Real Polynomial Function is Differentiable

Theorem

Let $n \in \Z_{\ge 0}$ be a positive integer.

Let $f_n: \R \to \R$ be a real polynomial function.

Then $f_n$ is differentiable over $\R$.

Proof

Let $f_n$ be an arbitrary real polynomial function of degree $n$.

The proof proceeds by induction.

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:

$f_n$ is differentiable over $\R$.

$\map P 0$ is the case $f_0$, where $f_0$ is of zero degree.

Such a real polynomial function is a constant function.

From Derivative of Constant we have that $f_0$ is differentiable over $\R$.

Thus $\map P 0$ is seen to hold.

Basis for the Induction

From Derivative of Identity Function: Corollary we have that $f_1$ is differentiable over $\R$.

Thus $\map P 1$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis

Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:

An arbitrary real polynomial function of degree $k$ is differentiable over $\R$

from which it is to be shown that:

An arbitrary real polynomial function of degree $k + 1$ is differentiable over $\R$.

Induction Step

This is the induction step:

$\forall x \in \R: \map {f_{k + 1} } x = a_{k + 1} x^{k + 1} + \map {f_k} x$

for some real polynomial function of degree $k$, where $a_{k + 1} \in \R$

From the induction hypothesis $f_k$ is differentiable over $\R$.

From Power Rule for Derivatives, the real function $x \mapsto x^{k + 1}$ is differentiable over $\R$.

Hence from the Linear Combination of Derivatives:

$f_{k + 1}$ is differentiable over $\R$.

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:

$\forall n \in \Z_{\ge 0}: f_n$ is differentiable over $\R$.

$\blacksquare$