Let $X$ be a metrizable topological space, and $C(X)$ the space of continuous functions. Is there a continuous norm (as function to $\mathbb{R}$) on $C(X)$? The topology is given by the family of seminorms (locally convex): $$ ||f||_K = \sup_{x \in K} \left| f(x) \right| $$ where $K \in X$ is compact.

I know that if the space $X$ is compact, then we have the norm. But how to prove that it is not in other case?

  • $\begingroup$ You seem to be asking two questions, with one of them well-defined. The first being whether or not there is a norm which generates the same topology as the one furnished by the family of seminorms. The second being whether or not a particular function (which you have not given) is a norm. For this question, what function $\|\cdot\|:C(X)\to\mathbb R$ would you consider? $\endgroup$ – Aweygan Apr 30 '18 at 3:25
  • $\begingroup$ @Aweygan, I asked about the existence of any norm as a function. No matter what topology it would generate. $\endgroup$ – Ann Apr 30 '18 at 13:42

Let $X=\mathbb N$ with the usual metric. Then $C(X)$ can be identified with $\mathbb R ^{\mathbb N}$ and the topology you are considering is the topology of convergence at each point, i.e. the product topology. It is well known that there is no norm for this topology.

  • $\begingroup$ ...because every continuous seminorm on $\mathbb R^{\mathbb N}$ is dominated by a seminorm of the form $q(x)=\max\{|x_1,\ldots,|x_N|\}$ and thus has a huge kernel. $\endgroup$ – Jochen Apr 30 '18 at 6:33
  • $\begingroup$ Thank's, but I asked about the existence of any norm as a function from topological space $C(X) \rightarrow \mathbb{R} (\mathbb{C})$. No matter what kind of topology it generates $\endgroup$ – Ann Apr 30 '18 at 13:45
  • $\begingroup$ There is always a norm on any vector space. You mentioned a particular topology in the statement and I proved that there is no norm for that topology. Every vector space is isometric, as a vector space, to a subspace of a Hilbert space so there is a norm and even an inner product on every vector space. However, it may not be easy to define a norm explicitly. $\endgroup$ – Kavi Rama Murthy May 1 '18 at 7:27

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