterly trivial bound: one can bound the magnitude |\int_E f(x)\ d\mu(x)| of an integral by LM, where the "base" L is an upper bound for the measure \mu(E) of E, and M is an upper bound for the magnitude |f(x)| of f(x) on E. Thus for instance

 \left| \int_a^b f(x)\ dx \right| \leq |b-a| \sup_{a \leq x \leq b} |f(x)|.

If \mu is a signed or complex measure, one must ensure that L bounds the total variation |\mu|(E) and not just \mu(E). (Mistakes have been made because this issue was neglected!)

One can also get lower bounds this way: if \mu is unsigned and \mu(E) is bounded from below by L, and f(x) is bounded from below by M, then \int_E f(x)\ d\mu(x) is bounded from below by LM. Note however that it is not enough to have |f(x)| bounded from below by M to draw a non-trivial conclusion; one must lower bound f(x) itself.


Basic measure theory; calculus; complex analysis

Example 1

Suppose one wants to show that the contour integral

 \int_{C_r} \frac{e^{iz}}{1+z^2}\ dz

goes to zero as r \to \infty, where C_r is the upper semicircle  0 \leq \theta \leq \pi\}. We can use "base times height" here. For the "base", observe that the length |C_r| of C_r (which is also the measure of C_r with respect to |dz|) is just \pi r. For the "height", observe that on the upper semicircle, e^{iz} has magnitude at most 1, while |1+z^2| is bounded from below by r^2-1 (for r > 1), and so \frac{1}{|1+z^2|} is bounded from above by \frac{1}{r^2-1}, thus leading to a total height bound of \frac{1}{r^2-1}. This leads to a total bound for the magnitude of integral of \frac{\pi r}{r^2-1}, which goes to zero as r \to \infty, and the claim follows from the squeeze test.

(Note that this technique does not work if the denominator was linear instead of quadratic. To see how to deal with that case, see the article "bound by a Riemann sum".)

General discussion

The base times height bound is reasonably efficient as long as

  • f does not oscillate significantly (i.e. there is no significant opportunity for cancellation); and

  • the magnitude of f is reasonably uniformly distributed across E (no "spikes" or large "empty regions").

When the latter condition fails, then it is often advantageous to first subdivide the integral into regions (i.e. divide and conquer) where the magnitude of the integrand is reasonably uniform, so that this method can be applied. This leads to such methods as "bound by a Riemann sum" and "dyadic decomposition".

The base times height method can also be used as a heuristic to guess the right size of an integral. If a function f is "expected" to "concentrate" on a subset of measure about L, and is also believed to typically have a value of about M on this subset, then it is reasonable to expect that the final integral is of order LM (unless one expects lots of cancellation in the integral, in which case it could have a considerably smaller magnitude). For instance, consider the integral \int_0^\infty \frac{1}{a^7+x^7}\ dx for some parameter a>0. Graphing this function, we see that it attains a "height" of about a^{-7} and is concentrated on a "base" interval of length about a, so a reasonable first guess for this integral is that it should be about a^{-6}, which is indeed the case (see the article "bound the integrand by something simpler" for more discussion).

The following variant of the base times height bound is also useful: if a function f has height at most M almost everywhere on a set E of measure at most L, then the L^p(E) of f for any 0 < p \leq \infty is at most L^{1/p} M.


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