$C^n$ norm

One can define an extended norm on the space $C^n(I)$ where $I$ is a subset of $ℝ$ as follows:

$${\parallel f\parallel }_{C^n}=\underset{x\in I}{sup}\underset{k\le n}{sup}\left|\frac{d^{k}f}{d{x}^k}\right|$$

If $f$ is a function of more than one variable (i.e. lies in $C^n(D)$ for a subset $D\in {ℝ}^m$), then one needs to take the supremum over all partial derivatives of order up to $n$.

That

$$\parallel \cdot {\parallel }_{C^n}$$

satisfies the defining conditions for an extended norm follows trivially from the properties of the absolute value (positivity, homogeneity, and the triangle inequality) and the inequality

If we are considering functions defined on the whole of ${ℝ}^m$ or an unbounded subset of ${ℝ}^m$, the $C^n$ norm may be infinite. For example,

$${\parallel e^x\parallel }_{C^n}=\mathrm{\infty }$$

for all $n$ because the $n$-th derivative of $e^x$ is again $e^x$, which blows up as $x$ approaches infinity. If we are considering functions on a compact (closed and bounded) subset of ${ℝ}^m$ however, the $C^n$ norm is always finite as a consequence of the fact that every continuous function on a compact set attains a maximum. This also means that we may replace the “$sup$” with a “$\max$” in our definition in this case.

Having a sequence of functions converge under this norm is the same as having their $n$-th derivatives converge uniformly. Therefore, it follows from the fact that the uniform limit of continuous functions is continuous that $C^n$ is complete under this norm. (In other words, it is a Banach space.)

In the case of $C^{\mathrm{\infty }}$, there is no natural way to impose a norm, so instead one uses all the $C^n$ norms to define the topology in $C^{\mathrm{\infty }}$. One does this by declaring that a subset of $C^{\mathrm{\infty }}$ is closed if it is closed in all the $C^n$ norms. A space like this whose topology is defined by an infinite collection of norms is known as a multi-normed space.