Differentiability Class is Subset of Differentiability Class of Lower Order

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Theorem

Let $f$ be a real function.

Let $f$ be an element of differentiability class $C^n$.

Then:

$\forall k \in \set {0, 1, \ldots, n - 1}: f \in C^k$

That is, $f$ is in all differentiability classes of order less than $n$.


Proof

The proof proceeds by induction.

For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition:

$f \in C^n \implies \forall k \in \set {0, 1, \ldots, n - 1}: f \in C^k$


Basis for the Induction

$\map P 1$ is the case:

$f \in C^1 \implies f \in C^0$

By definition of differentiability class $C^0$:

$C^0$ is the class of continuous real functions $C$.

Hence $\map P 1$ is seen to hold from Function in Differentiability Class 1 is also in Continuity Class:


This is the basis for the induction.


Induction Hypothesis

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


So this is the induction hypothesis:

$f \in C^r \implies \forall k \in \set {0, 1, \ldots, r - 1}: f \in C^k$


from which it is to be shown that:

$f \in C^{r + 1} \implies \forall k \in \set {0, 1, \ldots, r }: f \in C^k$


Induction Step

This is the induction step:

Let $f \in C^{r + 1}$.

Let $g = \dfrac {\d^r} {\d x^r} \map f x$.

Then:

$\dfrac {\d g} {\d x} = \dfrac {\d^{r + 1} } {\d x^{r + 1} } \map f x$

Thus $g \in C^1$.

By the basis for the induction it follows that $g \in C^0$.

That is, $g$ is continuous.

Hence by definition of $g$, $\dfrac {\d^r} {\d x^r} \map f x$ is continuous.

That is:

$f \in C^r$

By the induction hypothesis:

$\forall k \in \set {0, 1, \ldots, r - 1}: f \in C^k$

It follows that:

$\forall k \in \set {0, 1, \ldots, r}: f \in C^k$


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


Therefore:

$\forall n \in \Z_{\ge 0}: f \in C^n \implies \forall k \in \set {0, 1, \ldots, n - 1}: f \in C^k$

$\blacksquare$


Sources

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