Answer to Question #170316 in Real Analysis for Prathibha Rose

Question

Show that a polynomial f is of bounded variation in every compact interval [a,b]

Answer 1

Solution:

Let $f$ be a polynomial function of degree n.

Then for $a_{0}, a_{1}, \ldots, a_{n} \in \mathbb{R}$ with $a_{n} \neq 0$ we have:

$f(x)=a_{0}+a_{1} x+a_{2} x^{2}+\ldots+a_{n} x^{n}$

$\because$ $f$ is a polynomial, $f$ is continuous on any interval [a, b].

Now consider the derivative of $f$ :

$f^{\prime}(x)=a_{1}+2 a_{2} x+\ldots+n a_{n} x^{n-1}$

Notice that $f'$ is itself a polynomial. Therefore $f'$ exists and is continuous on any interval [a, b]. Since $f'$ is continuous on the closed and bounded interval [a, b] we must have by the Boundedness Theorem that $f'$ is bounded on [a, b] and hence bounded on (a, b).

Hence $f$ is continuous on [a, b], $f'$ exists, and $f'$ is bounded on (a, b), so by the Boundedness theorem, we must have that $f$ is of bounded variation on [a, b].

Hence, proved.