Let $(X,\|\cdot\|)$ be a normed space. It is known that the norm $\|\cdot\|$ induces a topology, known as the norm topology $\tau$ on $X$. Then the pair $(X,\tau)$ is a locally convex topological vector space. With this, $\tau$ can be generated by a family $P$ of semi-norms on $X$.

Question: Is it necessary that $\|\cdot\|$ is in $P$?


No, it is not necessary that $\lVert\,\cdot\,\rVert$ belongs to the family $P$. Unless $X = \{0\}$, when there is only one seminorm on the space - we could work around that if we allow $P = \varnothing$. And if $\dim X = 1$, then all seminorms on $X$ are constant multiples of the norm, and we must put at least one non-zero multiple of the norm in the family. Of course we could always take $P = \{ 2\cdot \lVert\,\cdot\,\rVert\}$ for $\dim X > 0$ and obey the letter of the law but violate its spirit.

More interesting, we can always take a family consisting of seminorms that are not norms if $\dim X > 1$. For a closed linear subspace $F$ of $X$, the norm induces a quotient norm on $X/F$,

$$\lVert \pi(x)\rVert_{X/F} := \inf \{ \lVert x-y\rVert : y \in F\} = \operatorname{dist}(x,F),$$

where $\pi\colon X \to X/F$ is the canonical projection. We can pull that back to a seminorm on $X$, by defining $p_F(x) := \lVert \pi(x)\rVert_{X/F}$.

Then $p_F$ is a continuous (with respect to $\tau$ of course) seminorm on $X$, with kernel $F$. Hence, taking a family $\mathscr{F}$ of nontrivial closed subspaces of $X$, we obtain a family $P = \{ p_F : F\in \mathscr{F}\}$ of continuous seminorms on $X$. Since all seminorms are continuous, the topology induced by $P$ is coarser than $\tau$.

It remains to see that one can arrange it so that the induced topology coincides with $\tau$. One way is to include two complementary subspaces in $P$. If $G,H$ are two closed subspaces such that

$$\alpha\colon G\times H \to X,\quad \alpha(x,y) = x+y$$

is a topological isomorphism, then $\{ p_G, p_H\}$ induces $\tau$. To see that, we note that the restriction of $p_H$ to $G$ is a norm which is equivalent to the restriction of $\lVert\,\cdot\,\rVert$ to $G$. Since $G \cap \ker p_H = G\cap H = \{0\}$, it is a norm, and it is continuous as the restriction of a continuous seminorm. Since $\alpha$ is a topological isomorphism, there is a $c > 0$ such that $\lVert x+y\rVert \geqslant c(\lVert x\rVert + \lVert y\rVert)$ for all $(x,y) \in G\times H$, hence

$$p_H(x) = \inf \{ \lVert x+y\rVert : y \in H\} \geqslant c\lVert x\rVert$$

for all $x\in G$. The same holds when we interchange the roles of $G$ and $H$, whence

\begin{align} \max \{ p_G(x+y), p_H(x+y)\} &= \max \{ p_G(y), p_H(x)\}\\ & \geqslant \frac{1}{2}(p_G(y) + p_H(x))\\ &\geqslant \frac{c}{2}(\lVert y\rVert + \lVert x\rVert)\\ & \geqslant \frac{c}{2}\lVert x+y\rVert \end{align}

for $x\in G, y\in H$, so $\max \{p_G, p_H\}$ is a norm that induces a finer topology than $\tau$ on $X$. Since it is also a $\tau$-continuous norm, it is equivalent to $\lVert\,\cdot\,\rVert$.

Finally, we need to see that in every normed space of dimension $\geqslant 2$ there are complementary subspaces. Let $\lambda \neq 0$ be a continuous linear form, and $x_0\in X$ with $\lambda(x_0) = 1$. Then $\ker \lambda$ and $\operatorname{span} \{x_0\}$ are complementary subspaces. $\ker \lambda$ is closed since $\lambda$ is continuous, and $\operatorname{span} \{ x_0\}$ is finite-dimensional, hence closed. The map $x \mapsto (\lambda(x)\cdot x_0, x - \lambda(x)\cdot x_0)$ is continuous, and its inverse $\alpha \colon (\operatorname{span} \{x_0\}) \times (\ker \lambda) \to X,\; (x,y) \mapsto x+y$ is also continuous, hence a topological isomorphism.

  • $\begingroup$ I did not expect a detailed answer sir. Thank you very much. $\endgroup$ – Juniven May 18 '15 at 21:25

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