Polynomial Functions as Functions of Bounded Variation

# Polynomial Functions as Functions of Bounded Variation

Recall from the Continuous Differentiable-Bounded Functions as Functions of Bounded Variation page that if $f$ is continuous on the interval $[a, b]$, $f'$ exists, and $f'$ is bounded on $(a, b)$ then $f$ is of bounded variation on $[a, b]$.

We will now apply this theorem to show that all polynomial functions are of bounded variation on any interval $[a, b]$.

 Theorem 1: Let $f$ be a polynomial function. Then $f$ is of bounded variation on any interval $[a, b]$.
• Proof: Let $f$ be a polynomial function of degree $n$. Then for $a_0, a_1, ..., a_n \in \mathbb{R}$ with $a_n \neq 0$ we have that:
(1)
\begin{align} \quad f(x) = a_0 + a_1x + a_2x^2 + ... + a_nx^n \end{align}
• Since $f$ is a polynomial, we have that $f$ is continuous on any interval $[a, b]$. Now consider the derivative of $f$ is therefore:
(2)
\begin{align} \quad f'(x) = a_1 + 2a_2x + ... + na_nx^{n-1} \end{align}
• 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 theorem referenced earlier, we must have that $f$ is of bounded variation on $[a, b]$. $\blacksquare$