Function Spaces and Their Norms

$C^\infty(\overline{\Omega})$ is not a Banach space because no single norm can capture its topology. It is instead a Fréchet space: a complete, metrizable, locally convex topological vector space whose topology is defined by a countable family of seminorms (here, $\|\cdot\|_{C^k}$ for $k = 0, 1, 2, \dots$).

Why can’t a single norm work? For any fixed $k$, there exist sequences in $C^\infty$ that converge in $C^k$ but diverge in $C^{k+1}$. For example, on $[0,1]$:

converges to 0 in $C^0$ ($\|f_n\|_\infty = 1/n \to 0$), but $f_n'(x) = \cos(nx)$ does not converge at all — the $C^1$ norms satisfy $\|f_n\|_{C^1} \geq 1$ for all $n$. No matter which $C^k$ norm you pick, there is always a “higher” derivative direction that it fails to control.

A Banach space norm would have to simultaneously control all derivatives, but controlling the $k$-th derivative imposes constraints that are strictly independent of controlling the $(k-1)$-th. Formally, $C^\infty$ is not normable: by Kolmogorov’s criterion, a locally convex space admits a norm if and only if it has a bounded neighborhood of zero, and in $C^\infty$ every neighborhood of zero contains functions with arbitrarily large $k$-th derivatives for sufficiently large $k$.

Despite not being Banach, $C^\infty$ is complete in its Fréchet topology (a Cauchy sequence in every $C^k$ norm converges in every $C^k$ norm), and much of the Banach space theory — including versions of the Open Mapping Theorem and Closed Graph Theorem — extends to Fréchet spaces.

Let $\{f_n\}$ be Cauchy in $C^k$. Then for each $|\alpha| \leq k$, the sequence $\{D^\alpha f_n\}$ is Cauchy in $C^0$ (sup-norm). Since $(C^0, \|\cdot\|_\infty)$ is complete, each $D^\alpha f_n$ converges uniformly to some continuous function $g_\alpha$. A standard argument (integrating and differentiating under the limit) shows $g_\alpha = D^\alpha g_0$, so $g_0 \in C^k$ and $f_n \to g_0$ in $C^k$.

Strictly speaking, elements of $L^p$ are not functions but equivalence classes of functions that agree almost everywhere. This quotient is essential: without it, $\|\cdot\|_p$ is only a seminorm, not a norm. The problem is that $\|f\|_p = 0$ does not imply $f = 0$ pointwise—only $f = 0$ a.e. For instance, the function that equals 1 at a single point and 0 elsewhere has $L^p$ norm zero but is not the zero function. Passing to equivalence classes identifies all such functions with the zero class, making $\|f\|_p = 0 \implies f = 0$ in $L^p$.

This is a fundamental difference from $C^k$ spaces, where functions are determined by their pointwise values and the sup-norm is a genuine norm without any quotienting.