Example

— Let E be a vector subspace of the space of real-valued functions defined on $I = [0, 1]$; let $\| f \|$ be a norm on E, under which E is complete, and such that its topology is finer than the topology of simple convergence. Suppose further that E contains the set $\mathscr{C}^\infty(I)$ of functions infinitely differentiable on I; we shall show that there exists an integer $k \geq 0$, such that E contains the set $\mathscr{C}^k(I)$ of all functions with a continuous $k$-th derivative in I.

For every pair of integers $m > 0,\ n \geq 0$, let $V_{mn}$ be the set of functions $f \in \mathscr{C}^\infty(I)$ such that $|f^{(h)}(x)| \leq 1/m$ for $0 \leq h \leq n$ and for all $x \in I$. The $V_{m,n}$ form a fundamental system of neighbourhoods of 0 for a metrisable topology compatible with the vector space structure of $\mathscr{C}^\infty(I)$, further $\mathscr{C}^\infty(I)$ is complete in this topology (FVR, II, p. 2, th. 1). Let $u$ be the canonical mapping of $\mathscr{C}^\infty(I)$ in E; we show that $u$ is continuous. From cor. 5 above it is sufficient to prove that if a sequence $(f_n)$ converges to 0 in $\mathscr{C}^\infty(I)$ and to a limit $f$ in E then necessarily $f = 0$. But this is immediate since, by hypothesis, $f$ is the simple convergence limit of $(f_n)$. Hence there exists an integer $k \geq 0$ and a number $a > 0$ such that the relation

implies $\| f \| \leq 1$ for all $f \in \mathscr{C}^\infty(I)$.

But $p_k$ is a norm on the space $\mathscr{C}^k(I)$ and $\mathscr{C}^\infty(I)$ is a subspace that is everywhere dense in $\mathscr{C}^k(I)$ for this norm (the set of polynomials being already everywhere dense in $\mathscr{C}^k(I)$, an immediate consequence of the Weierstrass-Stone theorem). By what has gone before, the identity mapping of $\mathscr{C}^\infty(I)$ (carrying the norm $p_k$) in E, is continuous, and so it can be extended continuously to the whole space $\mathscr{C}^k(I)$ (since E is complete). This proves our assertion.