# Single norm criterion

Let $E$ be a metrizable locally convex space whose topology is defined by an increasing sequence $\{p_n\}$ of seminorms. Show that the topology of $E$ can be defined by a single norm iff there exists an integer $N$ such that for all $n \geq N$, there exists a non-negative real number $k_n$ for which $p_n \leq k_np_N$.

This is a problem from Dieudonne.

Suppose the condition is satisfied for $N$ and constants $k_n$ for $n \geq N$. Then $p_{N}$ is a norm on $E$ because for every $x \in E$ there is $n$ such that $p_n(x) \gt 0$. If $n \lt N$ then $0 \lt p_n(x) \leq p_N(x)$ and otherwise $0 \lt p_n(x) \leq k_n p_N(x)$. In either case it follows that $p_N(x) \gt 0$. The topology $\tau_N$ on $E$ induced by $p_N$ is clearly coarser than the topology induced by the family $\{p_n\}$. It is also finer because all $p_n$ are continuous with respect to $\tau_N$. Thus, the norm $p_N$ induces the given topology on $E$.
Suppose there is a single norm $\lVert \cdot \rVert$ inducing the topology on $E$. Then all the semi-norms $p_n$ are continuous with respect to the norm $\lVert \cdot \rVert$, so for each $n$ there is $C_n$ such that $p_n(x) \leq C_n \lVert x\rVert$ for all $x \in X$. On the other hand, $\lVert \cdot \rVert$ is continuous with respect to the topology induced by the semi-norms $p_n$, therefore there is $N$ and a constant $C \gt 0$ such that $\lVert x\rVert \leq Cp_{N}(x)$ (since the family $\{p_n\}$ is increasing one semi-norm suffices). We can take $k_n = C_nC$.